# Thread: rational function

1. ## rational function

Hello,

I'm left with the instructions:

"A rational function, R(x) has the following characteristics: a vertical asymptote at x=3, a horizontal asymptote at y=2, and a hole at (2,-2). Sketch the function and determine what it could be."

The function I have created thus far is
Code:
2x(x-2) / (x-3)(x-2)
The problem is even though I have a hole for my function at x=2, the hole does not occur when y=-2.

My instructions are to "Describe what you must do in order for the hole to appear at (2,-2). Find a."

I suppose that a is a factor (or rather (x-a) or something of that format) which I need to apply to my function to create a hole at (2,-2), but I have never done this before and I've tried just plowing through math to figure this out, but I have no method!

Thanks a lot if you can help.

-Mike

2. Oh, I've found this:

http://www.mathhelpforum.com/math-he...functions.html

Quite sorry, I'll read through this thread and decide if I should delete!

EDIT: inability to delete threads?

3. Originally Posted by mikedwd
Oh, I've found this:

http://www.mathhelpforum.com/math-he...functions.html

Quite sorry, I'll read through this thread and decide if I should delete!

EDIT: inability to delete threads?
Moderators (and Ads) can delete threads if they feel it is necessary. I think this thread provides some good food for thought for someone who has similar questions, so I see no need to delete it.

-Dan

4. Originally Posted by mikedwd
Hello,

I'm left with the instructions:

"A rational function, R(x) has the following characteristics: a vertical asymptote at x=3, a horizontal asymptote at y=2, and a hole at (2,-2). Sketch the function and determine what it could be."

The function I have created thus far is
Code:
2x(x-2) / (x-3)(x-2)
The problem is even though I have a hole for my function at x=2, the hole does not occur when y=-2.

My instructions are to "Describe what you must do in order for the hole to appear at (2,-2). Find a."

I suppose that a is a factor (or rather (x-a) or something of that format) which I need to apply to my function to create a hole at (2,-2), but I have never done this before and I've tried just plowing through math to figure this out, but I have no method!

Thanks a lot if you can help.

-Mike
Well, the thread here: http://www.mathhelpforum.com/math-he...tml#post144408

goes a long way to solving the problem.

But doing it your way, you have
Code:
2x(x-2) / (x-3)(x-2)
but you get the wrong value of y for the hole. You get (2, -4) instead of (2, -2) .....

Anyway, persevering with your way:

The problem with your function is that you need
Code:
y = 2x / (x-3)
to be modified in some way so that when x = 2, y = -2. At the same time you need to make sure that y = 2 remains a horizontal asymptote .......

So why not consider finding the value of A such that
Code:
y = (2x + A) / (x-3)
gives y = -2 when x = 2 ......

Then your final function will be
Code:
(2x + A)(x-2) / (x-3)(x-2)
which can be simplified as necessary.

You will get the answer I found at the above hyperlink.