MATH 10 by YOUG JI INTERNATIONAL SCHOOL

YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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Functional Notation There is a special notation called functional notation that is frequently used in Mathematics when one variable is describe in terms of another. The equation y = 4x defines a function and can be written as f(x) = 4x. the notation f(x) “read as f of x” denotes the value of the function f at a given value of x or “f at x”. it is important to remember that f(x) does not mean f times x. the letter f stands for the function, and f(x) is the number paired with x. letters like f, g, P, G and others can be used to name functions. The variable x inside the parenthesis is called the independent variable and whose values are called domain (input) of the function. The result of f(x) at each domain x is called the range (output) of the function. For example, f(4) f(4) means replace x by 4 then simplify the resulting numerical expression. This process is called evaluating the function. Example: Given: f(x) = 2x + 3. Find the values of a. f(0) b. f(-5) a. f(0) means that x = 0. Replace the x with the value of 0 f(0) = 2 (0) + 3 f(0) = 3 b. f(-5) means that x = -5 replace the x with the value of -5 f(x) = 2x + 3 f(-5) = 2(-5) + 3 f(-5) = -10 + 3 f(-5) = -7 Given: f(x) = √ a. f(19) = √ b. f(10) = √

. Find the value of =√ =√

=9 =√

= 3√

Practice Exercise 1.1 A. Evaluate each expression. If f(x) = 10 x – 1. Find 1. 2. 3. B. Given 1. 2. 3. C. Given 1. g(-1) 2. g(√ ) 3. g(2 + x)

Relations and Functions YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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A relation is a set of ordered pairs. The domain of a relation is the set of first coordinates. The range is the set of second coordinates. A function is a relation in which each element of the domain corresponds to exactly one element of the range Example: determine whether the following relations are function or not. a. {(1,2),(2,5),(3,10),(4,17)}

b. {(1,2),(2,3),(2,0),(3,5)}

Solution: a. Each element in the domain, {1,2,3,4} is assigned no more than one value in the range;1 is assigned to only 2;2 is assigned only to 5; 3 is assigned only to 10 and 4 is assigned only to 17. Therefore it is a function. b. Since the domain element 2 is assigned to two different values in the range, 3 and 0, it is not a function. If we are given a set of ordered pairs, we can easily determine whether the relation is a function or not by simply looking if each first element is used only once in a given set. The following characteristics of a function will help us decide when we test for a function when two sets of numbers are given. 1. Each element in domain X must be matched with exactly on element in range Y. 2. Some elements in Y may not be matched with any element in X. 3. Two or more elements in X may be matched with same element in Y. Here are ways to describe the function 1. Mapping diagram Input x 1 2 3 4

output y 2 5 10 17

2. Table of values Input, x 1 2 3 4 Output, y 2 5 10 17

3. Graph â&#x20AC;&#x201C; the graph represents a function if an only if no vertical line intersects the graph in more than one point. 4. Rule or correspondence f:x x2 + 1, x = 1,2,3,4 YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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notice that the ordered pairs of numbers, the mapping diagram, the table of values, and the graph clearly shows that each value of y is obtained by adding 1 to the square of x. Hence, this is the rule or correspondence expressed in words for the said relation. 5. Equation: The rule or correspondence can be described by the equation y=x2 + 1 Example: Represent the rule â&#x20AC;&#x153;Add one to each x-coordinateâ&#x20AC;? with a table of values, an equation, a graph, or a diagram. Solution: a. Table of values Input, x Output,y 9 10 -45 -44 8.28 9.28 x x+1 b. Equation: y = x + 1 c. Graph x-y coordinate d. diagram Domain Range -45 -44 8.28 9.28 9 10

Vertical line test A graph represents a function if and only if no vertical line intersects the graph in more than one point Example: use the vertical line test to identify graphs in which y is a function of x. a.

a. b. c. d.

y y y y

is is is is

not a function of x not a function of x. Two values of y correspond to an x-value. not a function of x. Two values of y correspond to an x-value. a function of x

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Activity 1 A. Determine whether or not each relation is a function. Give the domain and range of each relation. 1. {(2,3),(4,5),(6,6)} 2. {(4,5),(4,6),(5,5),(5,6)} B. State whether or not each relation is a function 1. Input, D output, R 3 4 9 -3 2. x 4 4 4 4

y 4 8 12 16

C. Determine whether or not each equation defines y as a function of x. if it is a function, find the domain and range. 1. x + y = 9 2. x + y3 = 8 D. find the domain and range of each relation 1. f(x) = 4x2 + 3x + 1 2. g(x) =

Lesson 2 Operations on Functions Functions like numbers, can be added and subtracted. Functions can also be multiplied or divided. Because functions are usually given in equation form, we perform these operations by performing operations on algebraic expression that appear on the right side of the equations. For example we combine the following two functions using addition: Add the terms on the right side of the equal sign of equal sign of . Thus,

to the terms on the right side of the

= (3x + 1) + ( x2 – 3) = 3x – 2 + x2 = x2 + 3x – 2 Sum, Difference, Product, and Quotient of Functions Let be any two functions. YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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The sum f + g, difference f – g , product fg, and quotient are functions whose domains are the set of all real numbers common to the domain of f and g, and defined as follows: 1. Sum: 2. Difference 3. Product 4. Quotient

( )

Example: if a.

(x) = = = - x2 + x + 1

b. = 3x = = Answer the following: 1. Find

. Determine the domain of each function.

a. b. c.

The function of y = f(x) tells you that y is a function of x. if there is a rule relating to x, such as y = 3x+ 1, then you can write: F(x) = 3x + 1 The name of the function is f. other letters may be used to name functions, especially g and h. F(x) is read as “f of x” and this represents the value of the function at x. Domain of a function f is the set of all values of x for which f is defined. The range of a function f is the set of all values of f(x), where x is the element of the domain as the set of the function’s input values and the range as the set of the function’s output special notation f(x), represents the value of the function at the number x. Example: if f(x) = x + 8, evaluate f(4) Solution: Replacing x with 4, we have f(4) = 4 + 8 or 12 Even and odd functions The function f is an even function if and only if YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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F(-x) = f(x),

for all x in the domain of f.

The right side of the equation of an even function does not change even if x is replaced with –x. The function f is an odd function if and only if F(-x) = -f(x),

for all x in the domain of f.

Every term in the right side of the equation of an odd function changes sign if x is replaced by –x. Example: Identify each function as even or odd, or neither. a. F(x)m= x5 Solution: In each case, replace x with –x and simplify. If the right side of the equation stays the same, the function is even. If every term on the right changes sign, the function is odd. a. Use f(x) = x5 F(-x) = (-x)5 Replace x with -x = (-x)(-x)(-x)(-x)(-x) = -x5 Activity 2 1. Fill in the missing number represented by (?) in each given function a. F(x) = 12 – 7(x) f(-3) = 12 – 7(?) b. F(x) = √ f(-1) = √ 2. Evaluate each function at the indicated values of the independent variable and simplify the result. a. F(x) =

f(2)

b. h(x) = √ h(10) 3. Determine whether or not each function is even, odd, or neither. a. F(x) = x3 – 1 b. H(x) = x2 √

Lesson 3 Operations on Functions Functions, like numbers, can be added and subtracted. Functions can also be multiplied or divided. Because functions usually given in an equation form, we perform these operations by performing operations on algebraic expressions that appear on the right YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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side of the equations. For example, we can combine the following two functions using addition: F(x) = 3x + 1

and

g(x) =

–3

Add the terms on the right side of the equal sign of f(x) to the terms on the right side of the equal sign of g(x). Thus, (f + g) x = f(x) + g(x) = (3x + 1) + (x2 – 3) = 3x – 2 + x2 = x2 + 3x – 2

Sum, difference, product and quotient of functions Let f and g be any two functions. The sum f + g, difference f-g, product fg, and the quotient

are functions whose domains

are the set of all real numbers common to the domain of f and g, and defined as follows: 1. Sum: 2. Difference: 3. Product:

(f + g) (x) = f(x) + g(x) (f-g)(x) = f(x) – g(x) (fg)(x) = f(x). g(x)

4. Quotient:

(x) =

where g(x)

Example: Let f(x) = x2 – 5 and g(x) = 5x + 4 a. (f + g) (x) = f(x) + g(x) = (x2 – 5) + (5x + 4) = x2 + 5x – 1 b. If f(x) = 3x – 2 and g(x) = x2 + 2x – 3 (x) =

Activity 3 A. Find f + g, f – g, fg, and

. Determine the domain of each function.

1. F(x) = 3x + 4, g(x) = 2x – 1 2. F(x) = 2x – 5, g (x) = 4x2 3. F(x) = x – 4, g(x) = √ YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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4. F(x) =

, g(x) =

Lesson 3 Composite and Inverse Functions The composition of the function f with g is denoted by The domain of the composition function 1. X is in the domain of g and 2. g(x) is in the domain of f.

and is defined by the equation ( ) is a set of all x such that

Example: Given: f(x) = 4x – 5 and g(x) = x2 + 4 find: a. ( (x) = f(g(x)) = 4(g(x)) – 5 = 4(x2 + 4) – 5 = 4x2 + 16 – 5 = 4x2 + 11 b. ( = means g(f(x)) = (f(x))2 + 4 = (4x – 5)2 + 4 = 16x2 – 40x + 25 + 4 = 16x2 – 40x + 29 Inverse of a function Let f and g be two functions such that f(g(x)) = x for every x in the domain of g and g(f(x)) = x for every x in the domain of f. The function g is the inverse of the function f, and is denoted by f-1 ( read “f-inverse). Thus, f(f-1(x)) = x and f-1 (f(x)) = x the domain of f is equal to the range of f-1 and vice versa. Example: show that each function is an inverse of the other. f(x) = 6x and g(x) = f(g(x)) = x and g(f(x)) = x f(x) = 6x = f(g(x))= 6( ) = x g(x) = = = x Because g is the inverse of f(and vice versa), we can use inverse notation and write f(x) = 6x and f-1 (x) = the definition of the inverse of a function tells the domain is equal to the range of f-1, and vice versa. This means that if the function f is the set of ordered pairs (x,Y), then the inverse of f is the set of ordered pairs (y,x). If a function is defined by an equation, we can obtain the equation for f-1, the inverse of f, by interchanging the role of x and y in the equation for function f. Finding the inverse of a function The equation for the inverse of a function f can be found as follows: 1. Replace f(x) with y in the equation for f(x). 2. Interchange x and y. 3. Solve for the new y from the equation in step 2. If this equation does not define y as a function of x, the function f does not have an inverse function and this procedure ends. If this equation does define y as a function of x, the function f has an inverse function. -1 -1 4. If f has an inverse function, replace y in step 3 by f (x). the result can be verified by showing that f(f (x)) = x -1 and f (f(x)) = x

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Example: Find the inverse of f(x) = x3 + 3 Replace f(x) with y: y = x3 + 3 x = y3 + 3 Solve for y: x – 3 = y3

√

=√

√

Replace y with f-1(x) = √ Verify this result by showing that f(f-1(x))=x and f-1 (f(x))=x f-1(x) = √

f(x) = x3 + 3

f-1(f(x)) = √

f(f-1(x)) = (f-1 (x))3 + 3

=√

= (√

=√

=x–3+3

) +3

The horizontal line test for inverse functions A function f has an inverse that is a function, f-1, there is no horizontal line that intersects the graph of the function f at more than one point. Practice: 1. Find: a.

(x) b. ( f(x) = 2x, g(x) = x + 5 2. Find the inverse of each function. f = { (5,4),(0,3),(1,0)} 3. Find the inverse of each function. f(x) = 2x2 – 3

(

Lesson 4 Composite and Inverse Functions The composition of the function f with g is denoted by f .g and is defined by the equation F .g (x) = f (g(x)). The domain of the composition function f .g is the set of all x such that YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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1. X is in the domain of g and 2. g(x) is in the domain of f. Example: given f(x) = 4x – 5 and g(x) = x2 + 4, find: a. (f ) (x) b. (g )(x) Because (f means f (g(x)), we must replace each occurrence of x in the function f by g(x). f(x) = 4x – 5 given equation for f f (x) = f(g(x) = 4(g(x)) – 5 Replace x by g(x) 2 = 4(x + 4) – 5 Replace g(x) by x2 + 4 = 4x2 + 16 – 5 Multiply = 4x2 + 11 Thus, (f (x) = 4x2 + 11. b. (g (x) means g(f(x)). Hence, we must replace each occurrence of x in the function g by f(x) g(x) = +4 given equation for g 2 (g (x) = g(f(x)) = (f(x)) + 4 replace x by f(x) 2 = (4x – 5) – 4 replace f(x) by 4x -5 = 16x2 – 40x + 25 + 4 square the binomial 2 = 16 x – 40x + 29 simplify 2 Thus, (g (x) = 16x – 40x +29. Notice that (f (x) is not the same as g(f(x)). Inverse of a function Let f and g be two functions such that f (g(x)) = x for every x in the domain of g and g(f(x))= x for every x in the domain of f. The function g is the inverse of the function f, and is denoted by f-1 ( read “ f-inverse”.) Thus, f ( f -1(x)) = x and f -1 (f(x)) = x The domain of f is equal to the range of f -1, and vice versa. Example: show that each function is an inverse of the other. a. F(g(x)) = x and g(f(x)) = x F(x) = 6x

g(x) =

F(g(x)) = 6 (g(x))

g(f(x)) = (

=6( )

)

Finding the inverse of a function f can be found as follows: 1. Replace f(x) with y in the equation for f(x). 2. Interchange x and y. YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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3. Solve for the new y from the equation in step 2. If this equation does not define y as a function of x, the function f does not have an inverse function and tis procedure ends. If this equation does define y as a function of x, the function of x, the function f has an inverse function. 4. If f has an inverse function, replace y in step 3 by f -1(x). The result can be verified by showing that f(f-1(x)) =x and f -1(f(x)) = x. Example: find the inverse of the relation described by the set of ordered pairs { } Solution: the inverse of a relation is the set of ordered pairs obtained by switching the coordinates of each ordered pair in the relation Switch the coordinates of each ordered pair. {

}

{

Original relation }

Inverse relation

Find the inverse of f(x) = x3 + 3 Step 1:

replace f(x) with y :

y = x3 + 3

Step 2:

interchange x and y:

x = y3 + 3

Step 3:

solve for y:

x – 3 = y3

Step 4:

replace y with f -1(x):

√

=√

√

f-1(x) = √

Verify this result by showing that f(f -1(x)) = x and f -1(f(x)) = x f-1(x) = √

f(x) = x3 + 3

f-1(f(x)) = √

f(f -1(x)) = (f -1(x))3 + 3

=√

= √

=√

=x–3+3

Since f(f-1(x)) = f -1(f(x)) = x, then f -1 is the inverse of f. The graph of a function can tell us if it represents a function with an inverse. YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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The horizontal line test for inverse functions A function f has an inverse that is a function, f -1, if there is no horizontal line that intersects the graph of the function f at more than one point. Example: functions?

which of the following graphs represent functions that have inverse

a. .

