Determine the number of lines that are tangent to the curve y= and pass through the point (0,-2).

Question

Determine the number of lines that are tangent to the curve y = ((e^2x+1)*(x-2))/2 and pass through the point (0,-2).

Answer 1

There are 2 lines. You don't actually have to find them.

The key facts are that f(0) = -2 and the function is concave, then convex. x = 0 will, of course, have one of the tangent lines. The tangent lines while the curve is concave will cross the y-axis higher and higher as x moves away from 0, then increasingly lower where the function is convex, eventually crossing at y = -2 again. This will be the second line.

f'(x) = xe^2x -2e^2x + 1/2e^2x + 1/2 = xe^2x -3/2e^2x + 1/2

f''(x) = 2xe^2x - 2e^2x. This has the pattern mentioned above (concave, then convex).

FYI, the second line is tangent to the curve at 1.2790,-5.0145