Suppose that the functions $$ g $$ and $$ h $$ are defined as follows. $$ \begin{array}{l}g(x)=x-5 \ h(x)=(x-6)(x+4) \ \text { (a) Find }\left(\frac{g}{h}\right)(-6) \text {. } \ \text { (b) Find all values that are NOT in the domain of } \frac{g}{h} \text {. } \ \text { If there is more than one value, separate them with commas. } \ \text { (a) }\left(\frac{g}{h}\right)(-6)=\square \ \text { (b) Value(s) that are NOT in the domain of } \frac{g}{h}: \square\end{array} $$

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$$
\begin{aligned}
\text{(a) } \left(\frac{g}{h}\right)(-6) &= \frac{g(-6)}{h(-6)}\[8pt]
g(-6) &= -6 - 5 = -11,\[8pt]
h(-6) &= (-6-6)(-6+4) = (-12)(-2) = 24,\[8pt]
\therefore \left(\frac{g}{h}\right)(-6) &= \frac{-11}{24}.
\end{aligned}
$$

$$
\begin{aligned}
\text{(b)} \quad \text{Values not in the domain:} \quad h(x) &= 0\[8pt]
(x-6)(x+4) &= 0\[8pt]
x-6 = 0 \quad &\Longrightarrow \quad x = 6,\[8pt]
x+4 = 0 \quad &\Longrightarrow \quad x = -4.
\end{aligned}
$$

$$
\boxed{
\begin{aligned}
\text{(a) } \left(\frac{g}{h}\right)(-6) &= -\frac{11}{24},\[6pt]
\text{(b) Domain exclusion: } & x = 6,\, -4.
\end{aligned}
}
$$
$$ \begin{aligned} \text{(a) } \left(\frac{g}{h}\right)(-6) &= -\frac{11}{24},\[6pt] \text{(b) Domain exclusion: } & 6,\, -4. \end{aligned} $$

Suppose that the functions and are defined as follows.

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