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

##### 1 Answer

#### Explanation:

First, find the point on the curve that the normal line will pass through:

#f(-1)=(-4)^3(1)=-64#

So the normal line passes through

Now recall that the normal line is perpendicular to the line tangent to the curve. So, if we know the slope of the tangent line at this point, we take its opposite reciprocal to find the slope of the perpendicular normal line.

To find the slope of the tangent line we take the derivative of the function. We could expand

#f'(x)=(d/dx(x-3)^3)(x+2)+(x-3)^3(d/dx(x+2))#

We use the chain rule on

#f'(x)=3(x-3)^2(d/dx(x-3))(x+2)+(x-3)^3(1)#

Moreover,

#f'(x)=3(x-3)^2(x+2)+(x-3)^3#

#f'(x)=(x-3)^2[3(x+2)+(x-3)]#

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

Then the slope of the tangent line at

#f'(-1)=(-4)^2(-1)=-16#

Thus, the slope of the normal line is

The line passing through

#y+64=1/16(x+1)#