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

Jan 17, 2016

$y - 100 = \frac{1}{40} \left(x + 3\right)$

Explanation:

First, find the point the normal line will intercept.

$f \left(- 3\right) = {\left(4 + 6\right)}^{2} = 100$

The normal line will pass through the point $\left(- 3 , 100\right)$.

To find the slope of the normal line, we must first know the slope of the tangent line. The slope of the tangent line can be found through calculating $f ' \left(- 3\right)$.

First, find $f ' \left(x\right)$. Through the chain rule,

$f ' \left(x\right) = 2 \left(4 - 2 x\right) \cdot \left(- 2\right) = - 4 \left(4 - 2 x\right)$

The slope of the tangent line is

$f ' \left(- 3\right) = - 4 \left(4 + 6\right) = - 40$

However, the normal line is perpendicular to the tangent line. Perpendicular lines have opposite reciprocal slopes. Thus, the slope of the normal line is the opposite reciprocal of $- 40$ which is $\frac{1}{40}$.

Relate the point the normal line intercepts, $\left(- 3 , 100\right)$, and the slope of the line, $\frac{1}{40}$, in an equation in point-slope form:

$y - 100 = \frac{1}{40} \left(x + 3\right)$