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

##### 1 Answer

#### Explanation:

First, we take the derivative of the function:

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

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

Therefore, at *tangent* line is:

#f'(color(red)2) = 2(color(red)2) - e^(color(red)2+2)#

#f'(color(red)2) = 4 - e^4#

Since the normal line is perpendicular to the tangent line, its slope will be the negative reciprocal of the tangent line's slope. Let's call that slope

#m = (-1)/(f'(color(red)2)) = (-1)/(4-e^4) = 1/(e^4-4)#

We know the slope of the line; now, let's find the point where it intersects the curve. This is at

#f(color(red)2) = (color(red)2^2-4) - e^(color(red)2+2)#

#f(color(red)2) = 0 - e^4#

#f(color(red)2) = -e^4#

So the curve passes through the point

#(y-y_1) = m(x-x_1)#

#(y - (color(blue)(-e^4))) = 1/(e^4-4) (x - color(red)2)#

#y + e^4 = (x-2)/(e^4-4)#

You can manipulate this however you want to get whatever form you need.

*Final Answer*