Given function g defined by g(x) = -x³ + 5x² + 8x - 12 7.1 Write down the y-intercept of g - NSC Technical Mathematics - Question 7 - 2023 - Paper 1

To find g(-2):

$g(-2) = -(-2)^3 + 5(-2)^2 + 8(-2) - 12$

Calculating this gives:

$g(-2) = -(-8) + 5(4) - 16 - 12$

$g(-2) = 8 + 20 - 16 - 12 = 0$

Therefore, g(-2) = 0.

The x-intercepts occur where g(x) = 0. Given that we found g(-2) = 0, one x-intercept is x = -2.

To find the other x-intercepts, we factor the polynomial:

$g(x) = -x^3 + 5x^2 + 8x - 12 = 0$

Factoring yields:

$(x + 2)(x^2 - 2x + 6) = 0$

The quadratic $x^2 - 2x + 6$ has no real roots (as its discriminant $(-2)^2 - 4(1)(6) < 0$). Thus, the only x-intercept is at:

$x = -2$

To find the turning points, we first calculate the derivative g'(x):

$g'(x) = -3x^2 + 10x + 8$

We set the derivative to zero to find critical points:

$-3x^2 + 10x + 8 = 0$

Using the quadratic formula, where $a = -3$, $b = 10$, and $c = 8$:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-10 \pm \sqrt{10^2 - 4(-3)(8)}}{2(-3)}$

Calculating this yields:

$x = \frac{-10 \pm \sqrt{100 + 96}}{-6} = \frac{-10 \pm \sqrt{196}}{-6} = \frac{-10 \pm 14}{-6}$

Thus, we find:

1. $x = \frac{4}{-6} = -\frac{2}{3}$
2. $x = \frac{-24}{-6} = 4$

Now, substituting back into g(x) for y-coordinates:

• For $x = -\frac{2}{3}$:

$g\left(-\frac{2}{3}\right) = -\left(-\frac{2}{3}\right)^3 + 5\left(-\frac{2}{3}\right)^2 + 8\left(-\frac{2}{3}\right) - 12$

This evaluates to:

$g\left(-\frac{2}{3}\right) = -\frac{-8}{27} + \frac{20}{9} - \frac{16}{3} - 12$

Calculating gives coordinates approximately at:

$\left(-\frac{2}{3}, -\frac{400}{27}\right)$

• For $x = 4$:

$g(4) = -4^3 + 5(4^2) + 8(4) - 12 = 36$

Thus, turning points are approximately at:

$(-\frac{2}{3}, -\frac{400}{27})$ and $(4, 36)$.

Make sure to plot the y-intercept at (0, -12) and the x-intercept at (-2, 0). Indicate turning points at approximately $(-\frac{2}{3}, -\frac{400}{27})$ and $(4, 36)$. Ensure the curve reflects the behavior of a cubic function with one local maximum and one local minimum.

From the graph, it can be observed that g(x) is negative for:

$x \in (-2, 4)$

This interval indicates where the function is beneath the x-axis.