Question

What is the equation of the line normal to `f(x)=3(x-1)^2-6x+4` at `x=-2`?

Answer 1

`y-43=1/24(x+2)`

Explanation:

Find the point the normal line will intercept.

`f(-2)=3(-2-1)^2-6(-2)+4=43`

The normal line will intercept the point `(-2,43)`.

To find the slope of the normal line, know that the normal line is perpendicular to the tangent line. The slope of the tangent line can be found through calculating `f'(-2)`. The normal line, since it's perpendicular, will have an opposite reciprocal slope.

Find `f'(x)`:

`f'(x)=2*3(x-1)*d/dx[x-1]-6`

`=6(x-1)-6`

`=6x-12`

The slope of the tangent line is:

`f'(-2)=6(-2)-12=-24`

Thus, the slope of the normal line will be `1/24`.

Relate the slope of `1/24` and point `(-2,43)` in a line in point-slope form.

`y-43=1/24(x+2)`