Graph the function. Label all intercepts, critical points, and inflection points.

*f* (*x*) = *x*^{2}*e*^{x}

Answer

*f* (*x*) = *x*^{2}*e*^{x}

• Find dots.

The function *f* (*x*) is only zero when *x* = 0.

The derivative is

which is zero when *x* = 0 and when *x* = -2, so these are our critical points. When *x* = 0, we have *y* = 0 also.

*f*(-2) = 4*e*^{-2}

so the complete coordinates of the critical points are

(0,0) and (-2,4*e*^{-2}).

For the sake of graphing,

4*e*^{-2} ≅ 0.54.

The second derivative (using the first definition of *f*', not the factored version) is

This is zero when (*x*^{2} + 4*x* + 2) is zero. We need the quadratic equation to solve this one:

We need to find the corresponding *y*-values.

• Find shapes.

Set up the numberline:

The function *f* is never negative.

The function *f *'(*x*) = *xe*^{x}(2 + *x*) is negative when exactly one of the quantities *x* and (2 + *x*) is negative, and positive otherwise. Therefore *f *' is only negative when -2 < *x* < 0.

The function *f *"(*x*) = *e*^{x}(*x*^{2} + 4*x* + 2) is negative when *x* is between the roots of the polynomial (*x*^{2} + 4*x* + 2). That is, when

These are the shapes of *f *:

• Play connect-the-dots.

We play connect-the-dots, keeping in mind that *f* can't go below the *x*-axis: