Question

give an example of a rational function that has a horizontal asymptote at y=0 and a vertical asymptote at x=2 and x=1

Answer 1

Oh my goodness +_+ That picture is adorable!
I need him to knit me up some kitten mittens ASAP!

Answer 2

haha, I know right, I had to find a cute picture to match up with the username.

Answer 3

Normally when we have a rational function: (example):\[\Large\rm y=\frac{(x-2)(x-3)}{(x-4)(x-5)}\]The `denominator` is telling us where the `vertical asymptotes` are located.

Answer 4

So in the example I gave, we would have vertical asymptotes at x=4, x=5.

Answer 5

For a horizontal asymptote as y=0,
I think all we need is for the denominator to be of `larger` degree than the numerator.

Answer 6

that would mean that the denominator would be (x-2)(x-1)?

Answer 7

\[\Large\rm y=\frac{?}{(x-2)(x-1)}\]Good good good.

Answer 8

And what is the `largest degree` of x in the denominator.
If we were to expand it all out, we would end up with a square term, yes?

Answer 9

the largest degree of x is 1?

Answer 10

Nooo silly :O
It's a degree of 1 in each set of brackets.
But we're talking about the denominator as a whole.

So we would want to multiply the brackets out.
You don't really need to do all that work though.
Just notice that you have `two` brackets containing x's,
so when you multiply them out, you would end up with an \(\Large\rm x^2\) somewhere.

Answer 11

So in order to get an asymptote of y=0,
we want our numerator to have x's of `degree less than 2`.

Answer 12

Here's a nice easy polynomial to throw into the numerator.
It's a 0 degree polynomial.\[\Large\rm y=\frac{1}{(x-2)(x-1)}\]That should work just fine! c:

Answer 13

\[\Large\rm x^0=1\]

Answer 14

Ohhh, now I get it! thank you so much

Answer 15

yay team \c:/