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 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.