Question

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

Answer 1

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

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 `x=color(red)2`, the slope of the 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`.

`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 `x=2`, so:

`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 `(color(red)2, color(blue)(-e^4))`. Therefore, using the point-slope formula, the equation of the normal line is:

`(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