Solution a. b. c. d.

has an inverse function no inverse function no inverse function has an inverse function

Activity 4 I. II.

III.

IV.

Find (f (x) in f(x) = 2x, g(x) = x + 5 Find the inverse of a function and state whether f -1 is a function or not. } a. f = { 2 b. f(x) = 2x â&#x20AC;&#x201C; 3 Find the f(g(x)) and g(f(x)) and determine whether each pair of functions f and g are inverses of each other a. f(x) = 2x, g(x) = b. f(x) = - x, g(x) = -x Find the equation for f-1(x). then verify if your equation is correct by showing that f(f -1(x)) = x and f-1 (f(x)) = x in f(x) =

Lesson 5 Linear Functions A linear function is a function that can be written in the form F(x) = ax + b Where a and b are real numbers with a and f(x) not both equal to zero The graph of the linear function(x) = ax + b is exactly the same as the graph of the linear equation y=ax + b. If a = 0, then we get f(x) = b, which is called the constant function. If YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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a = 1 and b = 0, then we get the function f(x) = x, which is called the identity function. When we graph a function notation, we usually label the vertical axis as f(x) rather than y. Example: f(x) = 3x + 7 is a linear function because it is written in the form f(x) = ax + b with a=3 and b=7 g(x)= x3 – 1 is not a linear function because x has an exponent other than 1. You graph a linear equation in two variables by making a table of values, graphing enough ordered pairs to see a pattern, and connecting the points with a line or smooth curve. Graphing Linear Equation by the Point Plotting Method The Point Plotting Method of graphing y = mx + b 1. Make as table of values showing three or four ordered pairs that are solutions of the equations. 2. Plot these points on a rectangular coordinate system 3. Since the graph of y =mx + b is a line, connect the points with a line. Example: Graph the linear equation y = 2x – 3 Solution: choose several values for x and construct a table of values x -1 0 1 2

y=2x – 3 y=2(-1) – 3 y=2(0) – 3 y=2(1) – 3 y=2(2) – 3

y -5 -3 -1 1

(x,y) (-1,-5) (0,-3) (1,-1) (2,1)

Plot the points given in the table and complete the graph by drawing a line through them.

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Graphing the Linear Equations by the Intercepts Method X-intercept and Y-intercept The x-intercept is the x-coordinate of the point where the graph intersects the x-axis. The y-intercept is the y-coordinate of the point where the graph intersects the y-axis. Finding Intercepts 1. To find the x-intercept (x,0), let y = 0 and solve for x. 2. To find the y-intercept (0,y), let x = 0 and solve for y. Graphing Linear Equations by the Intercepts Method To graph a linear equation in two variables using the intercepts method: 1. 2. 3. 4. 5.

Determine the x-intercepts. Determine the y-intercepts. Plot the intercepts and label their coordinates. Connect the intercepts with a straight line. Check the graph by locating a third point on the graph and determine if it is a solution of the equation.

Graphing Linear Equations using the slope-intercept Method Slope of a line The slope of a line is the ratio of the amount of rise to the amount of run. Slope = m = The slope-intercept form of a line equation in two variables is given by Y = mx + b Where m is the slope of the line and b is the y-coordinate of the y-intercept of the graph Graphing Linear Equations Using the Slope-intercept Method To Graph a linear equation in two variables using the slope-intercept method: 1. 2. 3. 4.

Solve the equation for y. Determine the slope and the y-intercept from the equation. Plot the y-intercept and label its coordinates. Locate the next point on the line by using the slope (rise and run). Label its coordinates. 5. Draw the straight line through the points.

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Graphing Inverse Functions If we graph the one-to-one function and its inverse function on the same coordinate plane as well as the graph of y = x, we obtain an interesting picture. For example Graph the function f = {(2,4), (0,5),(-2,1)} Graph the inverse function f -1 = {(4,2),(5,0),(1,-2)} Graph the equation: y = x Example Determine the inverse function f -1(x) f(x) = 2x – 6 Solution f(x) = 2x – 6 y = 2x – 6 x = 2y – 6 x + 6 = 2y y= f-1= + 3

Practice I.

II.

III.

Find three solutions of the given equation a. 3x + 2y = 15 b. 15x – 5y = 30 c. 2x + 5 = 4y Rewrite each equation in slope intercept form a. 3x + 2y = 10 b. 2x -7y = 21 c. 2x – 7 + 2y = 0 Determine f -1(x) a. f(x) = 3x + 5 b. f(x) = 4x – 5

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c. f(x) = 2(x -1) Lesson 1.5 Linear Functions A linear function is a function that can be written in the form f(x) = ax + b, where a and b are real numbers with a and f(x) not both equal to zero. The graph of a linear function f(x) = ax + b is exactly the same as the graph of the linear equation y = ax +b. if a = 0, then we get f(x) = b, which is called constant function. If a =1 and b=0, then we get the function f(x)= x, which is called identity function. When we graph a function given in function notation, we usually label the vertical axis as f(x) rather than y. Graphing linear equations by point plotting method The point-plotting method of graphing y = mx + b 1. Make as table of values showing three or four ordered pairs that are solutions of the equations. 2. Plot these points on a rectangular coordinate system. 3. Since the graph of y=mx+b is a line, connect the point with a line. Graphing linear equations by the intercepts method x-intercept is the x-coordinate of the point where the graph intersects the x-axis y-intercept is the y-coordinate of the point where the graph intersects the y-axis Finding intercepts 1. To find the x-intercepts(x,0), let y = 0 and solve for x. 2. To find the y-intercept (0,y), let x=0 and solve for y Graphing linear equation by intercepts method To graph a linear equation in two variables using the intercepts method: 1. 2. 3. 4. 5.

Determine the x-intercept Determine the y-intercept. Plot the intercepts and label their coordinates Connect the intercepts with a straight line. Check the graph by locating a third point on the graph and determine if it is a solution of the equation.

Graphing linear equations using the slope-intercept method Slope of a line is the ratio of the amount of rise to the amount of run. Slope = m = The slope-intercept form of a linear equation in two variables is given by Y= mx + b Where m is the slope of the line and b is the y-coordinate of the y-intercept of the graph. YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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Graphing linear equations using the slope-intercept method To graph a linear equation in two variables using the slope-intercept method 1. 2. 3. 4.

the slope and the y-intercept from the equation Plot the y-intercept and label its coordinates. Locate next point on the line by using the slope (rise and run). Label its coordinates. Draw a straight line through the points.

Lesson 1.6 Quadratic Functions A quadratic function is any equation of the form f(x) = ax2 + bx + c Where a, b, and c are real numbers and a Note that y = x2 or f(x) = x2 is a quadratic function where b and c are both equal to 0 and a is not equal to 1. The graph of quadratic function is a parabola. Parabolas are shaped like cups.

Vertex form of a quadratic function The quadratic function F(x) = a(x â&#x20AC;&#x201C; h)2 + k , a is written in a vertex form. The graph is a parabola whose vertex is the point (h,k). the parabola is symmetric to the line x = h. if a > 0, the parabola opens upward; if a <0, the parabola opens downward. Line of symmetry

Example Find the equation of the parabola whose vertex is (2,3) and passes through the point (0,0) Solution The parabola opens downward since (2,3) is the highest point. Since the parabola has vertex (h,k) or (2,3), the equation must have the form. YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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y = -a (x – h)2+ k y = -a(x – 2)2 + 3 because the parabola passes through the point(0,0) it follows that f(0) = 0. Thus, we have 0 = -a(a – 2)2 + 3 0 = -4a + 3 a= This implies that the equation is Y=

(x – 2 )2 + 3

(x2 – 4x + 4) + 3

Y= Y= Practice Find two positive real number that satisfy the condition 1. Their sum is 26 and their product is a maximum. 2. Their difference is 6 and their product is a minimum.

3. What is the shape of a rectangle with a given perimeter when its area is a maximum? Quadratic Functions A quadratic function is any equation in the form of f(x) = ax2 + bx + c, where a,b,and c are real numbers and a . Note that y = x2 or f(x) = x2 is a quadratic function where b and c are both equal to zero and a is equal to 1. The graph of any quadratic function is called a parabola. Hence, the graph of y = x2 is a parabola. Parabolas are shaped like cups

The parabola has the line of symmetry. This means that part of the parabola on one side of the line is the reflection of the part of the other side.

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If the coefficient of x2 ( the value of a in ax2 + bx + c) is positive, the parabola opens upward. When the coefficient of x2 is negative, the graph opens downward. The vertex of the parabola is the minimum point of the graph when it opens upward. The vertical line passing through a vertex is the axis of symmetry. This line divides the parabola into two symmetric parts which is the mirror images of each other. The vertex form of a quadratic function f(x) = a (x – h)2 + k , a is written in vertex form. The graph is parabola whose vertex is the point (h,K). The parabola is symmetric to the line of x = h. if a>0, the parabola opens upward; if a<0, the parabola opens downward. The general procedure for graphing a parabola whose equation is in vertex form 1. Determine whether the parabola opens upward or downward 2. Determine the vertex of the parabola. The vertex is (h,k) 3. Find any x-intercept by replacing f(x) with 0. Solve the resulting quadratic equation for x. 4. Find the y-intercept by replacing x with 0. Solve the resulting equation for y. 5. Plot he intercepts and vertex. Connect these points with a smooth curve. Practice: 1. Determine algebraically the intercepts of the graph of each parabola a) Y=x2 -5x + 4 b) Y=-15x + 3x2 c) y = -5 + 3x - 2x2

Chapter 2 Polynomial functions

Lesson 2.1 Polynomial Functions

A function defined by f(x) = anxn + an-1xn-1 + an-2xn-2 +…+ a1x + a0 Where a0, a1…..anare real numbers and an

and n is a nonnegative integer.

A first degree polynomial function in x is generally called a linear function defined by P(x) = mx + b Where m and b are constants and m a slope m and y-intercept b.

. The graph of a linear function is a line having

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A second degree polynomial function in x is generally called a quadratic function defines by P(x) = ax2 + bx + c Where a b, and c are constants representing real numbers and a parabola whose line of symmetry is a vertical line.

. Its graph is a

In general, if a function P is defined by P(x) = anxn + an-1xn-1 + an-2xn-2 + … + a1x + a0 Where a0,a1,…an are real numbers and an and n is anonnegative integer, then P is called polynomial function in x of degree n, hence the functions P(x) = 3x 6+ 5x4 – 2x2 + x – 3,is a polynomial function of degree 6. Practice Determine the degree of polynomial in x. 1. 4x5 – 51x2 + 62 2. 2x2 – 4x5 +

-8

3. 4x3 - √

Lesson 2.2 The value of a polynomial function To evaluate a function using direct substitution, for example for the function f(x) = 3x 2 – 5x + 7, f(4) represents the value of the function when x is replaced by 4. F(x) = 3x2 – 5x + 7 F(4) = 3(4)2 – 5(4) + 7 = 48 – 20 + 7 = 35 To evaluate a function using synthetic substitution Procedure Step 1:

multiply the coefficient of first term by x

Step 2:

add the coefficient of second term

Step 3:

multiply by x

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Step 4:

add the coefficient of third term

Step 5:

multiply by x

Step 6:

add the last term

Example Evaluate 2x3 + 3x2 + 4x + 9 at x = 3 Solution Write the coefficient of the polynomial in a row, perform 2x3 + 3x2 + 4x + 9 Value of x = 3

2 Bring

step 2 step 4

coefficients

step 6

Down

102

the answer is 102

When writing the row of coefficients in synthetic substitution, you must include a zero coefficient for each “ missing tem of the polynomial. For example, you would write these coefficients for the polynomial 5x6 – 3x4 + 2x3 – x + 15 5

This zero represents The missing x6 term

-3

-1

this zero represents the missing x2term

Practice Use synthetic substitution to find the value of the given polynomial function at the indicated value. 1. P(x) = x4 + x3 – 4x2 – 8x – 10; P(-1) 2. P(x) = 2x3 – 6x + 1:P(3) 3. P(x) = -3x4 + 6x3 + x2 + 5x -2; P(-3) Lesson 2.3 Synthetic Division Synthetic division is a numerical method used for dividing a polynomial by a binomial of the form x – c. YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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To divide a polynomial by a – c 1. Arrange the polynomial in descending powers, insert a 0 coefficient for any missing term. 2. Write c for the divider, also the subtrahend in ( x- c). to its right, write the coefficients of the dividend. 3. Write the leading coefficient of the dividend on the bottom row. 4. Multiply the divider c (in this case,3) by the value just written on the bottom row. Write the product in the next column in the second row. 5. Add the values in the new column, writing the sum in the bottom row. 6. Repeat this series of multiplication and addition until all columns are filled in. 7. Use the numbers in the last row to write the quotient and the remainder in fraction form. The degree of the first term of the quotient is one less than the degree of the first term of the dividend. The final value in this row is the remainder.

