# How do you find the equation of the line tangent to y=x^3 at the point (-1,-1)?

Sep 22, 2016

$y = 3 x + 2$

#### Explanation:

$\left({x}_{0} , {y}_{0}\right) = \left(- 1 , - 1\right)$

$y = k x + n$ where $k = f ' \left({x}_{0}\right)$

$f ' \left(x\right) = 3 {x}^{2} \implies k = f ' \left({x}_{0}\right) = 3$

The tangent line goes through $\left(- 1 , - 1\right)$ so:

$- 1 = 3 \cdot \left(- 1\right) + n \implies n = 2$

Finally, the equation of the tangent line is:

$y = 3 x + 2$