Question

where is the hole for the following function located?
$f(x) = \frac{x + 3}{(x - 4)(x + 3)}$
$x = -3$
$y = 3$
$y = -3$
$x = 3$

Explanation:

Step1: Identify common factors

The function is \( f(x)=\frac{x + 3}{(x - 4)(x + 3)} \). The numerator and denominator have a common factor of \( x + 3 \).

Step2: Determine the hole

A hole in a rational function occurs where the common factor is zero (as long as it doesn't make the remaining function undefined). Set \( x+3 = 0 \), so \( x=-3 \). We check that when we cancel \( x + 3 \), the remaining function \( \frac{1}{x - 4} \) is defined at \( x=-3 \) (since \( -3-4=-7
eq0 \)), so the hole is at \( x = -3 \).

Answer:

\( x=-3 \)