Example 1.

x– 3

2x3 – 3x2 – x + 4

-3

-1

-3

-1

Bring down 2

2 4. 3

-3

-1

6 2 5. 3

2 2

6. 3

2 2

multiply by 3

-3

-1

6 3

add

-3

-1

6 3

9 add 8

Multiply by 3

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7. 3

-3

-1

6 3

9 24 add 8 28

Multiply by 3

Written from the last row of the synthetic division 2x2 + 3x + 8 + 2x3 – 3x2 – x + 4

x– 3

Practice Use synthetic division to divide the first polynomial by the second. 1. 5x3 + 6x2 – 8x + 2; x – 5 2. –x7 – x5 – x3 – x – 7; x + 1 3. x5 – 1; x – 1

Lesson 2.4 The Remainder Theorem If a polynomial P(x) is divided by x-c the quotient is Q(x) and the remainder is R. the relationship is defined by P(x) = (x-c) Q(x) + R Thus, R = P(c) Consider the division algorithm when the dividend, P(x), is divided by x – c. in this case, the remainder must be a constant because its degree is less than one, the degree of x – c

P(x) = D(x) Q(x) + R(x) the division algorithm

Dividend Divisor Quotient Remainder P(x) = (x – c)Q(x) + R

the divisor is x – c. called the constant remainder R.

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Evaluate P(x) at c. Thus, P9C) = (c – c) Q (c ) + R

to find P ( c ) , set x = c

P(c) = o Q( c ) + R P( c ) = R Therefore the value of P( c ) is the same as the remainder when P (x) is divided by x – c. The remainder theorem can be used to evaluate a polynomial function at c. rather than substituting c of x in P ( x), we divide the function by x – c. the remainder is P (c ). Example Find the remainder when P(x) = x3 + 5x – 4 is divided by x – 2 Solution When P(x) is divided by x – 2, remainder R = P ( 2) P ( 2) = (2)3 + 5(2) – 4 = 8 + 10 – 4 = 14

Practice Use the remainder theorem to find the remainder when a polynomial is divided by the binomial 1. x3 + 3x2 + 2x – 9; x – 1 2. 3x3 + 12 x2 + 3x + 8; x – 4 3. 2x3 + 4x2 + x – 6; x + 2

Lesson 2.5 The factor Theorem Let P(x) be a polynomial a. If P(c) = 0, then x – c is a factor of P(x). b. If x – c is a factor of P(x), then P(c) = 0 Consider the division algorithm when the divisor is of the form x – c. YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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P(x) = (x – c)Q(x) + R By the remainder theorm, the remainder R is P( c), so we can substitute P ( c ) for R. thus, P(x) = (x – c)Q(x) + P(c ) Note that if P(c ) = 0, then P(x) = ( x - c) Q (x) So that x – c is a factor of P(x). This means that for the polynomial function P(x), if P( c ) = 0, then (x – c ) is a factor of P ( x) Reverse the process and see what happens when x – c is a factor of P(x). this means that P(x) = ( x - c) Q (x) If we replace x with c, we have P( c ) = ( c - c) Q (c ) = 0 Thus, if x – c is a factor of P( x ), then P ( c ) = 0 Example Determine whether x – 4 is a factor of x3- 2x2 – 11x + 12 Solution Let P ( x) = x3- 2x2 – 11x + 12 P ( 4 ) = (4)3 – 2(4)2 – 11(4) + 12 = 64 – 32 – 44 + 12 =0

Practice Use the factor theorem to determine whether the given binomial of the form x – c is a factor of the given polynomial. If it is not, give the remainder when P( x ) is divided by x – c. 1. X + 1; 2x3 + 5x2 + 4x + 1 2. X – 2; 4x3 – 11x2 + 8x – 4 3. X + 3; x4 + 3x3 – 2x2 – 5x + 3 YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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Lesson 2.6 Finding rational Zero Multiple zero of a polynomials If a polynomial P(x) has (x-r) as factor exactly k times, then r is a zero of multiplicity k of the polynomial function y = P(x) For example, the polynomial P(x ) = ( x -3)2(x – 2)3( x + 1) has 3 as a zero of multiplicity 2 -2 as a zero multiplicity 3, and -1 as a zero of multiplicity 1. Finding Integral Zero of Polynomial Functions Let P(x) be a polynomial function in x with integral coefficients. Then the only possible integral zeros of P(x) are the divisors of the constant term. For example, if P(x ) = x3 + 2x2 – 8x + 10, then the possible integral zeros of P( x) are the same divisors of 10; . Find the zeros of P(x) = x3 + 2x2 – 5x – 6 List the integral zeros of P. The possible integral zeros are the divisors of -6, Use synthetic division to find one zero from among the divisors. Try x = 1 1

try x = -1; 2

-5

-6

-2

-8

Since P(1) = -8, 1 is not a zero

-5

-6

-1

-6

since P(-1) = 0 , -1 is a zero

Because -1 is a zero of P(x) thus we have P(x) = ( x + 1) ( x2 + x – 6) The remaining zeros of x3 + 2x2 – 5x – 6 are the factors of x2 + x – 6. Thus,

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X2 + x – 6 = 0 ( x + 3 ) ( x – 2) =0 X = -3 and x = 2 The cubic function has three zeros in the set of complex numbers.

Fundamental Theorem of Algebra A polynomial function P(x) of the degree n has exactly n complex zeros.

The Rational-Zero Theorem Given a polynomial function defined by P(x) = anxn + an-1xn-1 + ….+ a1x + a0 With integer coefficients and where n is a non-negative integer. If

is a rational number

in its simplest form and zero of y= P(x), then p and q are factors of a0 and an respectively. The factor theorem has implications for polynomial functions. That is, if x-c is a factor of P(x), then P( c) = 0. In other words, c is a solution ( or a root of P(x ) = 0 We call c a zero or root of a polynomial function P. hence, if P is a polynomial, then each solution of the equation P (x) = 0 is called a zero of P. to solve quadratic function or to find the zeros of a quadratic function, you may use the quadratic formula, completing the square, or factoring. For cubic function of the form P (x) = a(x-P) (x –q)(x-r), the zeros are p, q and r respectively. A zero polynomial function may be a multiple of zero. For example, the polynomial x2 + 4x + 4 can be expressed in factored form as (x+2)(x+2). Setting each factor equal to zero yields x = -2 in both cases. Thus, -2 is a zero multiplicity of P(x) = x2 + 4x + 4 since the factor ( x+ 2) occurs twice. Multiple zeros of a polynomials If a polynomial P(x) has (x – r) as a factor exactly k times, then r is a zero of multiplicity k of the polynomial function y = P (x). For example, the polynomial P (x) = (x-3)2 ( x+2)3 ( x+1) has 3 as a zero multiplicity of 2, -2 as a zero multiplicity 3, and YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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-1 as a multiplicity 1 There is a simple test that you can use to determine whether a given polynomial function has any integral zeros. Finding integral zeros of polynomial functions Let P (x) be a polynomial function in x with integral coefficients. Then the only possible integral zeros of P (x) are the divisors of the constant term. For example, if P(x) = x3 +2x2 – 8x + 10, then the possible integral zeros of P are the divisors of 10:

Example Find the zeros of P(x) = x3 + 2x2 – 5x – 6 Solution: List all possible integral zeros of P The possible integral zeros are the divisors of -6 Use synthetic division to find one zero from among the divisors. Since P(-1)= 0, -1 is a zero. Because -1 is a zero of P(x), (x+1) is a factor of P(x). Thus, we have P(x) = (x + 1) ( x2 + x – 6) The remaining zeros of x3 + 2x2 – 5x – 6 are the factors of x2 + x – 6. Thus, (x + 3)(x – 2) = 0, x = -3, x =2 The zeros of P(x) = x3 + 2x2 – 5x – 6 are -1, -3, and 2. Fundamental Theorem of Algebra A polynomial function P(x) of degree n has exactly n complex zeros. The Rational-Zero Theorem Given a polynomial function defined by P(x) = anxn + an-1xn-1 + … a1x + a0 With integer coefficients and where n is a non-negative integer. If is a rational number in its simplest form and a zero of y= P(x), then p and q are factors of a0 and an, respectively. For example P(x) = 4x3 + 8x2 + 5x – 3 the possible rational zeros are: : , Find the third degree polynomial function whose zeros are -4, 2, and The polynomial function can be expressed as P(x) = (x+4)(x-2)(3x-2) YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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Thus, the polynomial function whose zeros are -4, 2, is the third degree function P(x) = 3x3 + 4x2 – 28x + 16. Practice: 1. List all possible integral zeros a. x3 – 2x2 – 11x + 12 2. Find all zeros of f(x) = x2(x-3)2(x+4)2 3. List all possible, use synthetic division to test the possible rational zeros and find an actual zeros of P(x) = x3 – 3x – 2 4. Find the polynomial function with the following set of zeros: {-2, 3, -4}

Let P (x) be a polynomial function in x with integral coefficients. Then the only possible integral zeros of P (x) are the divisors of the constant term. For example, if P (x)= x3 + 2x2 – 8x + 10, then the possible integral zeros of P (x) are the divisors of 10: Example: find the zeros of P (x) = x3 + 2x2 – 5x – 6 Solution: List the possible integral zeros of P. The possible integral zeros are the divisors of -6, Use synthetic division to find one zero from among the divisors. Try x = 1 try = -1 1 1 2 -5 -6 -1 1 2 1 3 -2 -1 1 3 -2 -8 1 1

-5 -1 -6

-6 6 0

Since P (1) = -8, 1 is not a zero since P(-1) = 0, -1 is a zero Because -1 is a zero of P(x), ( x+ 1) is a factor of P ( x). thus we have P(x) = (x+1)(x2 + x – 6) The remaining zeros of x3 + 2x2 – 5x – 6 are the factors of x2 + x + 6. Thus, X2 + x – 6 = 0 (x + 3)( x – 2 )= 0 X = -3 and x = 2 Fundamental theorem of algebra A polynomial function P(x) of degree n has exactly n complex zeros. The rational- zero theorem Given a polynomial function defined by P(x) = anxn + an-1xn-1 + … a1x + a0 With integer coefficients and where n is a non-negative integer. If is a rational number in its simplest form and a zero of y = P (x), then p and q are factors of a0, and an, respectively. Example: find the zeros of each polynomial YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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P(x) = 2x2 – 5x – 3 Solution: this is only a simple quadratic function that can be solved by factoring 2x2 – 5x – 3 = 0 (2x + 1)( x – 3) = 0 2x + 1 = 0 or x – 3=0 x= thus,

x=3 and 3 are zeros of P (x) = 2x2 – 5x – 3

Practice: list all the possible integral zeros of each function 1. P (x) = x3 – 2x2 – 11x + 12 2. P(x) = x4 – 2x2 – 16 x – 15

Lesson 2.7 Descartes’ Rule of Signs Rule 1. the number of positive real zeros or roots of P(x) = 0 is either equal to the number of variations of sign of P(x) or is less than this number by a positive even number. 2. the number of negative real zeros or roots of P(x) = 0 is either equal to the number of variations in sign of P(-x) or is less than this number by a positive even integer. Example: determine the possible number of positive and negative real zeros of P(x) = x3 + 6x2 + 11x + 6 Because all the terms are positive, there are no variations in sign. Thus, there are no positive real numbers. To find the possibilities for the positive real zeros, count the number of sign changes in P(-x). replacing x with –x in the given function, we have P(-x) = (-x)3 + 6(-x)2 + 11(-x) + 6 or –x3 + 6x2 – 11x + 6 Thus, P(-x) has three variations in sign. By Descartes Rule of Signs, y=P(x) has either three negative zeros or 3 – 2 = 1 negative real zero. The Upper and Lower Bounds Theorem Let P(x) be a polynomial with real coefficients and with a positive leading coefficient. 1. If M 0 and the coefficients of the quotient and remainder obtain when dividing P(x) by x – M are all greater than or equal to 0. Then y = P(x) has no zeros greater than M. 2. If L and the coefficients obtained when dividing P(x) by x – L are alternately greater than or equal to 0 and less than or equal to 0, then y = P(x) has no zeros less than L. The numbers M and L are called the upper and lower bounds respectively, of the zeros of the given polynomial function. Zero entries may count as positive in between two negatives or negative in between two positives. Example: find the least nonnegative integral upper bound and the greatest non positive integral lower bound of the zeros of P(x) = x4 – x3 – 12x2 – 4x + 16

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Solution: find the greatest non positive integral lower bounds using synthetic substitution. 1 -1 -12 -4 16 -1 1 -2 -10 6 10 -2 1 -3 -6 8 0 -3 1 -4 0 -4 28 The division performed indicates that -2 is a zero and -3 is a lower bound. The possible zeros of P(x) = x4 – 12 x2 – 4x + 16 are . With – 3 as the greatest lower bound, there is no need to use synthetic division and test for the other possible zeros less than -3. Find the least upper bound using synthetic division. 1 -1 -12 -4 16 1 1 0 -12 -16 0 2 1 1 -10 -24 -32 3 1 2 -6 -22 -50 4 1 3 0 -4 0 5 1 4 8 36 196 Since the last numbers in the last row are all positive, 5 is the least upper bound. There is no need to use synthetic division to test for the other possible zeros greater than 5. Thus, the least Integral upper bound and greatest integral lower bound for the zeros of P(x) are 5 and -3, respectively. Practice: 1. Use the Descartes’ Rule of Signs to make a chart summarizing the possible combinations of positive, negative, and imaginary zeros of each of the following polynomial functions. a. P(x) = x2 – x – 5 2. Use Descartes’ Rule of Signs to state the possible number of positive and negative real zeros of P(x) = 2x3 + x2 – 25x + 12 Chapter 3 Exponential and logarithmic functions

Lesson 3.1 Exponential Functions An exponential function can be written as f(x) = bx Where b

, and x is any real number.

