Question

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

Answer 1

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(x)`

`f'(x^3-x)=3x^2-1`

`f'(-1)=3(-1)^2-1=2`

This is the slope of the line at point `(-1,0)`

The tangent line slope is the negative inverse of `2 => -1/2`

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

Using the formula: `y-y_1=m(x-x_1)`

We get: `y-(0)=(-1/2)(x-(-1))`

Therefore, our equation will look like:

`y=-1/2x-1/2`