# How do you graph f(x)=1/(x^2-16) using holes, vertical and horizontal asymptotes, x and y intercepts?

Jun 15, 2018

See answer below

#### Explanation:

Given: $f \left(x\right) = \frac{1}{{x}^{2} - 16}$

The given function is a rational (fraction) function of the form:

$f \left(x\right) = \frac{N \left(x\right)}{D \left(x\right)} = \frac{{a}_{n} {x}^{n} + {a}_{n - 1} {x}^{n - 1} + \ldots + {a}_{1} x + {a}_{0}}{{b}_{m} {x}^{m} + {b}_{n - 1} {x}^{n - 1} + \ldots + {b}_{1} x + {b}_{0}}$

First factor both the numerator and denominator. The denominator is the difference of squares: $\left({a}^{2} - {b}^{2}\right) = \left(a + b\right) \left(a - b\right)$

$f \left(x\right) = \frac{1}{{x}^{2} - 16} = \frac{1}{\left(x + 4\right) \left(x - 4\right)}$

Holes are found where linear terms can be cancelled from the numerator and denominator. In the given function, there are no holes .

Vertical asymptotes are found when $D \left(x\right) = 0$ (Don't include hole factors):

D(x) = (x +4)(x - 4) = 0; " " **x = -4** , " " **x = 4**

Horizontal asymptotes are related to $n \text{ and } m$:

$n < m$, horizontal asymptote is $y = 0$
$n = m$, horizontal asymptote is $y = \frac{{a}_{n}}{{b}_{n}}$
$n > m$, no horizontal asymptote

For the given function $n = 0 , m = 2 \implies n < m :$ horizontal asymptote is y = 0

Find intercepts:

Find $x$ - intercepts: set $N \left(x\right) = 0$

No x-intercepts

Find $y$ - intercepts: set $x = 0$:

$f \left(0\right) = \frac{1}{0 - 16} = - \frac{1}{16}$

$y$-intercept: (0, -1/16)