Example Evaluate exponential function for the given values of x

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F(x ) = 2x

x=3

If x = 3, then f( x) = 2x F(3 ) = 23 =8 F(x) = 2x + 1

x = -2

If x = -2 then f(x) = 2x + 1 F(-2) = 2 -2 + 1 = 2-1 =

Practice Evaluate the following for the indicated value(s) of x. 1. F(x) = 3x 2. F(x) = 4 – 3x 3. F(x) = ( )

Lesson 3.3 The natural Base e The natural Base e Most banks offer interest that is compounded more than once a year. If interest is compounded n times per year for t years at an interest rate r (expressed as a decimal, a principal p pesos grows to the amount A and is given by this formula. (

)

As the frequency of compounding increases, the amount in the account approaches ₱2 718. Because the number 2.718… is a special number in mathematics it is given a name e. The number e As n increases, (

) approaches e

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e = 2.718281828… The number e is called the natural number. The function exponential function. For the exponential function where x is a variable.

is called the natural , e is the constant 2.71828183…

Example 1 Use a calculator to calculate each expression. a.

press

ex0.01

it will display

1.01005

b. e0.05

press

ex0.05

it will display

1.64872

c. e0.1

press

ex0.1

it will display

1.10517

d. 2

press

2ex-0.24(-3) it will display

4.109

since e > 1, y = ex is an exponential function. Also, because e-1 = < 1 ,

is an

exponential function Example 2 Suppose a bank gives 4% interest compounded continuously. If an account starts with ₱10 000, what will be its value after 1 year, 5 years, and 10 years? Use the formula and 10 thus,

. Evaluate the formula

t=1

10 000e^(0.04 x 1) = 10 408.1

t=5

10 000e^(0.04 x 5) = 12 214.02

t=10

10 000e^(0.04 x 10) = 14 918.24

when P = 10 000, r = 0.04 and t 1,5

Practice: A. Use the calculator to evaluate each given that

1. f(4) 2. f(-2) 3.f(2.3) 4. f(-3.8) YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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5. f ( ) B. Graph each function. 1. f(x) = e-x 2. f(x) = e2x 3. f(x) =

Lesson 3.4 Logarithmic Functions The inverse of exponential function is logarithmic function. The inverse of the exponential function f(x) = bx can be written as f -1(x) = logbx. the number b, which must be positive and not equal to 1, is called the base of the logarithmic function. The expression log bx is called the base b logarithm of x. The logarithm of a positive number y to a positive base b where b is not equal to one is the exponent x to which the base must be raised to obtain the number y. Hence, if b>0, b and b, x, and y are related by the equation Y = bx Logby = x if and only if bx = y for b>0, b A logarithm is an exponent which b must have to produce y. In, either equation, b is called the base and must be a positive number, not equal to 1 The equations X = logby and y = bx are equal. The first equation is in logarithmic form and the second is in exponential form.

Logarithmic functions and their graphs For example, the inverse of y=2x is y=log2x. The graph of the function is

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The logarithm of a positive number y to a positive base b where b is not equal to one is the exponent x to which the base must be raised to obtain the number y. hence, if b>0, b and b, x, and y are related by the equation

Then x, which is the exponent of the base b is the logarithm of y to the base of b. thus the relation y = bx can also be expressed in the equation

We define logarithm as an exponent which is b must have to produce y In either equation, b is called the base and must be positive number not equal to 1. Logarithm form:

x = logby

Exponential form:

y = bx

Example 1 Write each logarithm equation in exponential form 1.) 3 = log5x

means 53= x

2.) log5125=3 means 53=125 3) 4 = logb81 means b4 = 81 = 4 means 10-4 =

4) Example 2

Write each exponential equation in logarithmic form. a. 92 = x means 2 = log9x b. b3 = 64 means 3 = logb64 c. ey = 9 means y = loge9 d.

= 64 means = log1664

Because logarithms are exponents, it is possible to evaluate some logarithms by inspection. The logarithm of x with base b, logb64, we ask, what power of 2 gives 64? Because 26 = 64, log264=6 YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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Example 3 Evaluate each a. log33 b. log216 c. log327 d.log255 e. log41 Solution Logarithmic Expression a. log33 b. log216 c. log327 d. log255

Question needed for evaluation 3 to what power gives 3? 2 to what power gives 16? 3 to what power gives 27? 25 to what power that gives 5?

e. log41

4 to what power gives 1?

Logarithmic Expression Evaluated Log33= 1 because 31 Log216= 4 because 24 = 16 Log327= 3 because 33 = 27 Log255= because =5 0 Log41 = 0 because 4 = 1

Because logarithms are exponents, they have properties that can be verified using properties of exponents. Properties of exponents If b>0 and b

, then

1. = 0 because b0 = 1 2. logbb = 1 because b1 = b 3. = x and = x (inverse properties) 4. If logbx = logby , then x = y ( one-to-one property) 5. Logac = 6.

Example 4 Evaluate each. YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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a) b) c) d) Solution: a) Because logbb = 1, we conclude that b) Because

= 0, we conclude that

c) Because

= x, we conclude that

d) Because

= x, we conclude that

= 15

Graphs of logarithmic functions To sketch the graph of y= log b x , you can use the fact that the graphs of inverse functions are reflections of each other along the line y = x Example 5 Graph f(x) = 3x and g(x) = log3x = in the same x-y plane Solution: Set up a table of coordinates for f(x) = 3x, reverse this coordinates to get the coordinates for the inverse function g(x) = log3x. x

-2 -1 0 1 2 1 3 9

1 3 9 -2 -1 0 1 2

Plot the ordered pairs in both sides. Connect each set with a smooth curve. The graph of the inverse can also be drawn by reflecting the graph of , about the line y = x

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Characteristics of the graphs of Logarithmic Functions    

The x – intercept is 1. There is no y-intercept. The y-axis is a vertical asymptote. If b>1, the function is increasing. If 0<b<1, the function is decreasing. The graph is smooth and continuous. It has no sharp corners or gaps.

The graph of logarithmic functions can be translated vertically or horizontally, reflected stretch or shrunk.

Lesson 3.5 Laws of Logarithms Let b be a positive number not equal to 1. Let x and y be any positive number and n be any real number then Law 1. Logbxy = logbx + logby Law 2. Logb = logbx logby Law 3. Logbxn = n logbx Law 1 Logarithm of a product. The logarithm of a product is equal to the sum of the logarithms of its factors Logbxy = logbx + logby Example: Expand the term a. Law 3 ( 8 5 ) = log3 8 + log35 Write as a single term b. Log3 3 + log3 27 = log3 ( 3 27 ) = log3 81 =4 Law 2. Logarithm of a Quotient. The logarithm of a quotient is equal to the logarithm of the numerator minus the logarithm of the denominator. Logb = logbx

logby

Law 3. Logarithm of a power. YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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The logarithm of a power xp is equal to the product of the exponent p and the logarithm of the base x. Logbxp = p logbx Example Use laws of logarithm to expand log3(8 = log38 + log35 log483 = 3 log48

law 1 law 3

Write the sum as a single logarithm Log33 + log327 = log3(3 = log381 =4 Practice Condense each logarithmic expression 1. Log4(12 2. Log9(9x) Evaluate each 1. Log296 + log232 2. Log464 â&#x20AC;&#x201C; log33

Lesson 3.6 Common logarithms The logarithmic function with base of 10 is called the common logarithmic function. To simplify the notation, we shall agree that when the base of a logarithm is not written, it is understood to be 10. That is, Log y = log10 y Common logarithms of the powers of 10 are easily determined by using the rule, logbbx = x For example; Log 1 000 = log 103 = 3 Log 100 = log 102 = 2 Log 10 = log 101 = 1 Log 1 = log 100 = 0 Log 0.1 = log 10-1= -1 Log 0.01 = log 10-2 = -2 For example we want to find the logarithm of 75, using a calculator to find log 75 , we press the key on the calculator 75 log 1.875061 The number displayed in the calculator is the approximate value of log 75. That is

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To check whether is correct, we must show that the equivalent exponential equation 101.875061 is true. to this, we press the following keys on the calculator: 10 xy 1.875061 = 74.999955 Note that 74.999955 can be rounded off to 75. If log x = y, then x is called the antilogarithm of y. in symbols, x = antilog y. Example Find x if logx = 4.32 Solution We can find x in two ways Using the inverse of the log function (antilog) we have 4.32 INV log 20 892.96 Using the equivalent exponential equation x = 104.32, we have 10 xy4.32 = 20 892.96 In both cases, x = antilog 4.32 = 20 892.96 Hence, log 20 892.96 = 4.32 Practice Use a calculator to determine the value of each common logarithm 1. Log 279 2. Log 123.4 3. Log 23 4. Log 78 256 Lesson 3.7 Change of Base Change of base formula Let a, b, and x be positive real numbers such that a different base as follows: Base a Logbx =

and b

. Then logbx can be converted to to a

Base 10 logbx =

Example Find log316 log316 =

The logarithmic function with base e is the natural logarithmic function and is denoted by the special symbol ln x, read as “ the natural logarithm of x” or “el en of x. YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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The function defined by F(x) = logex = ln x, x >0 Is called the natural logarithmic function Properties of natural logarithm 1. ln 1 = 0 because e0 = 1 2. ln e = 1 because e1 = e 3. ln ex = x and eln x = x 4. if in x = ln y, then x = y Example Use a calculator to evaluate ln 1953 ln 1953 = 7.57712 ln ln = ln 4 â&#x20AC;&#x201C; ln 5 = 1.386294 â&#x20AC;&#x201C; 1.609438 = -0.223144 Practice Evaluate each by using a calculator 1. ln 54 2. ln0.85 3. ln25.83 Find the exact value 1. log3(-27) 2. log91 3. log100

Lesson 3.8 Exponential and logarithmic Equations An exponential equation is an equation containing a variable in an exponent Example 3x = 38 Strategies for solving Exponential and logarithmic Equations 1. Rewrite the given equation in a form that allows the use of the one-on-one property of exponential or logarithmic functions. 2. Rewrite the exponential equation in logarithmic fform and apply the Inverse Property of logarithmic functions 3. Rewrite the logarithmic equation in exponential form and apply the Inverse Property of exponential functions

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Two basic strategies for solving Exponential and logarithmic functions One-on-one properties bx = by logbx = logby Inverse properties blogbx =x logbbx = x Example Solve the equation 4x = 15 Solution 4x = 15 Log44x = log415 X = log415 X= X = 1.9534

A logarithmic equation is an equation containing a variable in a logarithmic expression. If a logarithmic equations is in the form of logbx = c, we can solve the equation by rewriting it in its equivalent exponential form bc = x Example Solve log5(x + 8 ) = 3 Solution log5(x + 8 ) = 3 x + 8 = 53 x + 8 = 125 x = 117

Another example solve 5 + 2 lnx = 4 2 ln x = -1 ln x = eln x = e x=e x Practice Solve each exponential equation 1. 49x = 343 2. 5-x = 125 3. 8x -2 = 16

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Solve each logarithmic equation 1. Log3x = 5 2. ln x = 1 3. log6x2 = 4 Lesson 3.9 Exponential Growth The equation y = aebt is an effective model of many kinds of population growth. In this equation a is the initial population at time 0, y is the population after time t, and b is a positive constant that depends on the situation. The constant b is often called the exponential growth rate. The first we need to do is determine why some things grow at the rate that reflects their current size. Also in studying exponential growth it is common to refer to the starting point of the pattern as stage 0 or the initial value. The delta notation is often used in describing changes in quantities, with the symbol meaning ‘change in”. Steps in developing an experimental model 1. Write the information given as the value of the quantity of two points in time as two ordered pairs (t,y). 2. Substitute the first ordered pair into the model y = aebt and simplify to find a. 3. Substitute the second ordered pair into the model to find b. Example In the year 2000, town A has a population of 20 000. In the following year, there were 800 births and 200 deaths, for a net growth of 600 Predict town A’;s population in the year 2008 In the year 2008, t=8 and P = 20e0.0295688022)(8)

The population would be 25 335

Any quantity for which the rate of change is proportional to the amount present can be modeled by the formula Q = aebt 1. If b >0 the y is growing exponentially 2. If b < 0 then y is decaying exponentially

Chapter 4 Trigonometry

Lesson 4.1 Angles and Angle Measure YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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A directed angle is an ordered pair of rays with common endpoint, one ray called the initial side of the angle and the other called the terminal side of the angle, together with a rotation from the initial side to the terminal side.

Terminal side Initial side Positive angles are generated by counterclockwise rotations and negative angles are generated by clockwise rotations. The angles are often named by Greek letters such as (alpha),

A has a rotation of

of a revolution counterclockwise.

Negative angle (clockwise) If the terminal side of an angle in standard position coincides with a coordinate axis, then the angle is called a quadrantal angle.

Measuring angles using degrees Angles are measured by determining the amount of rotation from the initial side to the terminal side. by definition, the ray has rotated through 360 degrees. A degree, 1 , is of a complete rotation. Two angles with the same initial and terminal sides are called coterminal angles. An angle of x is coterminal with angles x

YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

where k is an integer.

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Example Find appositive angle less than 360 that is coterminal with 420 angle Solution 4200 â&#x20AC;&#x201C; 3600 = 600 A 600 angle is coterminal with the 4200 angle.

Measuring angles using radians One radian is the measure of a central angle the radius of the circle.

that intercepts an arc s equal in length to

Relationships between degrees and radians A measure of 1 degree (10) is equivalent to a rotation of

of a complete revolution about

a vertex. Since 2 radians correspond to one complete revolution, degrees and radians are related. 3600 = 2 1800 = Rules in conversion between degrees and radians 1. To convert degrees to radians, multiply the number of degrees by

2. To convert radians to degrees, multiply the number of radians by

Example Convert 600 to radians Solution 600 = 600

Convert =

rad =

to degrees =

= 300

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Practice A. Express each angle in radians 1. 400 2. -2400 3. 1300 B. Convert to degrees 1. 2. 3.

The radian measure formula,

can be used to find the non negative radian measure

of a central angle of the circle. The lengths of the arc intercepted by the cebtral angle is given by s = .

Example A circle has a radius of 20 centimeter find the length of the arc intercepted by a central angle of 1200 Solution Convert 1200 to radians. Thus, 1200 = 1200

Next, find the length of the arc where =

and r = 20 cm

s= = 20 ( ) = 41.89

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The formula used to measure arc lengths can be applied to analyze the motion of a particle moving at a constant speed along a circular path.

