The graph below represents the function defined by $h(x)=x^3 - 3x^2 - 9x - 5$ A and C are the turning points of $h$ - NSC Technical Mathematics - Question 7 - 2021 - Paper 1

To show that $x + 1$ is a factor of $h$, we can use polynomial long division or synthetic division.

1. Substitute $x = -1$ into the function:

   $h(-1) = (-1)^3 - 3(-1)^2 - 9(-1) - 5 = -1 - 3 + 9 - 5 = 0$

   Therefore, since $h(-1) = 0$, $x + 1$ is a factor of $h$.

Given that $h(x)$ can be factored as $h(x)=(x + 1)(x^2 + 2x - 5)$, we can find the roots of $x^2 + 2x - 5 = 0$:

Using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-2 \pm \sqrt{2^2 - 4(1)(-5)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 20}}{2} = \frac{-2 \pm \sqrt{24}}{2}$

Thus, the roots are:

$x = -1 \pm \sqrt{6}$

Therefore, D has coordinates (x-coordinate, 0), where x is one of the roots.

At turning point C, $h$ intercepts the x-axis. Considering the previously found roots, if we solve for:

$h(x) = 0$

We find that C is a point where either root from D intersects the x-axis. The coordinates of C will be the other x-intercept determined from the quadratic equation derived earlier.

To determine where $h$ is increasing, we find the first derivative of $h$ and analyze its sign:

$h'(x) = 3x^2 - 6x - 9$

Setting $h'(x) > 0$ and solving for $x$ will give us the intervals for increase. By finding critical points and testing intervals, we find that $h$ is increasing in the intervals derived from the solution of $h'(x) = 0$.