5. (8 pts) Compute the derivative of h(x) = x^3+x^2/e^3x
and simplify your answer.

  h'(x) = (-3x^3 +2x)/e^(3x)

Step-by-step explanation:

The formula for the derivative of a quotient is ...

  (u/v)' = (vu' -uv')/v²

Using u = (x^3 +x^2) and v = e^(3x), this becomes ...

  h'(x) = (e^(3x)(3x^2 +2x) -(x^3 +x^2)(3)(e^(3x))/e^(6x)

  h'(x) = (-3x^3 +2x)/e^(3x)

if the eqn of the parabola is f(x) = (x-4)(x+2), we need to mult. this out to find the coordinates of the vertex.

then f(x) = x^2 + 2x - 4x - 8, and f(x) = x^2 - 2x - 8.

one way in which you can find the vertex is to calculate x = -b / (2a).   here, that comes to

        )

x = = 1.   f(1) = 1^2 - 2(1) - 8 = - 9.   vertex is at (1, -9).

        2(1) 

find another point on the curve:   let x=0, for which y=-8.   then (0, -8) is on the curve.   plot the vertex and this point and draw a smooth parabola thru both.

hope this