Thread: Where would the asymptotes be located on a graph like this?

1. Where would the asymptotes be located on a graph like this?

The original graph:

Then the graph is transformed by

Where would the asymptotes go? Does each point on the graph represent a vertical asymptote?

2. Re: Where would the asymptotes be located on a graph like this?

vertical asymptotes appear when the function tends towards plus or minus infinity,

such as when a 0 would appear in a denominator somewhere.

In your transformed function $f(x)$ appears in a denominator. Are there any places where $f(x)=0$ ?

To find horizontal asymptotes you need to know the behavior of the function as $x \to \pm \infty$.

The graph given doesn't display this so you aren't able to determine any horizontal asymptotes from the information given.

3. Re: Where would the asymptotes be located on a graph like this?

Originally Posted by romsek
vertical asymptotes appear when the function tends towards plus or minus infinity,

such as when a 0 would appear in a denominator somewhere.

In your transformed function $f(x)$ appears in a denominator. Are there any places where $f(x)=0$ ?

To find horizontal asymptotes you need to know the behavior of the function as $x \to \pm \infty$.

The graph given doesn't display this so you aren't able to determine any horizontal asymptotes from the information given.
I'm not sure what you mean by the vertical asymptotes part. Can you explain further?

4. Re: Where would the asymptotes be located on a graph like this?

You were the one who asked about "asymptotes". Perhaps it would help if you told us what you think "asymptotes" are!

5. Re: Where would the asymptotes be located on a graph like this?

Originally Posted by HallsofIvy
You were the one who asked about "asymptotes". Perhaps it would help if you told us what you think "asymptotes" are!
I know what asymptotes are, however was confused as to how asymptotes work in graphs that are not linear or quadratic like this. I'm wondering where would I draw the asymptotes for the transformed functions?

6. Re: Where would the asymptotes be located on a graph like this?

Originally Posted by Latinized
I know what asymptotes are, however was confused as to how asymptotes work in graphs that are not linear or quadratic like this. I'm wondering where would I draw the asymptotes for the transformed functions?
O.K. let's look at the actual function:
$f(x)=\begin{cases}1 &: x\in [-4,-2] \\ -x-1 &: x\in [-1,0]\\2x-1 &: x\in[0,1]\end{cases}$ so let $g(x)=\dfrac{1}{f(x)}+1\begin{cases}2 &: x\in[-4,-2] \\ \dfrac{x}{x+1} &: x\in [-2,-1)\cup (-1,0]\\\dfrac{2x}{2x-1} &: x\in [0,0.5)\cup (0.5,1]\end{cases}$

Now you should see what you need.

7. Re: Where would the asymptotes be located on a graph like this?

Originally Posted by Latinized
I know what asymptotes are, however was confused as to how asymptotes work in graphs that are not linear or quadratic like this. I'm wondering where would I draw the asymptotes for the transformed functions?
You can take the approach Plato took and list everything out explicitly but I think that misses the point of the problem.

The asymptotes, now that $f(x)$ is in the denominator after the transform, will be where $f(x)=0$

These points are easily seen on the graph as $x=-1,~x=\dfrac 1 2$

and indeed if you look at the graph of $\dfrac {1}{f(x)}+1$

you see that this is indeed where the vertical asymptotes are.

8. Re: Where would the asymptotes be located on a graph like this?

Originally Posted by Latinized
I know what asymptotes are, however was confused as to how asymptotes work in graphs that are not linear or quadratic like this.
So you know what asymptotes are in linear and quadratic equations? That makes no sense because linear and quadratic equations don't have asymptotes!

9. Re: Where would the asymptotes be located on a graph like this?

Originally Posted by HallsofIvy
So you know what asymptotes are in linear and quadratic equations? That makes no sense because linear and quadratic equations don't have asymptotes!
OP probably means linear and quadratics once they've been transformed as $\dfrac {1}{f(x)}+1$