How do you graph f(x)=(x^2-16)/(x^2-6x+8) using holes, vertical and horizontal asymptotes, x and y intercepts?

Feb 28, 2018

First, factor:

$y = \frac{\left(x + 4\right) \left(x - 4\right)}{\left(x - 4\right) \left(x - 2\right)}$

As you can see, there is $x - 4$ in both the numerator and denominator, so there is a hole at $x = 4$

To find the vertical asymptote, set the simplest form of the denominator equal to zero:

$y = \frac{\left(x + 4\right) \cancel{\left(x - 4\right)}}{\cancel{\left(x - 4\right)} \left(x - 2\right)}$

So set $x - 2$ equal to zero:

$x - 2 = 0 \implies$

$x = 2$

To find a horizontal asymptote, find the degree the polynomials are. Let's say $f \left(x\right) = \frac{{x}^{a}}{{x}^{b}}$

If $a > b$, there are no horizontal asymptotes.
If $a = b$, $\frac{a}{b}$ is the horizontal asymptote.
If $a < b$, $y = 0$ is the horizontal asymptote.

Here, the degrees are the same so $\frac{1}{1}$ is our horizontal asymptote
($y = 1$).

To find the x-intercept, plug in $y = 0$:

$\frac{x + 4}{x - 2} = 0$

$x = - 4$

The x-intercept is at $\left(- 4 , 0\right)$

To find the y-intercept, plug in $x = 0$

$\frac{0 + 4}{0 - 2} = y$

$y = - 2$

The y-intercept is at $\left(0 , - 2\right)$

Graph it:

graph{(x^2-16)/(x^2-6x+8) [-15.26, 17.28, -6.31, 9.96]}