Calculus:Finding The Equation of The Line Tangent to The Curve

Equation of Tangent to The Curve Given the function

f (x) = x3 − 4x Find the equation Text of the tangent lineText of f at the point where

x=2 Derivative of a polynomial

dy dx

a = na n

n−1

f (x) = x3 − 4x dy dx

= 3x2 − 4

Evaluate the first derivative at

x=2 2 dy = 3(2) − 4 (2) dx dy dx

(2)= 8 Slope of the tangent line

f (x) = x3 − 4x x=2 3 f (2) = (2) − 4(2) f (2) = 8 − 8 Point located in f f (2) = 0 (2, 0) Slope= dy (2)= 8 dx Find the y value for

With slope and a point you can write the equation of the line tangent

y − y1 = m(x − x1 ) y + 0 =8(x − 2) y = 8(x − 2)