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

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Answer 1 · 2018-02-28T01:46:34

First, factor:

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

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

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

$y=((x+4)cancel((x-4)))/(cancel((x-4))(x-2))$

So set $x-2$ equal to zero:

$x-2=0=>$

$x=2$

To find a horizontal asymptote, find the degree the polynomials are. Let's say $f(x)=(x^a)/(x^b)$

If $a>b$ , there are no horizontal asymptotes.
If $a=b$ , $a/b$ is the horizontal asymptote.
If $a<b$ , $y=0$ is the horizontal asymptote.

Here, the degrees are the same so $1/1$ is our horizontal asymptote
( $y=1$ ).

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

$(x+4)/(x-2)=0$

$x=-4$

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

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

$(0+4)/(0-2)=y$

$y=-2$

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

Graph it:

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

How do you graph f(x)=(x^2-16)/(x^2-6x+8)f(x)=x2−16x2−6x+8f(x)=(x^2-16)/(x^2-6x+8) using holes, vertical and horizontal asymptotes, x and y intercepts?