Find the Holes in the Graph f(x)=(x^2-25)/(x^2-4x-45)

Rewrite ${\displaystyle 25}$ as ${\displaystyle 5^2}$.

Since both terms are perfect squares, factor using the difference of squares formula, ${\displaystyle a^2-b^2=(a+b)(a-b)}$ where ${\displaystyle a=x}$ and ${\displaystyle b=5}$.

Consider the form ${\displaystyle x^2+bx+c}$. Find a pair of integers whose product is ${\displaystyle c}$ and whose sum is ${\displaystyle b}$. In this case, whose product is ${\displaystyle -45}$ and whose sum is ${\displaystyle -4}$.

Write the factored form using these integers.

Cancel the common factor.

Rewrite the expression.

Set ${\displaystyle x+5}$ equal to ${\displaystyle 0}$.

Subtract ${\displaystyle 5}$ from both sides of the equation.

Substitute ${\displaystyle -5}$ for ${\displaystyle x}$ in ${\displaystyle \frac{x-5}{x-9}}$ and simplify.

The holes in the graph are the points where any of the cancelled factors are equal to ${\displaystyle 0}$.