Thread: 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!

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.

.
Now divide by .

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.

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 with quotient x+ 3 and remainder 4: y= x+ 3 + 4/(x-3). The (slant) asymptote is y= x+ 3.

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?

If you could explain the process, that would help a lot. Thanks!

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 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.

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!