Question

match the rational expression with its domain restriction(s). domain restrictions are excluded values from the denominator $f(x)=\frac{x + 9}{x - 16}$ $f(x)=\frac{2x + 9}{16}$ $f(x)=\frac{5}{x^{2}-16}$ no excluded value(s) $x
eq16$ $x
eq4,x
eq - 4$

Explanation:

Step1: Recall domain - restriction rule

The denominator of a rational function cannot be zero.

Step2: Analyze $f(x)=\frac{x + 9}{x - 16}$

Set the denominator $x-16 = 0$. Solving for $x$ gives $x = 16$. So the domain restriction is $x
eq16$.

Step3: Analyze $f(x)=\frac{2x + 9}{16}$

The denominator is a non - zero constant ($16$). So there are no values of $x$ that make the denominator zero, and there are no excluded values.

Step4: Analyze $f(x)=\frac{5}{x^{2}-16}$

Factor the denominator: $x^{2}-16=(x + 4)(x - 4)$. Set $(x + 4)(x - 4)=0$. Using the zero - product property, $x+4 = 0$ gives $x=-4$ and $x - 4=0$ gives $x = 4$. So the domain restrictions are $x
eq4,x
eq - 4$.

Answer:

$f(x)=\frac{x + 9}{x - 16}$ matches $x
eq16$
$f(x)=\frac{2x + 9}{16}$ matches no excluded value(s)
$f(x)=\frac{5}{x^{2}-16}$ matches $x
eq4,x
eq - 4$