Linear and angular speed If s is the length of the arc traveled in time t of a particle moving at a constant speed along a circular arc of radius r, then the linear speed of the particle is given by Linear speed = Moreover, if is an angle (in radian measure) corresponding to an arc length s, then the angular speed of the particle is given by Angular speed = A relationship between two kinds of speed can be established by dividing both sides of the arc length formula, s = r , by t

Thus, linear speed is the product of the radius r and the angular speed . Linear speed in terms of speed The linear speed, v, of a point a distance r from the center of rotation is given by v = rw where w is the angular speed in radian per unit of time. Example The second hand of a clock is 10.5 centimeters long. Find the linear speed of the tip of this second hand as it passes around the clock face. Solution In one revolution, the arc length traveled by the second hand is S=2 =2 = 21

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The time required for the second hand to travel this distance is t = 1 minute = 60 seconds. The linear speed of the tip of the second hand is Linear speed = = = 1.099 cm/sec.

Practice A. Complete the table by giving the radian measure of the central angle of a circle that intercepts an arc of length s. Radius, r 20 inches 16 ft 12.8 cm 75 km

Arc length, s Radian measure of the central angle, 5 in. 4 ft. 20 cm 100 km

B. Find the length of the arc on a circle of radius r intercepted by central angle Radius, r 12 inches 9 feet 8 yards 16 cm 30 feet

Central angle, 45 3150 2250 600 600

Arc length, s

C. Solve 1. A wheel is rotating at 250 revolutions per minute. Find the angular speed in radians per second. 2. A turntable turns at 33 revolutions per minute. Find the angular speed in radians per second. 3. A car with a tire of radius 16 inches is rotating at 450 revolutions per minute. Find the speed of the car to the nearest kilometers per hour.

Lesson 4.2 The Unit Circle YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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The standard equation of a circle with center (h,k) and radius r is given by (x-h)2 + (y-k)2 = r2 A Unit Circle is a circle whose center is a t (0,0) and whose radius is 1 unit. Consider a circle of radius 1 cm whose center is at the origin. This is called a unit circle represented by the equation x2 + y2 = 1. Since the circumference of the circle is 2 , then the unit circle has a circumference of 2 units. Thus, a circle of radius 1 centimeter has a circumference of 2 centimeters or approximately 6.28 centimeters. In the unit circle, the radian measure of the angle is equal to the measure of its intercepted arc. One-half of the circumference or equal to . ( of 2 ) Three fourths of 2 or circumference is equal to An arc length is the distance between two points along a circle expressed in linear units. Example Inscribe a square in a unit circle. Find the length of the arcs whose endpoints are the vertices of the square and find the coordinates of these vertices. Solution To do this, bisect each quadrant. The intersections of the angle bisectors and the circle are the vertices of the inscribed square. YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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Since the circle is divided into eight congruent arcs, then of 2 of 2 of 2 of 2

Practice Answer each How far will a point travel of it goes 1.

of the way around the unit circle?

Lesson 4.3

Trigonometric Functions of Real Numbers Let

be a real number and let (x,y) be a point on the unit circle corresponding to . Sin = y

csc = , y

Cos

sec = , x

Tan

=x ,x

cot =

Example Evaluate the six trigonometric functions of Solution The point P on the unit circle that corresponds to

, has coordinates (-1,0). We use x

= -1 and y = 0 to find the values of the trigonometric functions. YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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Sin

= y = -1

csc

Cos

=x=0

sec

Tan

cot

Domain and range of sine and cosine functions The domain of the sine and the cosine functions is the set of all real numbers The range of the sine and cosine functions is the set of real numbers from -1 to 1 inclusive.

Even and Odd Trigonometric Functions The cosine and secant functions are even, for any Cos (- ) = cos

and sec (-

in R,

= sec

The sine, cosecant and cotangent functions are odd. For any Sin ( Tan (

) = - sin , = - tan , and

in R,

csc (- ) = -csc cot (

= - cot

Periodic functions A function f is periodic if there exist a positive real number c, such that f( For all in the domain of f. the smallest number c for which the equation is satisfied is called the period of f. If we begin at any point P on the circle and travel a distance of 2 units along the circumference, we will return to the same point P. because the trigonometric functions are defined in terms of the coordinates of that point P we have the following results. Sin (

and cos

= cos

The sine and cosine functions are periodic functions and have a period 2 Trigonometric functions of any angle

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Sin

csc

Cos

sec

Tan

cot

Example Given tan

and sin

is positive, find the other trigonometric function values of .

Solution The tangent function is negative in the second and fourth quadrants, and the sine function is positive in the first and second quadrants. Therefore, P is in the second quadrant √ √ √ √ = 13

The other trigonometric function values of

are

Sin

csc

Cos

sec

Tan

cot

Reference Angles A reference angle is acute angle between the terminal side and the x-axis. Example Angle

is a reference angle of angle

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, 2200 and 3200

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In general, an acute angle

is the reference angles of angles 180 –

Practice A. Find the other trigonometric functions values of 1. Sin 2. Cos 3. Sec B. Find the reference angle 1. 1950 2. 3210 3. -1500

Lesson 4.4 Right triangle trigonometry Trigonometric functions in a Right Triangle Let

be an acute angle of a right triangle. The six trigonometric functions of angle

Sin

cos

Csc

tan

sec

cot

Example Find the remaining trigonometric function values of

if sin

and

an angle in a

right triangle. Solution By using the definitions of trigonometric functions and the Pythagorean Theorem, we have Sin

, hence opp = 24 and hyp = 25

Adj = √ =√

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=√ =7

Sines, Cosines and Tangents of Special Angles √

√

Notice that sin 300 =

√ = cos 600. Also sin 600 =

√

= cos 300 and sin 450 =

√

= cos 450.

This happens because 300 and 600 are complementary angles as well as 450 and 450. In general, it can be shown that the cofunctions of complementary angles are equal. This means that, if is an acute angle, then the following relationships are true. Sin ( 900 –

cos (900 –

Tan (900 –

cot (900 –

Sec (900 –

csc (900 –

Example Find the exact value of sin2 300 and cos2 450 Solution Substitute the values if sin 300 and cos 450 and simplify. sin2 300 + cos2 450 = ( )

√

( ) =

Test Answer the following A. Fill in the blank to make the statement true YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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1. If the legs of a right triangle are 9 cm and 40 cm long, the length of its hypotenuse is __________. 2. In a 30-60-90 triangle, if the side opposite the 300-angle is 2 cm long, then its hypotenuse is __________ cm long. 3. In an isosceles right triangle, if the hypotenuse is √ cm long, each leg is __________ cm long. B. Find the exact value of each expression. 1. Csc 300 – sec 300 2. sec cos 3. 2 sin 300 – sec 600 tan 450

Lesson 4.5 Inverse Trigonometric Functions We can define trigonometric functions whose inverse are functions. This can be done by restricting either the domain of the basic trigonometric function or the range of their inverse relation. This can be done in several ways. For the inverse sine function, we choose the range close to the origin that allows all inputs in the interval [-1, 1] to have function values. Thus, we choose the interval *

+for the range.

For the inverse cosine function, we choose the range close to the origin that allows all inputs in the interval [-1,1] to have function values. The interval [0, ] satisfies the condition. For the inverse tangent function, we choose the range close to the origin that allows all real numbers to have function values. The interval *

+satisfies this requirement.

Inverse Trigonometric Functions Function

Domain

y= y=

[

y= y = Arcos

[

y= y = Arctan

(

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Range

[ 0, ]

)

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The notation arcsin x arises because the function is the length of an arc on the unit circle for which the sine is x. The notation is not an exponential notation. It does not mean

. Either of the two kinds of notation can be read “the inverse sine of x” or the

arc sine of x” or “the number (or angle) whose sine is x. that is Read: y is a number whose sine is x Or y

where arc means “the angle whose” Read: y is the angle whose sine is x.

The graphs of the inverse trigonometric functions are as follows:

Domain: ⌈

⌉

Domain: ⌈

Range: *

Range:[ 0, ]

Domain: [

⌉

Range:(

)

The next diagrams show the restricted ranges for the inverse trigonometric functions on a unit circle. It is important that these ranges be memorized. The missing endpoints indicate inputs that are not in the domain of the original function.

The exact values of *

can be found by thinking of

as the angle in the interval

+ whose sine is x. For example, (

)

( Since we are associating

) with an angle, we will represent such an angle as .

Example 1 YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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Find the exact value of each of the following. ( )

(

√

)

Solution: a. Let

= ( )

Find the angle Sin

whose sine value is equal to . That is,

and

Because

= , then

= . Thus,

( )= . Verify the answer b. Find the angle Sin

√

whose sine value is equal to

√

and

Because sin (

√

)

(

√

)

We know that f( and ) ( )= . We apply these properties to functions sine, cosine, tangent, and their inverse functions to have the following properties. Inverse Properties 1. 2. 3. cos 4. 5. 6.

for every x in the interval [-1,1] for every x in the interval * + =x

for every x in the interval [-1, 1] for every x in the interval [0, for every real number x for every x in the interval * +

Calculators give the approximate values of inverse trigonometric functions. Use the keys marked , Example

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Use the calculator to find the value of each term 1. 2. 3. Solution Function

Mode

keystrokes

Display,rounded to five decimal places

radian

1รท 3 = sin-1

0.33984

radian

1 รท 4 = cos-1

1.31812

radian

9.43

-1.46515

tan-1

Practice Use a calculator to evaluate each expression. 1. Sin-10.2 2. Sin-1(-0.53) 3. Cos-1 4. Tan-1(-24)

Lesson 4.6 Solving Right Triangles Right triangles can be model many applications in the real world. In this type of application, you are given one side of a right triangle and one of the acute angles and asked to find the other side, or you are given two sides and asked to find one of the acute angles. We can solve a right triangle when we are given any of the following: 1. 2. 3. 4. 5.

An acute angle and the adjacent leg An acute angle and the opposite leg An acute angle and the hypotenuse. A leg and the hypotenuse Two legs.

Steps in solving right triangles YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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1. 2. 3. 4.

Draw a right triangle Label the sides and the vertices of the right triangle Enter the given data in the figure Use the appropriate trigonometric function to find the unknown parts of the triangle

Example Solve the right triangle given a = 30, and c = 65 Solution Find A by using the sine function Sin A =

(0.4615 INV SIN) m⦟A = 27.5 find B m⦟B = 90 = m⦟A m⦟B

90 – 27.5

m⦟B

62.5

Solve b by using the Pythagorean Theorem b=√ =√ =√

Definition The angle between a horizontal and a line of sight above the horizontal is called an angle of elevation Line of sight

Angle of elevation

Horizontal line YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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The angle between a horizontal and a line of sight below the horizontal is called an angle of depression. Horizontal line Angle of depression Line of sight

Example Suppose an airplane begins it descent to a runway from an altitude of 12,000 ft. a. If the airplane is 78,000 ft. from the runway as measured along the ground, what is the angle of depression? b. What is the length of the runway if the angle of elevation from the far end of the runway to the airplane is 6.20?

Solution Make a drawing of a situation

Angle of depression 6.20 Runway

12 000 ft 78 000 ft

a. Use the fact that the measures of the angle of elevation and the angle of depression are the same. Use the tangent function to find

The angle of depression is 8.750

b. The distance from the airplane to the far end of the runway as measured along the ground is 78,000 + x, where x is the length in feet of the runway. Thus, Tan 6.20 = X = YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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X The runway is about 32 461.87 ft.

Bearings Navigation is the art and science of directing craft or vehicle from one position to another. In surveying and navigation, directions are generally given in terms of bearings. A bearing measures the acute angle a path or a line of sight makes with a fixed northsouth line as shown in the figures. For instance, the bearing S 400 E in ( a ) means 40 degrees east of south.

N W

E 400 S 400 degrees east of south

N 550 W

S 55 degrees west of north

N 650 W

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S 65 degrees east of north

E 330 S 33 degrees west of south

Practice I. Give the lengths of the sides and the angles 1. A = 750 a = 8.45 2. b= 90, c = 200 3. a = 235, b = 85 II.

Solve the problem 1. A 15-foot ladder is resting against a wall and made an angle of 480 with the ground. Find the height to which the ladder will reach on the wall. 2. If the sun is 280 above the horizon, find the length of a shadow cast by a building that is 495 feet tall.

Lesson 4.7 Fundamental Identities Basic Identities Fundamental Identity A. Reciprocal Identities 1. ; 2. ; 3. ; B. Ratio Identities 4. ;

Alternative ; ; ;

5. YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

Restriction on Not coterminal with 0, Not coterminal with Not a quadrantal angle Not coterminal with Not coterminal with 0, Page 63

C. Pythagorean Identites 6.

none

Not coterminal with

Not coterminal with 0,

Trigonometric identities can be used to evaluate trigonometric function values. Example 1 √

Use the values

and

to evaluate the remaining trigonometric

functions. Solution: Using reciprocal identity, we have

√

√ √

Example 2 Perform the indicated operations and simplify

Practice and Application I. Use trigonometric identities to simplify each expression. 1. 2. II. Multiply then simply each product. 1.

) Factor each expression then simplify the result.

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IV. Perform the indicated operations and apply fundamental identities to simplify each expression. 1.

Lesson 4.8

Verifying Trigonometric Identities Guidelines for verifying Trigonometric Identities 1. Work with one side of the equation at a time. It is often better to work with the more complicated side first. 2. Look for opportunities to factor an expression, add fractions, square a binomial, or create a monomial denominator. 3. Look for opportunities to use fundamental identities. Note which functions are in the final expressions you want. Sines and cosines pair up well, as do cosecants and cotangents, and do secants and tangents. 4. If there is no direct identity that can be substituted to reduce one side to the other, change everything into sines and cosines. 5. Always try something. Even paths that lead to dead ends give you insights. Example 1 Verify the identity: Solution: The left side is more complicated, so start with it. Apply a quotient identity:

and

A reciprocal identity: Divide out common factor Apply the reciprocal identity

Example Verify the identity :

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Solution:

Practice and Application Verify each identity. 1. cos x csc x 2. 3.

