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

##### 1 Answer

#### Explanation:

First, find the point the normal line will intercept.

#f(-3)=(4+6)^2=100#

The normal line will pass through the point

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

First, find

#f'(x)=2(4-2x)*(-2)=-4(4-2x)#

The slope of the tangent line is

#f'(-3)=-4(4+6)=-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

Relate the point the normal line intercepts,

#y-100=1/40(x+3)#