How do you graph #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=((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]}