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 $h(x)$, we will perform polynomial long division or use direct substitution.

Substituting $x = -1$ into $h$:
$h(-1) = (-1)^3 - 3(-1)^2 - 9(-1) - 5 = -1 - 3 + 9 - 5 = 0$
Since $h(-1) = 0$, it shows that $x + 1$ is indeed a factor.

Using synthetic division, we can divide $h(x)$ by $x + 1$ to find the other factors:

$h(x) = (x + 1)(x^2 - 4x - 5)$
Factorizing further, we get: $h(x) = (x + 1)(x - 5)(x + 1)$
So the x-intercepts are $x = -1$ and $x = 5$. The coordinates of D are (5, 0).

From our factorization, we know that $x = -1$ and $x = 5$. To determine the coordinates of C, we examine the point where $h$ changes from increasing to decreasing. The coordinates of C can be found as C(-1, 0).

To determine where $h$ is increasing, we find the derivative $h'(x)$ and set it to zero to find critical points. We know:

$h'(x) = 3x^2 - 6x - 9$
Setting $h'(x) > 0$ gives the intervals of increase. The values of $x$ for which $h$ is increasing are $x otin (- ext{∞}, -1]$ and $x > 5$.