Answer:
Explanation:
Remark
I have to assume that you know calculus. It is the only way the problem can be done that I know of. If you don't, I'm not sure how you will do this.
The curve is of y = e^(-2x) + x^2 - 3
The curve crosses the y axis when x = 0. The y value is
y = e^0 + x^2 - 3
yint = 1 + 0 - 3
yint = -2
The slope at point (0,-2) is
y' = -2e^(-2x) +2x
y' = -2 at point A
Therefore the normal will have a slope
m1 * m2 = - 1
The slope of the curve C at A = -2
The equation of the tangent line at A = -2x - 2
Call this m1
m2 = slope of the normal
-2 * m2 = -1
m2 = 1/2
Equation of the line (l) =
y = 1/2 x - 2
The graph is shown below. Notice the two lines actually look like they are at a 90 degree angle.