Lesson 4.9 Sum and Differences Formulas Trigonometric identities and formulas Formula Addition and Subtraction Formulas For any 1. 3.

5. 6.

The cosine subtraction formula can be stated in words. The cosine formula The cosine of the difference of two angles is the sum of the product of their cosines and the product of their sines Example Find the exact value of sin 750 Solution Write 750 as 450 + 300 and use the sum formula for sine. Sin 750 = sin(450 + 300) YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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= sin 450 cos 300 + cos 450 sin 300 =

√

Practice Apply trigonometric identities to find the exact value of each expression. 1. Cos 2850 2. Tan 750 3. Cos 1050

Lesson 4.10 The Law of Sines, Cosines and Tangents For any triangle ABC,

Note: 1. The Law of Sines can also be written in the reciprocal form. 2. The Law of sines can also take several forms. a=

Example Solve each triangle ABC using the given information 1. a=12 cm, b = 23 cm, and A = 340 Solution Find B using the Law of sines

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Sin B = =

Practice Solve each triangle 1. If A = 420, B = 960, and b = 12 2. If A = 420, B = 480, and c = 12 3. If A = 330, C = 1280, and a = 16

Chapter 5 Statistics and probability Lesson 5.1 Statistics Statistics is branch of mathematics that deals with the collection organization presentation analysis and interpretation of data. Statistics involves much more than simply drawing graphs and computing averages. A population is a complete collection of all elements (scores, peopleâ&#x20AC;Ś) to be studied A census is a collection of data from every element in a population A sample is a subcollection of elements drawn from a population. Types of samples 1. In a random sample, each member of the population has an equally likely chance of being selected. The members of the sample are chosen independently of each other. 2. A convenience sample is a sample that is chosen so that it is easy for the researcher.

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3. In a stratified random sample, the population is divided into subgroups, so that each population member is in only one subgroup. In here, individuals are chosen randomly from each subgroup. 4. A cluster sample is a sample that consists of items in a group such as a neighborhood or a household. The group maybe chosen at random. 5. A systematic sample is obtained using an ordered list of the population, thus selecting members systematically from the list. The nature of data 1. Quantitative data consists of numbers representing counts or measurements. 2. Qualitative data can be separated into different categories that are distinguished by some nonnumeric characteristics. Quantitative data can either be discrete and continuous. 1. Discrete data result from either a finite number of possible values of possible values or countable number of possible values as 0, or 1, or 2, and soon. 2. Continuous data result from infinitely many possible values that can be associated with points on a continuous scale in such a way that there are no gaps or interruptions. Another way to classify data is to use four levels of measurement 1. Nominal level of measurement is characterized by data that consists of names, labels, or categories only. 2. Ordinal level of measurement involves data that maybe arranged in some order but differences between data values either cannot be determined or are meaningless. 3. Interval level of measurement is like the ordinal level. But meaningful amounts of differences between data can be determined. It has no inherent (natural) zero starting point (where none of the quantity is present. 4. Ratio level of measurement is the interval level modified to include the inherent zero starting point (where zero indicates that none of the quantity is present). Test Identify each 1. A collection of data from every element in a population. 2. A complete collection of all elements to be studied. 3. A branch of mathematics that deals with the collection, organization, presentation, analysis, and interpretation of data. 4. Can be separated into categories that are distinguished by some nonnumeric characteristics 5. Consist of numbers representing counts or measurements. YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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6. A sample that over-represent or under-represent parts of the population 7. Each member of the population has an equally likely chance of being selected. 8. A sample that consist of items in a group such as a neighborhood or a household. 9. It is characterized by data that consist of names, labels, or categories only. 10. A sample that is chosen so that it is easy for the researcher.

Lesson 5.2

Presentation of Data A. Graph 1. The line graph is used to represent changes in data over a period of time. In line graph, data are represented by points and are joined by line segments. A line graph may be curved, broken or straight. Generally, the horizontal axis is used as the time axis and the vertical axis is used to show the changes in the other quantity. Example The table shows the number of students in high schools in Region 1 from 1970 to 2000. Draw a line graph to represent the data. State a conclusion that you can make from the graph. Year 1970 1980 1990 2000

Enrollment in millions 1.3 1.9 1.8 1.5

Solution Use a vertical line axis for the number of enrollees and the horizontal axis for the year.

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2 1.8 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0 1970

1980

1990

2000

It appears that enrollment peaked in 1980 and has been declining since.

2. The bar graph is a graph which uses horizontal or vertical bars to represent data. When a bar graph has bars which extend from left to right, it is called horizontal bar graph. On the other hand, if the bars extend from bottom to top, it is called a vertical bar graph. In a vertical bar graph, the vertical line is called the scale of the bar graph while the horizontal bar graph, the horizontal line is the scale of the bar graph. Example Arnold surveyed a sample of people at a basketball game to find out their favorite drink. The results are shown in the table. Represent the data using a bar graph. Type of drinks Cola Root beer Lemon Fruit Iced tea

Frequency 25 20 10 15 12

Solution

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30 25

frequency

20 15 10 5 0 cola

root beer

lemon

fruit

iced tea

Type of drink

3. The pie graph or pie chart is another visual representation of data. It is used to show how all the parts of something are related to the whole. It is represented by a circle divided into slices or sectors of various sizes that show each part’s relationship to the whole and to the other parts of the circle. To create a circle graph of pie graph 1. First express the number in each category as a percentage of the total. 2. Convert this percentage to an angle in the circle. Multiply the percent by 360 to find the number of degrees for each category. 3. Use a protractor to construct a circle graph.

Example The first month operation expenses of frequency Enterprises are shown as follows in the table. categories salaries rent advertising Materials and supplies amount Ᵽ60 000 Ᵽ30 000 Ᵽ20 000 Ᵽ10 000 Construct a circle graph showing the expenses of frequency Enterprises. Solution YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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a. Write the ratio of each entry to the total entries, as a percent. This is done by finding the total ( âąŁ120,000) and then dividing each entry by the total. Salaries: Rent : Advertising:

= 0.50 = 50% = 0.25 = 25% = 0.167 = 0.17= 17%

Materials and supplies:

= 0.083 or 0.08 = 8%

b. A circle has 3600, so the next step is to multiply each percent by 3600 to get the angle of the sectors. Salaries: 360 x 0.50 = 1800 Rent: 360 x 0.25 = 900 Advertising: 360 x 0.17 = 600 Materials and supplies: 360 x 8% = 300

Expenses for Frequency Enterprises

materials and supplies 25% advertising 8%

salaries 50%

rent 17%

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4. A pictograph is a graph that uses pictures to illustrate data. To construct a pictograph, the following steps are to be followed: 1. Collect the necessary data. 2. Round off the numerical data if necessary. 3. Choose an appropriate symbol to represent the subject. 4. Indicate the quantity each symbol represent. Example Draw a pictograph for the given facts below The annual budget of School during 2011 â&#x20AC;&#x201C; 2015 Year 2011 2012 2013 2014 2015 Budget P3 000 000 P4 000 000 P5 000 000 P5 000 000 P5 000 000 Solution Choose a symbol to represent the annual budget Let = P1 000 000 Annual budget of school during 2011 â&#x20AC;&#x201C; 2015

2011

2012

2013

2014

2015

B. Tables A statistical table contains four main components: table heading; body; stubs and boxheads or column captions. The table heading shows the table number and the table title. The main part of the table is the body. It contains the quantitative information that we are looking for the table. Stubs are the labels or categories which presented as values of a variable, usually found at the left portion of the table. The boxheads are the captions that appear above the columns. In addition to the four components, a footnote and a source note may be placed immediately below the main part of the table.

Example

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Number of cars registered in the key cities of Metro Manila by type of diesel used: 2001 – 2007 Box Heads

Stubs

cities Manila Quezon City Caloocan Pasay San Juan Makati Paranaque Pasig

gas 27 300 32 583 29 123 15 876 22 336 28 645 15 223 17654

diesel 10 428 13 647 8 762 8 634 11 428 9 978 8 351 4 378

body

Practice Present the data given 1. A group of 200 students were asked to name their favorite food. Show the result given in the table on bar graph.

Chicken and rice Beef and rice Pork and rice Salad French Fries Noodles

80 30 35 10 15 40

2. A man inherited Ᵽ2 000 000 from his parents. He invested it in Housing Ᵽ1 000 000 Money Market Ᵽ 500 000 Beauty Product Ᵽ 300 000 Food Production Ᵽ 200 000 Show this information on a pie graph

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Lesson 5.3 Frequency Distribution 11

Frequency is the total number of times each score appears in the distribution. To fill the f, we go back to the original scores and enter each score starting from the first score in the list by placing a slash alongside the score 11. Continue until all the score have been recorded. The total frequency in this distribution is 60 (â&#x2C6;&#x2018; scores 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10

f | 0 || | | || || | |||| || |||| || || |||| - | |||| | |||| |||| |||| - |||| |||| ÎŁf = 60

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A class interval is made up of the lower limit and an upper limit. In the class interval of 10-11, the lower limit is 10 and the upper limit is 11, hence the left member of the class interval is the lower limit while the right member is the upper limit. To construct a grouped frequency distribution 1. Get the difference between the highest score and the lowest score. Add 1 to the difference to arrive at the total number of scores or potential scores. The difference between the highest score and the lowest score is called the range. 2. Decide on the number of class intervals which is appropriate to the given set of data. Divide the final number in step 1 by the desired number of the class intervals, then at the width of class interval (i). 3. Write the lowest score in the set of raw scores as the lower limit in the lowest class interval. Add to this value i – 1 to obtain the upper limit in the lowest class interval. The lowest score is 10. Thus, the lowest class interval is 10 – 11 since 10 + i – 1 = 10 + 2 – 1 = 11. 4. The next lower limit can be obtained by adding I to the lower limit of the previous class interval. To get the corresponding upper limit is 12 and the corresponding upper limit is 13 since 10 + 2 = 12 and 11 + 2 = 13. 5. Continue the step 4 until all the scores are included in their corresponding class intervals. 6. Fill in the f column by following what we have done in the frequency distribution.

Class interval 28 – 29 26 – 27 24 – 25 22 – 23 20 – 21 18 – 19 16 – 17 14 – 15 12 – 13 10 - 11

f 1 3 3 3 6 6 8 6 10 14 N = 60

The true limits of a class interval The true limits or class boundaries of a given score is the score plus or minus one half of the unit of measure or the place value of the given score. Example 5,

or 4.5 – 5.5

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Class interval 28 – 29 26 – 27 24 – 25 22 – 23 20 – 21 18 – 19 16 – 17 14 – 15 12 – 13 10 - 11

True limits 27.5 – 29.5 25.5 – 27.5 23.5 – 25.5 21.5 – 23.5 19.5 – 21.5 17.5 – 19.5 15.5 – 17.5 13.5 – 15.5 11.5 – 13.5 9.5 – 11.5

f 1 3 3 3 6 6 8 6 10 14 N = 60

Cumulative Frequency Distribution A cumulative frequency distribution can be obtained by adding the frequency starting from the frequency of the lowest class interval up to the frequency of the highest class interval. It is also possible to do the reverse, that is, we start to cumulate in the other direction. This is the greater than cumulative frequency. Cumulative frequency distribution Apparent limits 28 – 29 26 – 27 24 – 25 22 – 23 20 – 21 18 – 19 16 – 17 14 – 15 12 – 13 10 - 11

True limits 27.5 – 29.5 25.5 – 27.5 23.5 – 25.5 21.5 – 23.5 19.5 – 21.5 17.5 – 19.5 15.5 – 17.5 13.5 – 15.5 11.5 – 13.5 9.5 – 11.5

f 1 3 3 3 6 6 8 6 10 14 N = 60

<cf 60 59 56 53 50 44 38 30 24 14

>cf 1 4 7 10 16 22 30 36 46 60

The graph of a cumulative frequency distribution is represented by a line graph known as Ogive. In graphing the “less than cumulative frequency”, the cumulative frequencies are plotted against the true limits or upper limits or class boundaries. While graphing the “greater than cumulative frequency”, the cumulative frequencies are plotted against the true lower limits or lower class boundaries

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less than ogive of the frequency distribution 70 60 50 40 30 20 10 0 11.5

13.5

15.5

17.5

19.5

21.5

23.5

25.5

27.5

29.5

25.5

27.5

Greater than ogive of the frequency Distribution 70 60 50 40 30 20 10 0 9.5

11.5

13.5

15.5

17.5

19.5

21.5

23.5

Cumulative Percentage Distribution To convert a cumulative frequency distribution to a cumulative percentage distribution, simply divide each term in the <cf by N and multiply by 100. The cumulative percentage Distribution falls below or above a class boundary. Apparent limits 28 – 29 26 – 27 24 – 25 22 – 23 20 – 21 18 – 19

f 1 3 3 3 6 6

<cf 60 59 56 53 50 44

<cpf 100.00 98.33 93.33 88.33 83.33 73.33

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16 14 12 10

– 17 – 15 – 13 - 11

8 6 10 14 N = 60

38 30 24 14

63.33 50.00 40.00 23.33

The cumulative Percentage distribution enables us to determine what percent of the distribution fall below or above a class boundary. The table shows that 88.33 percent of the students got a score lower than 23.5. Test The following table shows the marks a gained in an arithmetic examination by 100 pupils. The maximum possible mark was 100. marks 91 – 100 81 – 90 71 – 80 61 – 70 51 – 60 41 – 50 31 – 40 21 – 30 11 – 20 1 - 10

frequency True limits <cf 5 11 14 17 23 14 8 4 2 2

<cpf

1. Complete the table 2. Identify: a. i b. Total frequency c. Lower limit of the interval 31 – 40 d. True limit of the interval 71 – 80 3. How many pupils got marks less than 70.57? 4. How many pupils got marks greater than 50.5? 5. What percent of the pupils gor score lower than 61.5?

