The graphs below represent the functions defined by $g(x) = - (x + 2)(x - 1)(x - 3)$ and h(x) = 2x + p$ E and F are the turning points of g - NSC Technical Mathematics - Question 7 - 2022 - Paper 1

To find the value of p, we utilize the function $h(x) = 2x + p$ and evaluate at point C where $x = 3$:

$h(3) = 2(3) + p = 0$

Thus, solving gives:

$6 + p = 0 \Rightarrow p = -6$

The length of segment AC can be calculated using the distance formula:

$AC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ where A(0; 0) and C(3; 0):

$AC = \sqrt{(3 - 0)^2 + (0 - 0)^2} = \sqrt{3^2} = 3 \text{ units}$

Expanding the expression:

1. First, expand $(x + 2)(x - 1)$: $= x^2 + 2x - x - 2 = x^2 + x - 2$

2. Then, multiply by $(x - 3)$: $= (x^2 + x - 2)(x - 3)$ $= x^3 - 3x^2 + x^2 - 3x - 2x + 6$ $= x^3 - 2x^2 - 3x + 6$

3. Finally, multiply by -1: $g(x) = - (x^3 - 2x^2 - 3x + 6) = -x^3 + 2x^2 + 3x - 6$.

To find the turning points E and F, we need to identify the local maxima and minima of $g(x)$. This can be done by setting the first derivative to zero and solving:

$g'(x) = -3x^2 + 4x + 3 = 0$ Using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-4 \pm \sqrt{4^2 - 4(-3)(-3)}}{2(-3)}$

Calculate this gives two x-values which can then be substituted back into $g(x)$ for the corresponding y-values. Coordinate points E and F will be derived accordingly.

To find the values of x for which $g(x) > 0$, identify the segments of the graph where the curve lies above the x-axis. From analyzing the roots of the polynomial, determine the intervals between roots where the polynomial is positive. Thus, the values of x are found by analyzing these intervals.