Question

find an equation for the line that is tangent to the curve y = 2x^3 - 2x at the point (-1,0). the equation is y = . (type an expression using x as the variable.)

Explanation:

Step1: Find derivative

$y' = 6x^{2}-2$

Step2: Evaluate slope at $x = - 1$

$m=y'(-1)=6(-1)^{2}-2 = 4$

Step3: Use point - slope form

$y - 0=4(x + 1)$

Answer:

$y = 4x+4$