# End-Behavior Asymptote

• Jan 31st 2010, 11:39 PM
iyppxstahh
End-Behavior Asymptote
Hi, guys. This is my first thread on here, so bare with me if I make some mistakes.
I'm doing these end behavior problems and need some help...

1. Find the end-behavior asymptote(s) in the graph of f(x)=sqrt(4x^2-9)/(x+5).
I can't figure out a way to divide the square root.

2. Find the end-behavior asymptote(s) in the graph of f(x)=(4x-15x^2)/(x^2-121).
This is how I solved for it:
(4x-15x^2)/(x^2-121)=-15+(4x-1815)/(x^2-121)
y=-15
I'm not sure if I did it right. I just followed example problems.

3. Same for f(x)=(x^2-5)/(x-3).
My answer this time was y=4+((x+4)/(x-3)), so if I did the last problem correctly, the asymptote should be y=4.

4. Find the polynomial end-behavior asymptote in the graph of f(x)=(2x^3+6x-1)/(x+2).
My answer was y=2x^2-4x+14

If you could explain the process, that would help a lot. Thanks!
• Feb 1st 2010, 04:01 AM
HallsofIvy
Quote:

Originally Posted by iyppxstahh
Hi, guys. This is my first thread on here, so bare with me if I make some mistakes.
I'm doing these end behavior problems and need some help...

1. Find the end-behavior asymptote(s) in the graph of f(x)=sqrt(4x^2-9)/(x+5).
I can't figure out a way to divide the square root.

$\displaystyle \frac{\sqrt{4x^2- 9}}{x+5}= \frac{\sqrt{4x^2- 9}}{\sqrt{(x+5)^2}}= \sqrt{\frac{4x^2- 9}{x^2+ 10x+ 25}}$.
Now divide $\displaystyle 4x^2- 9$ by $\displaystyle x^2+ 10x+ 25$.

Quote:

2. Find the end-behavior asymptote(s) in the graph of f(x)=(4x-15x^2)/(x^2-121).
This is how I solved for it:
(4x-15x^2)/(x^2-121)=-15+(4x-1815)/(x^2-121)
y=-15
I'm not sure if I did it right. I just followed example problems.
Yes, since the remaining fraction has lower degree in the numerator than in the denominator, it will go to 0 and so f(x) will go to -15.

Quote:

3. Same for f(x)=(x^2-5)/(x-3).
My answer this time was y=4+((x+4)/(x-3)), so if I did the last problem correctly, the asymptote should be y= 4.
No, you did not divide correctly. Since the numerator is of degree 2 and the denominator of degree 1, you would expect a quotient of degree 2-1= 1. x-3 divides into $\displaystyle x^2- 5$ with quotient x+ 3 and remainder 4: y= x+ 3 + 4/(x-3). The (slant) asymptote is y= x+ 3.

Quote:

4. Find the polynomial end-behavior asymptote in the graph of f(x)=(2x^3+6x-1)/(x+2).
My answer was y=2x^2-4x+14
You appear to have divided incorrectly again. Could you show exactly how you got that?

Quote:

If you could explain the process, that would help a lot. Thanks!
• Feb 1st 2010, 08:47 PM
iyppxstahh
Quote:

No, you did not divide correctly. Since the numerator is of degree 2 and the denominator of degree 1, you would expect a quotient of degree 2-1= 1. x-3 divides into $\displaystyle x^2- 5$ with quotient x+ 3 and remainder 4: y= x+ 3 + 4/(x-3). The (slant) asymptote is y= x+ 3.
Oh, thank you. I accidentally switched the quotient and the remainder.

Quote:

You appear to have divided incorrectly again. Could you show exactly how you got that?
Well, the answer was was y=2x^2-4x+14 R-29, and I took out the remainder...I haven't figured out how to post equations yet, but I've checked and re-checked the problem, and it is divided correctly. Isn't it?

Thank you, so much!
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