Intercepts of Rational Functions II

Problem 271: For the function

$f(x) = \frac{x^2 - 16x + 63}{x^2 - 25}.$

Find the $y$ – and $x$ – intercepts.

Solution: To find the $y$ intercept, we let $x = 0$ .

$\begin{align*} f(x) = y & = \frac{x^2 - 16x + 63}{x^2 - 25} \\ & = \frac{(0)^2 - 16(0) + 63}{(0)^2 - 25} \\ & = \frac{63}{-25}. \end{align*}$

That is, the $y$ -intercept is $(0, -63/25)$ .

To find the $x$ intercept, we let $y = 0$ .

$\begin{align*} \frac{x^2 - 16x + 63}{x^2 - 25}= 0 & \implies x^2 - 16x + 63= 0 \\ & \implies (x -9)(x-7) = 0 \\ & \implies x = 7, 9.\end{align*}$

Therefore, the $x$ -intercepts are $(9, 0)$ and $(7,0)$ .