Analyze function h(x) = x3 − 2x2 − 15x: find intercepts and stationary points.

Let's analyze function (c) $h(x) = x^3 - 2x^2 - 15x$.

Step 1: Find the intercepts with the axes.
To find the y-intercept, set $x=0$:

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

The y-intercept is $(0,0)$.

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 $(0,0)$, $(5,0)$, and $(-3,0)$.

Step 2: Find the stationary points.
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 $x=3$:

$$h(3) = (3)^3 - 2(3)^2 - 15(3) = 27 - 2(9) - 45 = 27 - 18 - 45 = 9 - 45 = -36$$

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

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

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

To combine these, use a common denominator of 27:

$$h\left(-\frac{5}{3}\right) = -\frac{125}{27} - \frac{50 \times 3}{9 \times 3} + \frac{25 \times 27}{27}$$ $$$$