Question

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

Answer 1

First, factor:

`y=((x+4)(x-4))/((x-4)(x-2))`

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=((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]}