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

For q(x), we also set it to zero:

$\frac{12}{x} - 2 = 0$ Solving this yields:

$\frac{12}{x} = 2 \implies 12 = 2x \implies x = 6$ Therefore, the x-intercept of q is (6, 0).

To find the turning point of k, we first find the derivative:

$k'(x) = 2x - 2$ Setting the derivative to zero:

$2x - 2 = 0 \implies x = 1$ Now, substituting back to find k(1):

$k(1) = (1 - 5)(1 + 3) = (-4)(4) = -16$ Thus, the turning point is at (1, -16).

The function q has a vertical asymptote where the denominator is zero, hence:

$x = 0$ There’s also a horizontal asymptote determined by the limit as x approaches infinity:

$y = -2$ Thus, the equations of the asymptotes for q are:

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

To sketch the graphs of k and q, plot the x-intercepts and turning points on the axes. The graph of k is a parabola opening upwards, with intercepts at (5, 0) and (-3, 0), and the turning point at (1, -16). For q, draw the asymptotes and note that it approaches the x-axis as it heads towards negative and positive infinity, with an x-intercept at (6, 0). Ensure to show the vertical asymptote at x = 0 and horizontal asymptote at y = -2.