Write down the equations of the asymptotes of g. Determine the value(s) of x for which: g(x) = h(x), -4 < -2x-2/3 <= 0, g is increasing. Determine the equation of the line of symmetry of g with a negative x-intercept.

Here, $p=0$ and $q=-1$.

The vertical asymptote is $x=0$.
The horizontal asymptote is $y=-1$.

5.2.2 Determine the value(s) of x for which:

a) $g(x) = h(x)$
Given $g(x) = \frac{3}{x} - 1$ and $h(x) = -2x - 2$.
Setting them equal:

Multiplying by $x$ (assuming $x \neq 0$):

Rearranging into a standard quadratic form $ax^2 + bx + c = 0$:

To find the solutions, we use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
Here, $a=2$, $b=1$, $c=3$.

Since the discriminant ($\Delta = -23$) is negative, there are no real solutions for $x$.

Note: The provided graph clearly shows two intersection points between $g(x)$ and $h(x)$. This indicates a likely typo in the definition of $g(x)$. If $g(x)$ was intended to be $g(x) = -\frac{3}{x} - 1$ (which matches the graph's appearance), the solutions would be:

Factoring the quadratic equation:

This gives two solutions:

Given the contradiction, I will provide the answer based on the algebraic definition as written, and then note the discrepancy.

b) $-4 < \frac{-2x-2}{3} \le 0$
This is a compound inequality. We solve each part separately.

$-4 < \frac{-2x-2}{3}$.
Multiply both sides by 3:

Add 2 to both sides:

Divide by -2 and reverse the inequality sign:

$\frac{-2x-2}{3} \le 0$.
Multiply both sides by 3:

Add 2 to both sides:

Divide by -2 and reverse the inequality sign:

We need $x < 5$ AND $x \ge -1$.

c) g is increasing.
The given function is $g(x) = \frac{3}{x} - 1$.
To determine where $g(x)$ is increasing, we can examine its derivative or the behavior of the hyperbola.
The derivative of $g(x)$ is $g'(x) = -\frac{3}{x^2}$.
Since $x^2$ is always positive for $x \neq 0$, $g'(x) = -\frac{3}{\text{positive number}}$ will always be negative.
This means $g(x)$ is always decreasing for all $x \neq 0$.

Note: If we assume $g(x) = -\frac{3}{x} - 1$ (as suggested by the graph), then $g'(x) = \frac{3}{x^2}$. In this case, $g'(x)$ would always be positive, meaning $g(x)$ is increasing for all $x \neq 0$. Based on the written function, $g(x)$ is never increasing.

5.2.3 Determine the equation of the line of symmetry of g with a negative x-intercept.
The lines of symmetry for a hyperbola of the form $y = \frac{k}{x-p} + q$ pass through the intersection of its asymptotes $(p, q)$ and have gradients of $\pm 1$.
From 5.2.1, the asymptotes of $g(x) = \frac{3}{x} - 1$ are $x=0$ and $y=-1$. Their intersection is $(0, -1)$.

For $m=1$:

For $m=-1$:

For $y = x - 1$: Set $y=0 \implies 0 = x - 1 \implies x = 1$. (Positive x-intercept)
For $y = -x - 1$: Set $y=0 \implies 0 = -x - 1 \implies x = -1$. (Negative x-intercept)

The line of symmetry with a negative x-intercept is $y = -x - 1$.