Question

match the rational expression with number of excluded values it will have.
$f(x)=\frac{2x + 9}{16}$
0 excluded values
$f(x)=\frac{5}{x^{2}-16}$
2 excluded values
$f(x)=\frac{x + 9}{x - 16}$
1 excluded value

Explanation:

Step1: Analyze first rational - expression

For $f(x)=\frac{2x + 9}{16}$, the denominator is a non - zero constant 16. Since the denominator is never zero for any real value of $x$, there are 0 excluded values.

Step2: Analyze second rational - expression

For $f(x)=\frac{5}{x^{2}-16}$, set the denominator equal to zero: $x^{2}-16 = 0$. Using the difference of squares formula $a^{2}-b^{2}=(a + b)(a - b)$, we have $(x + 4)(x - 4)=0$. Solving for $x$, we get $x=-4$ and $x = 4$. So there are 2 excluded values.

Step3: Analyze third rational - expression

For $f(x)=\frac{x + 9}{x - 16}$, set the denominator equal to zero: $x-16=0$. Solving for $x$, we get $x = 16$. So there is 1 excluded value.

Answer:

$f(x)=\frac{2x + 9}{16}$ matches 0 excluded values.
$f(x)=\frac{5}{x^{2}-16}$ matches 2 excluded values.
$f(x)=\frac{x + 9}{x - 16}$ matches 1 excluded value.