Sketched below is the graph of h(x) = \frac{1}{x + p} + q - NSC Mathematics - Question 4 - 2022 - Paper 1

To find the x-intercept, we set h(x) = 0:

\frac{1}{x + p} + q = 0
\Rightarrow \frac{1}{x - 1} + 2 = 0
\Rightarrow \frac{1}{x - 1} = -2
\Rightarrow 1 = -2(x - 1)
\Rightarrow 1 = -2x + 2
\Rightarrow 2x = 1
\Rightarrow x = \frac{1}{2}.

The coordinates of the x-intercept are (\frac{1}{2}, 0).

The x-intercept of h is at x = \frac{1}{2}. For g(x) = h(x + 3), we find the x-intercept by solving:
h(x + 3) = 0, thus:

x + 3 = \frac{1}{2}
\Rightarrow x = \frac{1}{2} - 3 = -\frac{5}{2}.

The axis of symmetry for the graph of h is located at y = 2 (the y-coordinate of the asymptote). Setting t + 1 = 2 gives:

t + 1 = 2
\Rightarrow t = 1.

To solve -2 ≤ \frac{1}{x - 1}, we rearrange the inequality:

1 \geq -2(x - 1)
\Rightarrow 1 \geq -2x + 2
\Rightarrow -2x \leq -1
\Rightarrow x \geq \frac{1}{2}.

Additionally, we must consider the domain restrictions:

x - 1 \neq 0 \Rightarrow x \neq 1.

Thus, the solution is:
[ x \in \left(\frac{1}{2}, 1\right) \cup (1, \infty) ] for which -2 ≤ \frac{1}{x - 1} holds.