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

##### 1 Answer

The vertical line

#### Explanation:

The normal line passes through the point

The normal line is perpendicular to the tangent line. So, using the derivative, we can find the slope of the tangent line at

To find this derivative, let's use the quotient rule:

#f'(x)=((d/dx(2x-4)^2)e^x-(2x-4)^2(d/dxe^x))/(e^x)^2#

We'll need the chain rule for

#f'(x)=(2(2x-4)^1(d/dx(2x-4))e^x-(2x-4)^2e^x)/e^(2x)#

Note that

#f'(x)=(4(2x-4)e^x-(2x-4)^2e^x)/e^(2x)#

We can simplify, if we want:

#f'(x)=((2x-4)e^x(4-(2x-4)))/e^(2x)#

#f'(x)=((2x-4)(8-2x))/e^x#

#f'(x)=(4(x-2)(3-x))/e^x#

So we see that