Analyze the function h(x) = x3 − 2x2 − 15x, finding intercepts and stationary points.

To find the y-intercept, set $x=0$:

$$h(0) = (0)^3 - 2(0)^2 - 15(0) = 0$$

The y-intercept is $\boxed{\text{(0,0)}}$.

To find the x-intercepts, set $h(x)=0$:

$$x^3 - 2x^2 - 15x = 0$$

Factor out $x$:

$$x(x^2 - 2x - 15) = 0$$

Factor the quadratic expression:

$$x(x-5)(x+3) = 0$$

This gives the x-intercepts: $x=0$, $x=5$, and $x=-3$.
The x-intercepts are $\boxed{\text{(0,0), (5,0), and (-3,0)}}$.

First, find the first derivative $h'(x)$:

$$h'(x) = \frac{d}{dx}(x^3 - 2x^2 - 15x) = 3x^2 - 4x - 15$$

Set $h'(x)=0$ to find the x-coordinates of the stationary points:

$$3x^2 - 4x - 15 = 0$$

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

$$x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(3)(-15)}}{2(3)}$$ $$x = \frac{4 \pm \sqrt{16 + 180}}{6}$$ $$x = \frac{4 \pm \sqrt{196}}{6}$$ $$x = \frac{4 \pm 14}{6}$$

The x-coordinates are:

$$x_1 = \frac{4 + 14}{6} = \frac{18}{6} = 3$$ $$x_2 = \frac{4 - 14}{6} = \frac{-10}{6} = -\frac{5}{3}$$

Now, find the corresponding y-coordinates by substituting these values back into $h(x)$:
For $x=3$:

$$h(3) = (3)^3 - 2(3)^2 - 15(3) = 27 - 18 - 45 = -36$$

Stationary point 1: $(3, -36)$.

For $x=-\frac{5}{3}$:

$$h\left(-\frac{5}{3}\right) = \left(-\frac{5}{3}\right)^3 - 2\left(-\frac{5}{3}\right)^2 - 15\left(-\frac{5}{3}\right)$$ $$h\left(-\frac{5}{3}\right) = -\frac{125}{27} - 2\left(\frac{25}{9}\right) + 25$$ $$h\left(-\frac{5}{3}\right) = -\frac{125}{27} - \frac{50}{9} + 25$$ $$h\left(-\frac{5}{3}\right) = \frac{-125 - 150 + 675}{27} = \frac{400}{27}$$

Stationary point 2: $\left(-\frac{5}{3}, \frac{400}{27}\right)$.

Find the second derivative $h''(x)$:

$$h''(x) = \frac{d}{dx}(3x^2 - 4x - 15) = 6x - 4$$

Evaluate $h''(x)$ at each stationary point:
For $x=3$:

$$h''(3) = 6(3) - 4 = 18 - 4 = 14$$

Since $h''(3) > 0$, $(3, -36)$ is a local minimum.

For $x=-\frac{5}{3}$:

$$h''\left(-\frac{5}{3}\right) = 6\left(-\frac{5}{3}\right) - 4 = -10 - 4 = -14$$

Since $h''\left(-\frac{5}{3}\right) < 0$, $\left(-\frac{5}{3}, \frac{400}{27}\right)$ is a local maximum.

The stationary points are $\boxed{\text{(3, -36) (local minimum) and } \left(-\frac{5}{3}, \frac{400}{27}\right) \text{ (local maximum)}}$.

Set $h''(x)=0$:

$$6x - 4 = 0$$ $$6x = 4$$ $$x = \frac{4}{6} = \frac{2}{3}$$

Now, find the corresponding y-coordinate by substituting $x=\frac{2}{3}$ into $h(x)$:

$$h\left(\frac{2}{3}\right) = \left(\frac{2}{3}\right)^3 - 2\left(\frac{2}{3}\right)^2 - 15\left(\frac{2}{3}\right)$$ $$h\left(\frac{2}{3}\right) = \frac{8}{27} - 2\left(\frac{4}{9}\right) - 10$$ $$h\left(\frac{2}{3}\right) = \frac{8}{27} - \frac{8}{9} - 10$$ $$h\left(\frac{2}{3}\right) = \frac{8 - 24 - 270}{27} = \frac{-286}{27}$$

The point of inflection is $\boxed{\text{} \left(\frac{2}{3}, -\frac{286}{27}\right) \text{}}$.