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.