Question

Algebra pleaseeeeeee help

Answer 1

Answer:

Step-by-step 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.