Given functions $k$ and $q$ defined by $k(x) = (x - 5)(x + 3)$ and $q(x) = \frac{12}{x} - 2$ respectively - NSC Technical Mathematics - Question 4 - 2019 - Paper 1

To find the x-intercept of $q(x)$, we set $q(x) = 0$. Therefore, we solve:

$\frac{12}{x} - 2 = 0$

Rearranging gives:

$\frac{12}{x} = 2 \Rightarrow 12 = 2x \Rightarrow x = 6$

Thus, the x-intercept of $q$ is $x = 6$.

To find the turning point of $k(x)$, we first find the derivative:

$k'(x) = 2x - 2$

Setting $k'(x) = 0$ gives:

$2x - 2 = 0 \Rightarrow x = 1$

Next, we substitute $x = 1$ into $k(x)$ to find the y-coordinate:

$k(1) = (1 - 5)(1 + 3) = (-4)(4) = -16$

Therefore, the coordinates of the turning point are $(1, -16)$.

To determine the asymptotes of $q(x)$, we analyze its expression:

$q(x) = \frac{12}{x} - 2$

The vertical asymptote occurs where the function is undefined, which is at:

$x = 0$

The horizontal asymptote can be found by considering the behavior as $x \to \infty$:

$\lim_{x \to \infty} q(x) = 0 - 2 = -2$

Thus, the equations of the asymptotes are:

1. Vertical: $x = 0$
2. Horizontal: $y = -2$

To sketch the graphs of $k$ and $q$, the following features should be included:

• For the graph of $k(x)$:

  • X-intercepts at $(5, 0)$ and $(-3, 0)$
  • Y-intercept at $k(0) = (0 - 5)(0 + 3) = -15$
  • Turning point at $(1, -16)$

• For the graph of $q(x)$:

  • X-intercept at $(6, 0)$
  • Vertical asymptote at $x = 0$
  • Horizontal asymptote at $y = -2$

On the same axes, plot points for both functions and clearly indicate the intercepts, turning points, and asymptotes.