Find the equation of the line that is tangent to the graph of y = x3 at the point (-2 , -8).

Question

1. Answer in the form: Ax + By + C = 0, where A, B and C are relatively prime integers and A > 0.

Answer 1

y = x^3

y' = slope = 3x^2 at x=-2 = 3(-2)^2 = 12

y'(-2) = 12

for the tangent line,

y=mx + b where m=12

y=12x + b plug in the point (-2,-8) to solve for b

-8 =12(-2) + b

b= -8 +24 = 16

y=12x +16 is the line tangent to y=x^3 at the point (-2,-8)