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

##### 1 Answer

#### Explanation:

First, simplify

#f(x)=x^2-4x-2x+8+2x+4#

#f(x)=x^2-4x+12#

Find the point the tangent line will intercept.

#f(-2)=(-2)^2-4(-2)+12=24#

The tangent line passes through the point

To find the slope of the tangent line, find

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

The slope of the tangent line is:

#f'(-2)=2(-2)-4=-8#

However, the question asks for the **normal line**. The normal line is simply the line perpendicular to the tangent line at the point on the function.

Since perpendicular lines have opposite reciprocal slopes and the slope of the tangent line is

We know the slope is

#y-24=1/8(x+2)#

Check a graph:

graph{((x-4)(x-2)+2x+4-y)(y-24.125(x+2))=0 [-15, 20, -29.3, 200]}