Lesson 5.4 Measures of Central Tendency Central tendency refers to that numerical value in the central region of a distribution of scores. YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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There are three measure of central tendency: 1. The Mean (commonly called the average) of a set of n numbers is the sum of all the numbers divided by n. it is the part of the distribution around which the values balance. 2. The Median is the middle number when the number in a set of data is arranged in descending order. It provides the necessary information about the value of the middle position in the distribution. 3. The Mode is the number that occurs most often in a set of data which means the score with the highest frequency. A set of data can have more than one mode.

Formula Mean (grouped data) ̅=∑ Where ̅ = mean F = frequency Xm = class mark (average of lower interval and upper interval) ∑ = sum of the product of frequencies and class marks N = total frequency Example Calculate the mean of 40 students given scores f 98 – 100 2 95 – 97 1 92 – 94 1 89 – 91 6 86 – 88 6 83 – 85 5 80 – 82 9 77 – 79 2 74 – 76 3 71 - 73 5 Solution In order to use the formula, we still have to add the columns for class mark (Xm) and fXm

scores 98 – 100 95 – 97 92 – 94 89 – 91 86 – 88 83 – 85 80 – 82

f 2 1 1 6 6 5 9

Xm 99 96 93 90 87 84 81

fXm 198 96 93 540 522 420 729

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77 – 79 74 – 76 71 - 73

̅=∑

2 78 156 3 75 225 5 72 360 N = 40 ΣfXm = 3339

= 83.475

The mean score of 40 students in a Math Quiz is approximately equal to 83.475

The median of a Grouped Data The first step in the computation of the median of a grouped data is to determine the class interval which contains the ( ) score. This can be located under the column <cf of the cumulative frequency distribution. The class interval that contains the ( ) score is called the median class of the distribution. To calculate median, we use the Formula The Median of a grouped data ̃=

)

Where ̃ = Mean = the lower boundary or true lower limit of the median class N = total frequency = cumulative frequency before the median class = frequency of the median class I = size of the class interval

Cumulative Frequency Distribution of a 30-point Math Quiz

Scores 28 – 29 26 – 27 24 – 25 22 – 23 20 – 21 18 – 19 16 – 17 14 – 15 12 – 13

f 1 3 3 3 6 6 8 6 10

<cf 60 59 56 53 50 44 38 30 24

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10 – 11 14 14 N = 60

Median class

Solution

score = ( )

= 30 th score

The class interval that contains the 30th score is 14 – 15 . XLB = 13.5 Cfb = 24 Fm = 6 I=2 ̃=

) = 13.5 + (

) = 13.5 + 2 = 15.5

This means that 50% of the students got scores below 15.5 or if the passing score is 50 percentof the total number of points, almost one-half of the class failed in that particular quiz. The Mode of a Grouped Data In the computation of the mode given a frequency distribution, the first step is to get the modal class. The modal class is that class interval with the highest frequency. To compute for the mode, we use the formula ̂ = XLB + *

Where XLB = lower boundary of the modal class = difference between the frequency of the modal class and the frequency of the class interval preceding it = difference between the frequency of the modal class and the frequency of the class interval succeeding it I = size of the class interval Example Find the mode of the data scores 98 – 100 95 – 97

f 2 1

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92 – 94 89 – 91 86 – 88 83 – 85 80 – 82 77 – 79 74 – 76 71 - 73

1 6 6 5 9 2 3 5 N = 50

Modal class: 80 – 82 XLB = 79.5 =9–2=7 =9–5=4 I=3 ̂ = XLB + *

+ = 79.5 + *

Practice Find the mean, median, and mode for each set of data. x 65 – 69 60 – 64 55 – 59 50 – 54 45 – 49 40 – 44 35 – 39 30 – 34

f 7 8 9 15 12 7 4 8

Lesson 5.5 Other Measures of Position The Quartiles (Q1, Q2, Q3) The quartiles are points that divide a distribution into four equal parts. 1. 1st quartile (Q1). There are 25% of the observations below Q 1 and 75% of the observations above Q1. Q1 is sometimes called the lower quartile. 2. 2nd quartile (Q2). There are 50% of the observations below and 50% of the above Q2. The second quartile is also the median, hence, the former name is seldom used. 3. 3rd quartile (Q3). There are 75% of the observations below Q3 and 25% of the observations above Q3. Q3 is sometimes called the upper quartile.

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There are n observations in a set of data, then Q1 can be identified as the observation, and Q3 as the

[

observation.

Formula Q1 and Q3 (grouped data) Q1 = Xlb + * Where

Xlb = lower boundary of the Q1 class N = total frequency Cfb = cumulative frequency before Q1 class Fq1 = frequency of the Q1 class i = size of the class interval

Q3 = Xlb + * Where

Xlb = lower boundary of the Q3 class N = total frequency Cfb = cumulative frequency before Q3 class Fq3 = frequency of the Q3 class i = size of the class interval

Example Use the table to calculate the Q1

class interval 134 - 139 128 â&#x20AC;&#x201C; 133 122 - 127

f 10 9 8

<cf 50 40 31

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116 – 121 110 – 115 104 – 109 98 – 103 92 – 97 86 – 91

1 5 2 9 5 1 N = 50

23 22 17 15 6 1

Q1 class: Class interval: 98 – 103 XLB = 97.5 cfb = 6 fq1 = 9 i=6 Q1 = Xlb + *

Q1 = 97.5 + (

)

= 101.83

The Percentiles The percentiles are the score-points that divide a distribution into 100 equal parts. For example, the 10th percentile (P10) separates the lowest 10 % from the other 90%; 25th percentile (P25) separates the lowest 25 % from the other 75% while the 80th percentile (P80) separates the lowest 80% from the other 20%. P5 = Xlb + *

Example Use the frequency distribution to calculate P10 class interval 134 - 139 128 – 133 122 - 127

f 10 9 8

<cf 50 40 31

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116 – 121 110 – 115 104 – 109 98 – 103 92 – 97 86 – 91

1 5 2 9 5 1 N = 50

P10 class:

23 22 17 15 6 1

Class interval: 92 – 97 XLB = 91.5 Cfb = 1 Fp10 = 5 i=5 P10 = Xlb + * = 91.5 + *

+ +

= 96.3

The Deciles( D1,D2, ….D9) The deciles are the score –points that divide a distribution into ten equal parts. The deciles are computed in the same way that the median, the quartile, and the percentiles are calculated.

Formula D1 = Xlb + *

D7 = Xlb + *

Example YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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Us use the frequency distribution to calculate D5

Class interval 134 – 139 128 – 133 122 – 127 116 – 121 110 – 115 104 – 109 98 – 103 92 – 97 86 - 91 N=

f 10 9 8 1 5 2 9 5 1 50

<cf 50 40 31 23 22 17 15 6 1

Solution D5 class: Class interval: 122 – 127 XLB = 121.5 Cfb = 23 FD5 = 8 i=6 D5 = Xlb + *

= 121.5 + *

= 123

Test Use the table to answer item 1 – 10 x 29,000 – 31,999 26,000 – 28,999 23,000 – 25,999 20,000 – 22,999

f 2 2 5 7

<cf

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17,000 – 19,999 14,000 – 16,999 11,000 – 13,999 8,000 – 10,999 5,000 – 7,999

10 18 6 7 3 N - 60

1. Complete the table 2. The class interval containing the Q1 class 3. The frequency of the Q1 class 4. The class interval containing the Q3 class 5. The lower boundary (XLB) of the Q3 class 6. The class interval containing the D4 class 7. The frequency of D4 class 8. The class interval containing the P30 class. 9. The lower boundary of the P90 class 10. The width of the class interval.

Lesson 5.6 Measures of Variation and Dispersion The measures of spread are commonly termed as measures of dispersion or measures of variation. There are five measures of variation 1. 2. 3. 4. 5.

Range Quartile deviation Mean deviation Variance Standard deviation

The Range In the measures of variation, it is the range which gives us the quickest measure. The range is simply defined as the difference between the highest score (h.s.) and the lowest score (l.s.). In the group data, the range is the difference between the upper boundary (u.b.) of the highest class interval and the lower boundary (l.b.) of the lowest class interval. Example YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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The range of the set of scores:{11,12,13,15} is 15 – 11 = 4 The range of the frequency distribution

scores 98 – 100 95 – 97 92 – 94 89 – 91 86 – 88 83 – 85 80 – 82 77 – 79 74 – 76 71 - 73

f 2 1 1 6 6 5 9 2 3 5

Range = u.b. – l.b. = 100.5 = 70.5 = 30

The Mean Deviation (M.D.) The Mean Deviation is a measure of variation that makes use of all the scores in a distribution. This is more reliable than the range. Formulas Mean Deviation (ungrouped data) M.D. =

∑

̅̅̅

Where X represents the scores in the distribution ̅ is the mean N is the number of observation

Grouped Frequency Distribution

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M.D. =

̅̅̅

∑

or M.D. =

∑

̅̅̅

Where f represents the frequency observation X. Steps in calculating Mean Deviation (M. D.) 1. Compute the mean of the distribution. 2. Add the entries in column |X - ̅|. this means, subtract the mean from each score and get the absolute value (Deviation from the mean). 3. Get the total of the score under the heading |X - ̅|. 4. Divide the sum obtained in step 3 by N. Example Find the mean deviation of ungrouped frequency distribution x 10 12 12 14

Calculate the mean ̅=∑ = Add the column |x - ̅| X 10 12 12 14

|x - ̅| 2 0 0 2 ̅ =4

M.D. =

= =1

Example Calculate the mean deviation of grouped frequency distribution Add the entries in column Xm, Class interval 134 – 139 128 – 133 122 – 127

f 10 9 8

Xm 136.5 130.5 124.5

, fXm 1365 1174.5 996

̅ . Add the entries under column fXm and ̅ 18.6 12.6 6.6

YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

̅.

̅ 186 113.4 52.8 Page 91

116 – 121 110 – 115 104 – 109 98 – 103 92 – 97 86 - 91

1 5 2 9 5 1 N = 50

118.5 112.5 106.5 100.5 94.5 88.5

118.5 562.5 213 904.5 472.5 88.5 Σ = 5895

Calculate the mean using the formula ̅ = ̅=∑

∑

0.6 5.4 11.4 17.4 23.4 29.4

0.6 27 22.8 156.6 117 29.4 Σ

. Then fill the column

̅ = 705.6 ̅ and

̅.

= 117.9

Next, find the mean deviation Using the formula M.D. =

̅̅̅

∑

= 14.11

The Variance and the Standard Deviation The variance is defined as the quotient of the sum of the squared deviations from the mean divided by N – 1 while the standard deviation is the square root of the variance. Formula Variance (S2) and Standard Deviation S2 =

∑

and

S=√

∑

Steps 1. 2. 3. 4. 5. 6.

Calculate the mean. Get the difference between each score and the mean. Get the square of the difference. Get the sum of the squared deviations in step 3. Divide the sum of the squared differences by n – 1. The number obtained is called the variance. Take the square root of the variance, the standard deviation.

Example Find the standard deviation of the prices of 250-gram powdered soap YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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Solution X 90 73 78 79 83 95 77 79 74 82 ΣX = 810 ̅= = 81

(X-̅ 9 -8 -3 -2 2 14 -4 -2 -7 1 Σ( X - ̅

Variance = S2 = S2 =

∑

̅ 81 64 9 4 4 196 16 4 49 1 Σ

̅ = 428

= 47.55

Standard Deviation = S = √ S=√

∑

= 6.89

The variance and Standard Deviation of Grouped Data Formula of raw score method Variance ∑

(∑

)

Standard Deviation √

∑

(∑

)

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Example Find the variance and standard deviation of the following distribution. X 30 – 34 25 – 29 20 -24 15 – 19 10 – 14

F 4 5 6 2 3 N = 20

Xm 32 27 22 17 12 Σf xm = 465

Fxm 128 135 132 34 36

1024 729 484 289 144

f 4096 3645 2904 578 432 Σ = 11655

Substitute: Σf xm = 465 Σ

= 11655

In the formula: ∑

(∑

)

√

Practice A. 1. 2. B.

Find the range of each set of ungroup data 7, 23, 33, 38 28, 22, 27, 30 Find the range of each of the following grouped data x f 30 -34 9 25 -29 8

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20-24 7 15 -19 6 10-14 5 x 35 – 38 31 – 14 27 – 30 23 – 26 19 – 22 15 – 18 11 - 14

f 6 3 2 4 8 2 3

Find the variance and standard deviation x 45 -51 38 – 44 31 – 37 24 – 30 17 – 23 10 - 16

f 5 97 116 82 19 6

Lesson 5.7 Fundamental Principle of Counting The fundamental principle f counting states that we can find the total number of ways that two or more separate tasks can happen by multiplying the number of ways each task can happen separately. If one thing can occur in m ways and a second thing can occur in n ways, and a third thing can occur in p ways, and so on, then the sequence of things can occur in m x n x p x… ways Example Three coins are tossed. How many outcomes are possible? Solution Each coin can land in one of two ways. Head or tail [

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There are 8 possible outcomes Practice Answer each question 1. A multiple choice test has four questions. Each question can be answered with a, b, c, or d. how many outcomes are possible? 2. Four coins are tossed. How many possible outcomes are possible? 3. Five dice are rolled. How many outcomes are possible?

Experiment – an activity with observable results. Outcomes – the result of an experiment. Sample space – the set of all possible different outcomes of an experiment Event – a collection of outcomes When a coin is tossed, getting a head is an event. The said event consist of the outcome “head”. In throwing a die, getting a “4” is an event. This means that the outcome is 4. Any activity that involves a chance such as rolling a die is an experiment. The result of an experiment is an outcome. The set of all different outcomes of an experiment is called a sample space of the experiment. In describing the sample space of an experiment, you will write all the possible outcomes of the experiment. Example Find the possible outcomes for each experiment a. Tossing a coin once Solution The sample space for this experiment can be written as S = {H,T}. Thus, n(S) = 2 There are two possible outcomes in the sample space b. Throwing a coin and a die together

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If the coin comes up head, there are six possible outcomes for the coin and the die. H1,H2,H3,H4,H5,and H6. If the coin comes up tail, there are six possible outcomes for the coin and the die: T1, T2, T3, T4, T5, and T6 Thus, the sample space for the experiment can be written as S = { H1,H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6} Then, n(S) = 12 The following rules are to be followed whenever we are to analyze an experiment 1. Identify an outcome. 2. Find all outcomes. 3. Count the total number of outcomes. Practice Answer each question 1. A multiple choice has four questions. Each question is answered with a, b, c, or d. how many outcomes are possible? a. How many outcomes show a? b. How many outcomes show b? c. Suppose the multiple-choice test has five questions. How many sets of answers are possible? d. Suppose e were included as a possible answer. How many sets of answers are possible on a four test/. On a five test?

