# rational function

• May 14th 2008, 01:43 PM
mikedwd
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
• May 14th 2008, 01:50 PM
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!

• May 14th 2008, 06:45 PM
topsquark
Quote:

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!

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
• May 14th 2008, 07:17 PM
mr fantastic
Quote:

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

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) .....

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.