1. ## rational functions

so i'm trying to help my little brother with his math homework, but i can't recall some of the things i ned to know. basically, the problem states that some rational function, r(x) has a vert. asymptote at x = 3, a vert. asymptote at y = 2, and a hole at (2, -2).
is there some sort of formula i'm supposed to know how to use? because the instructions request that i "put in the factor[s] that would account" for the asymptotes and "find a."

help?

2. Originally Posted by toscaandsage
so i'm trying to help my little brother with his math homework, but i can't recall some of the things i ned to know. basically, the problem states that some rational function, r(x) has a vert. asymptote at x = 3, a vert. asymptote at y = 2, and a hole at (2, -2).
is there some sort of formula i'm supposed to know how to use? because the instructions request that i "put in the factor[s] that would account" for the asymptotes and "find a."

help?
If there is a mathematical asymptote at $x=3$ then for the function $y=\frac{f(x)}{g(x)}$ then $g(3)=0\text{and}f(3)\ne{0}$

Since there is a horizontal asymptote at y=2 $\lim_{x\to\infty}\frac{f(x)}{g(x)}=2$ and since there is a hole at x=2 there cannot be a y-coordinate then $f(2)=0,g(2)=0$

3. sure but how do you make sure that the hole is when y = -2?

4. Originally Posted by toscaandsage
so i'm trying to help my little brother with his math homework, but i can't recall some of the things i ned to know. basically, the problem states that some rational function, r(x) has a vert. asymptote at x = 3, a vert. asymptote at y = 2, and a hole at (2, -2).
is there some sort of formula i'm supposed to know how to use? because the instructions request that i "put in the factor[s] that would account" for the asymptotes and "find a."

help?
Build it up:

"vert. asymptote at x = 3": $r(x) = \frac{A}{x - 3}$.

"horizontal asymptote at y = 2": $r(x) = \frac{A}{x - 3} + 2$.

"(2, -2)" would be a point on the curve if there was no hole, that is, r(2) = -2 if there was no hole:

$-2 = \frac{A}{2 - 3} + 2 \Rightarrow A = 4$. So $r(x) = \frac{4}{x - 3} + 2$.

"a hole at" x = 2: $r(x) = \frac{4(x - 2)}{(x - 3)(x - 2)} + 2$.

5. Originally Posted by mr fantastic
Build it up:

"vert. asymptote at x = 3": $r(x) = \frac{A}{x - 3}$.

"horizontal asymptote at y = 2": $r(x) = \frac{A}{x - 3} + 2$.

"(2, -2)" would be a point on the curve if there was no hole, that is, r(2) = -2 if there was no hole:

$-2 = \frac{A}{2 - 3} + 2 \Rightarrow A = 4$. So $r(x) = \frac{4}{x - 3} + 2$.

"a hole at" x = 2: $r(x) = \frac{4(x - 2)}{(x - 3)(x - 2)} + 2$.

Why does adding 2 to the function create an asymptote at y=2? I was under the idea that in order to create a horizontal asymptote, you must have the same degree in the numerator and denominator, and, in this case, the leading coefficient of the numerator must be twice the leading coefficient of the denominator?

your function seems correct, I'm just confused. Thanks a lot if you can explain!

I suppose I can see that 2/1 would meet those qualifications, I was just unsure of how it applies itself to the rest of the function...I'm no algebra wiz, so

6. Originally Posted by mikedwd
Why does adding 2 to the function create an asymptote at y=2? I was under the idea that in order to create a horizontal asymptote, you must have the same degree in the numerator and denominator, and, in this case, the leading coefficient of the numerator must be twice the leading coefficient of the denominator?

your function seems correct, I'm just confused. Thanks a lot if you can explain!

I suppose I can see that 2/1 would meet those qualifications, I was just unsure of how it applies itself to the rest of the function...I'm no algebra wiz, so
$\frac{4}{x - 3} + 2 = \frac{4}{x - 3} + \frac{2(x-3)}{x-3} = \frac{4 + 2(x-3)}{x-3} = \frac{2x - 2}{x-3}$.

Edit: 1. So the required function I found can also be written as $y = \frac{(2x - 2)(x-2)}{(x-3)(x-2)}$.

2. The thread http://www.mathhelpforum.com/math-he...tml#post145259 is of related interest.