Lesson 5.8 Permutations A permutation is an arrangement of things in a definite order or ordered arrangement of distinguishable objects without allowing repetitions among objects. In general, if n is a positive integer, then n factorial denoted by n! is the product of all integers less than or equal to n. n! = n . (n-1). (n-2)â&#x20AC;Ś 2.1 As special case, we define 0! = 1

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Most scientific calculators have a key that can calculate a factorial. The key will look like x! or n! depending upon the kind of calculator. For example, to find 4!, enter 4, press factorial key, and you obtain 24. Example 7! = 7 x 6 x 5 x 4 x 3 x 2 x 1 = 5040 4!5! = (4 x 3 x 2 x 1)(5 x 4 x 3 x 2 x 1) = 2880 =

= 15

nPr

= means the number of permutations that can be made with n things taken r things at a time. Formula The number of permutations of n things taken n at a time is given by nPn = n(n-1)(n-2) â&#x20AC;Ś(3)(2)(1) = n! The number of permutations of n taken r at a time is given by nPrn(n-1) (n-2) â&#x20AC;Ś (n-r +1) Example Compute 7P3 = Solution Use the formula nPr with n=7 and r = 3 7P3

= 7 x 6 x 5 = 210

Circular Permutations Circular Permutations is a special case of permutation where the arrangement of things is in circular pattern. It is called a circular permutation. The most common example of this type is the seating arrangement of people around a circular table. When people are seated around a circular table, we cannot say that one seated on the first seat as what we have on linear permutations (arrangement of things in a line). In general, there is no first place in the arrangement of things in a circular pattern. So when each moves his position by one place clockwise or counterclockwise, the relative positions are not changed. Formula The number of circular permutations of n different things is

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(n – 1)! = (n – 1)(n – 2)…(2)(1)

The number of permutations of n different things around a key ring and the like is –

–

Example In how many ways can 5 persons be seated around a circular table? Solution This is a circular permutation of 5 things Therefore, there are (5 – 1)! = 4! = 24 Distinguishable permutations Example In how many ways can be arranged in a key ring? Solution The number of permutations of 5 keys around a key ring is

Permutations with Repetitions The numbers of permutations of n things where a things are alike, b things are alike, and c things are alike, and so forth, is P=

Example Find the number of distinguishable permutations of the letters in each word. Heterogeneous Solution

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There are 13 letters in the word “heterogeneous” but there 4 e’s and 2 o’s. hence the number of distinguishable permutation is given by P=

= 129,729,600

Test Evaluate each 1. 2. 3. 4.

8! 10! 12! 4! + 5!

How many different ways can the letters of each word be arranged? 1. PARALLEL 2. GOOGOPLEX 3. MATHEMATICS

Lesson 5.9

Combinations A combination is a selection made from a group of items without regard to their order. The number of combinations of n things taken r at a time is given by nCr

Example Evaluate 5C2 Solution 5C2

= 10

Test 1. Evaluate each a. 8C4 b. 4C4

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2. From a class of 32 girls and 18 boys, how many study groups of 3 girls and 2 boys can be formed? 3. Describe the difference between the permutation and combination.

Lesson 5.10

Probability Experiment – an activity with observable results Outcomes – the results of an experiment Sample space – the set of all possible different outcomes of an experiment Example Example Find the possible outcomes for each experiment c. Tossing a coin once Solution The sample space for this experiment can be written as S = {H,T}. Thus, n(S) = 2 There are two possible outcomes in the sample space d. Throwing a coin and a die together If the coin comes up head, there are six possible outcomes for the coin and the die. H1,H2,H3,H4,H5,and H6. If the coin comes up tail, there are six possible outcomes for the coin and the die: T1, T2, T3, T4, T5, and T6 Thus, the sample space for the experiment can be written as S = { H1,H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6} Then, n(S) = 12 The following rules are to be followed whenever we are to analyze an experiment YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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4. Identify an outcome. 5. Find all outcomes. 6. Count the total number of outcomes.

Practice Answer each problem 1. There are three red pen, 4 blue pens, 2 black pens and 5 green pens in a drawer. Suppose you choose a pen at random. a. What is the probability that the pen chosen is red? b. What is the probability that the pen chosen is blue? c. What is the probability that the pen chosen is red or black? 2. Suppose you draw one card at random from a standard deck of 52 playing cards. Find the probability of each outcome. a. The king of hearts b. The green of spades c. A red card d. Not a red card 3. Two dice are rolled. What is the probability of getting a. A 4 and a 6? b. not 3 and a 5? c. The same number on both die? d. A sum of 8?

Simple and Compound Events The sample space of an experiment is a universal set whose elements are the outcomes of the experiment. Sometimes, we are only interested in looking at a particular outcome. For example, if we bet for the number 6 in a game of dice then we are interested only in the outcome 6. Or sometimes, we bet of outcomes {2,4,6} which is a part of the sample space {1,2,3,4,5,6}. Any set of outcomes derived from the sample space is called an event of the sample space. We denote a sample space by capital letter S. Similarly, we will use a capital letter to denote an event. Event- a subset of a sample space of an experiment Subset â&#x20AC;&#x201C; Let E be an event of the sample space S. since every outcome in E is an outcome in S, we say that event E is a subset of S, denoted by E S. YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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Since the events of an experiment are subsets of the sample space of the experiment, we can talk about the union of two events. We can also consider the complement of an event. Probability is the measure of how likely an event is to occur. The more likely an event is to occur, the higher its probability. The less likely an event is occur, the lower its probability. An event is either impossible, certain, or somewhere between these extremes. An event is mathematically impossible if it can never happen mathematically certain if it always happen. If an event is as likely as not, the probability that it will happen is the same as the probability that it will not happen. When all possible outcomes are equally likely, the probability of an event E written P( E ) is given by the formula. P(E)= A probability is a number from 0 through 1 and is often written as a fraction in lowest terms. Example The spinner may stop on any one of the eight numbered sectors of the circle. ( assume that the spinner will not stop on the line between two sectors). Use the spinner at the right to find each probability. a. P(2) b. P(9) Solution P(2) =

the number of favorable outcome is one.

P(9) = 0 there is no 9 in the sample space

Probability of Success and of failure If an event can succeed in s ways and fail in f ways, then the probability of success, P(s) and of failure,P(f) are as follows. P(s) =

P(f) =

Example

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A box contains 10 blue marbles, 8 red marbles, 6 yellow marbles, and 5 white marbles. What is the probability that if you choose a marble from the box without looking , or at random, will you choose a red marble? Solution The probability of choosing a red marble is written P(red) There are 8 ways to choose a red marble and there are 10 + 6 + 5 or 21 ways not to choose a red marble. So, s = 8 and f = 21 The probability of selecting a red marble is

Investigation Fair and Unfair Games In a fair game, each player has the same probability of winning. In an unfair game, the probabilities of winning are not equal. Experimental results or theoretical probability can be used to determine whether the game is fair and unfair. Rolling thunder is a game for two players who take turns rolling two numbers cubes. If sum of the numbers rolled is a multiple of 3, player X scores one point. If the sum rolled is multiple of 4, players scores one point. 1. 2. 3. 4. 5.

Play a game of Rolling Thunder with a partner. Does a player score for each roll of the number cubes? Can both players score on the same roll? Explain List the possible outcomes when two number cubes are rolled. How many outcomes are favorable to player x? how many are favorable to player Y? 6. Is Rolling Thunder a fair game?

Chapter 6 Introduction to Calculus Lesson 6.1

Limits of Functions If there is a number L to which the terms an of the sequence becomes closer as more terms are taken we say that that limit of an as n increases without bound is L. in symbols, YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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=L For the successive values of

, for n = 1, 2, 3, â&#x20AC;Ś, 10,100, 1000, as n gets larger,

approaches 1, hence, ( For an=

)

, we must have noted that

is always positive and gets smaller and smaller

as more and more terms are taken. Mathematically, this means that

( ) = 0.

For an = n, you must have noticed that an increases without approaching any number and can be greater than any number. In symbols,

Properties of Square Roots 1. =0

To evaluate algebraically certain limits of sequences, we assume the following properties of limits. If

= A, 1. 2. 3. 4. 5.

= B, and C is a constant, then =C + = =( =

=C + )(

= CA =A+B =A B

= , provided that B

Example: evaluate each a.

= using property 3 = Basic limit and property 1 = Hence there is no limit or the limit does not exist. = =

Distributive law

property 3

=3+0 =3

property 1 and basic limit

Let us extend the concept of limits to functions.

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f(x)

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Let us consider the values of the function, f(x) =

-2 -1.5 -1.1 -1.01 -1.001 -0.999 -0.99 -0.9 -0.5 0

2 2.5 2.9 2.99 2.999 3.001 3.01 3.1 3.5 4

as x approaches -1. Notice that -1 is not a domain of f(x) because the denominator is zero when x = - 1. The table shows the evaluation of f(x) for several x-values as x approaches -1. From the values in the table we guess that f(x) gets very close to 3 as x approaches 1. We write this as Generalizing the concept of a limit of a function, we have Let (a) denote a real number and let f(x) denote a function. We say that L is the limit of f(x) as x approaches (a) if f(x) gets closer and closer to L as x approaches (a). in this case, we write Limit theorems Let us suppose that both valid. Theorem 1 Example

and

exist. Then the following rules are

C is a constant. =4

Theorem 2 Example

Theorem 3 Example

[

Theorem 4 Example

[

for all positive numbers n. applying theorem 2 with a = 2 and n = 3

(3)4

for all constants of C. = 567

= )= = 2(4)3â&#x20AC;&#x201C; 3(4)2 = 128 â&#x20AC;&#x201C; 48

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= 80 [ Theorem 5 [ [ Theorem 5 states that the limit of the product is the product of the limits. Thus, Example

[

(x2 + 3) = = =[ [ =[ [ = 56

Another solution: =

theorem 4

= (2)5 + 3(2)3 = 32 + 24 = 56 Theorem 6

provided

Example

= =

theorem 4

= = Theorem 7

[

Example

â&#x2C6;&#x161;

for all positive number r.

=[ = = = =3 YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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Theorem 8

If f(x) is a polynomial function, then

Example

= 3(3)4 – 2(3)3 + 4(3) – 3 = 243 – 54 + 12 – 3 = 198

Rule

Rational Function Rule , where f(x) and g(x) are polynomials in x and g(a)

Example

for f(x) = 1 – 2x2 and a = 3, evaluate

Solution:

substituting f(x) = 1 – 2x2 and a = 3, into the expression and simplifying, we obtain = =

[

= =

combining like terms

factoring

Since the denominator is zero at x = 3, we factor the numerator in order to apply the function rule. =

= [

Thus, =

Hence, the desired limit is

Practice Evaluate each limit by using the limit theorems YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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1. 2. 3.

Lesson 6.2 Continuity A function is continuous at x = a if the following three conditions are satisfied; 1. The function is defined at x = a, that is, f(a) exist. 2. The function approaches a limit as x approaches a, that is, 3. The limit approached is the value of the function at point a, that is

exists.

If one or more of these three conditions fails to hold at a, then the function is said to be discontinuous at a. Example:

given f(x) = 3x – x2. Determine if f is continuous at x = 4

Solution:

f(4) = 3(4) – (4)2 = 12 – 16 =

(1) is satisfied

= = 12 – 16 =

(2) is satisfied (3) is satisfied

Therefore, the function f(x) = 3x – x2 is continuous at x – 4.

Practice Find the value of x for which the following functions are discontinuous. 1. f(x) = 2. f(x) = 3. f(x) =

lesson 6.3 YOUNG JI INTERNATIONAL SCHOOL/COLLEGE

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The Derivative The first derivative of a function at a point is the slope of the tangent to the curve of the function at that point. The slope m of the line in terms of tangent of M = tan =

is given by

The slope of the non-vertical line is a constant. That is, the rate of change as x changes is constant over the length of line. For any other curve, the slope is not constant and must be determined for each particular point of interest. f(x) =

by the definition of slope

but = f(x+ Hence we can write the slope as

– f(x)

by substitution

When ↔ , the secant line is moved in the manner mentioned earlier, we are actually approaching 0. Thus, the limiting position of the secant line PV is the line tangent to the curve at Q. Taking the slope

at the limiting position, we have

This clearly illustrates that the derivative of a function f(x) is the slope of the tangent to the curve at any value of x. the process of getting the derivative of a function is called differentiation. The derivative of a function can be denoted by any of the given notations. 1. (x), if the main function is given as f(x). 2. , if the function is given as y. 3. Dxy, read as “derivative of y with respect to x” 4.

if y is expressed as an explicit function.

f(x)

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The value of the derivative at (a)is given by any of these notations: 1.

(a)

|x = a

4. 5. The four step rule for finding the derivative of a function. 1. Consider the function y = f(x). Find (f(x function for f(x). 2. Find f(x + 3. Divide by

in the

to obtain

4. Find the limit as

approaches 0 to obtain

Example: Find the derivative of f(x) = Solution: 1. f(x +

+1=

2. f(x +

3. 4.

= = 2x +

= =

= 2x + 0 = 2x Because approaches 0, therefore,

Practice Find the derivative of the given function by using the formula

1. 2. 3. 4.

f(x) = 3 f(x) = 2x + 13 f(x) = -7x + 42 f(x) = 7x + 3

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