# How do you find the equation of a line tangent to the function y=x^3-x at (-1,0)?

Jan 25, 2017

Derive the function $y$; plug in your given point in that derivative function; find the negative inverse of that slope; use the point-intercept formula to find the equation of that line.

#### Explanation:

Let $y = f \left(x\right)$

$f ' \left({x}^{3} - x\right) = 3 {x}^{2} - 1$

$f ' \left(- 1\right) = 3 {\left(- 1\right)}^{2} - 1 = 2$

This is the slope of the line at point $\left(- 1 , 0\right)$

The tangent line slope is the negative inverse of $2 \implies - \frac{1}{2}$

So, we have our tangent line slope and our point.

Using the formula: $y - {y}_{1} = m \left(x - {x}_{1}\right)$

We get: $y - \left(0\right) = \left(- \frac{1}{2}\right) \left(x - \left(- 1\right)\right)$

Therefore, our equation will look like:

$y = - \frac{1}{2} x - \frac{1}{2}$