The graph below shows a cubic function defined by $h(x) = x^{3} + px^{2} + 9x - 2$ which cuts the x-axis at A, B(2; 0) and C - NSC Technical Mathematics - Question 7 - 2024 - Paper 1

We know that the cubic function cuts the x-axis at points A, B(2; 0), and C. To find $p$, we substitute $B(2; 0)$ into the function:

$h(2) = 2^{3} + p(2)^{2} + 9(2) - 2 = 0$

Calculating this gives:

$8 + 4p + 18 - 2 = 0$ $4p + 24 = 0 \Rightarrow 4p = -24 \Rightarrow p = -6$

From the graph, we gather that point B is $(2; 0)$ and point C lies on the x-axis at an undetermined coordinate yet to be found. To find the x-intercepts of $h(x)$, we solve:

$h(x) = 0 \Rightarrow x^{3} - 6x^{2} + 18x - 2 = 0$

Assuming C has coordinates $(x_C; 0)$, the length of BC is given by:

$BC = |x_C - 2|$

After calculation, the length BC is expressed as:

$BC = \sqrt{3}$

The turning points D and E can be found by setting the first derivative $h'(x) = 0$:

$h'(x) = 3x^2 - 12x + 9 = 0$

Solving this quadratic gives:

$(x - 3)(x - 1) = 0 \Rightarrow x = 1 \text{ and } 3$

Substituting these values into the original function $h(x)$ gives:

At $x = 1$:
$h(1) = 1 - 6 + 9 - 2 = 2 \Rightarrow D(1; 2)$

At $x = 3$:
$h(3) = 27 - 54 + 27 - 2 = -2 \Rightarrow E(3; -2)$

For $x > 2$, we analyze the sign of $h(x)$ and $h'(x)$:

From previous results, $h(x) > 0$ for $x ext{ in } (2; 3)$ and $h(x) < 0$ for $x > 3$. Hence:

• $h(x) \times h'(x) > 0$ when both $h(x)$ and $h'(x)$ are positive or negative.

This gives the intervals $x ext{ in } (2; 3)$ or $x > (2 + \sqrt{3})$.