# Math Help - aSYMPTOTES

1. ## aSYMPTOTES

I need help finding the asymptotes of the following equations:

#1: y = 2^(x^2 -1) +3

I believe that this has no asymptotes. I graphed it (even though I'm not supposed to use my calculator) and all I get is s parabala and I believe that there can be any x or y value so is it true that there are no asymptotes in this equation??

#2: y = log base 8 of (x^2 -4) + 1

I don't know how to find the asymptotes of this one??

#3 y = log(1/(x-6))

I think x=6 is an asymptote cause it is in the denominator of the fraction. Are there any others???

THANK SOOO MUCH!!!

2. Originally Posted by Mr_Green
I need help finding the asymptotes of the following equations:

#1: y = 2^(x^2 -1) +3
It is not a parabola. But it is a trick question. Note that 'x' can take on any value. So it cannot have vertical asymptotes. But what happens when you graph it? That is the trick question. This is an double exponential it has a kollasal rate of increase. Thus it seems as if it is a vertical asymptote at +2 and -2 but it really is not.

#2: y = log base 8 of (x^2 -4) + 1

I don't know how to find the asymptotes of this one??
What is the domain?
$x^2-4>0$
$x^2>4$
$x>2 \mbox{ or }x<-2$.
Thus at the point when we reach the break in the domain we have a vertical asymptote (mathematicians say the infinima and supremum )
Thus, at
$x=\pm 2$

3. Originally Posted by Mr_Green

#3 y = log(1/(x-6))

I think x=6 is an asymptote cause it is in the denominator of the fraction. Are there any others???
Yes you are right a vertical asymptote is at x=6.

Let us find the domain,
$\log (1/(x-6))$
That needs to be,
$1/(x-6)>0$
$x-6>0$
$x>6$.

So at the point where the domain breaks is at $x=6$ (going from the right). Or you can reason like you.

4. Originally Posted by ThePerfectHacker
It is not a parabola. But it is a trick question. Note that 'x' can take on any value. So it cannot have vertical asymptotes. But what happens when you graph it? That is the trick question. This is an double exponential it has a kollasal rate of increase. Thus it seems as if it is a vertical asymptote at +2 and -2 but it really is not.

What is the domain?
$x^2-4>0$
$x^2>4$
$x>2 \mbox{ or }x<-2$.
Thus at the point when we reach the break in the domain we have a vertical asymptote (mathematicians say the infinima and supremum )
Thus, at
$x=\pm 2$

does that mean that #1 has no asymptotes?

THANKS btw