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

Jan 11, 2016

$y - 43 = \frac{1}{24} \left(x + 2\right)$

#### Explanation:

Find the point the normal line will intercept.

$f \left(- 2\right) = 3 {\left(- 2 - 1\right)}^{2} - 6 \left(- 2\right) + 4 = 43$

The normal line will intercept the point $\left(- 2 , 43\right)$.

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 $f ' \left(- 2\right)$. The normal line, since it's perpendicular, will have an opposite reciprocal slope.

Find $f ' \left(x\right)$:

$f ' \left(x\right) = 2 \cdot 3 \left(x - 1\right) \cdot \frac{d}{\mathrm{dx}} \left[x - 1\right] - 6$

$= 6 \left(x - 1\right) - 6$

$= 6 x - 12$

The slope of the tangent line is:

$f ' \left(- 2\right) = 6 \left(- 2\right) - 12 = - 24$

Thus, the slope of the normal line will be $\frac{1}{24}$.

Relate the slope of $\frac{1}{24}$ and point $\left(- 2 , 43\right)$ in a line in point-slope form.

$y - 43 = \frac{1}{24} \left(x + 2\right)$