Question

What is the equation of the line normal to `f(x)=x ^3-x^2` at `x=-2`?

Answer 1

The equation of the normal line is `y=-1/14x-85/12`.

Explanation:

We have the function, `f(x)=x^3-x^2`.

At `x=-2`, `f(x)=(-2)^3-(-2)^2`

`=-8-4`

`=-12`

So, the point is at `(-2,-12)`.

First, we find the slope of the tangent line at `x=-2`, and that's the derivative of the function.

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

So at `x=-2`,

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

`=3*4+4`

`=14`

Now, the normal line is perpendicular to the tangent line, and we know that the slope of the normal line is the negative reciprocal of the tangent line, as

`m_1m_2=-1`, where `m_1,m_2` are the gradients of the tangent line and normal line, respectively.

And so, the slope becomes, `-1/14`.

The point-slope form states that,

`y-y_0=m(x-x_0)`

And so,

`y-(-12)=-1/14(x-(-2))`

`y+12=-1/14(x+2)`

`y+12=-1/14x-1/7`

`y=-1/14x-1/7-12`

`=-1/14x-85/12`

Here is a graph of the line: