How to find vertical and horizontal asymptotes of...

Question

suwhitney · Answer

$$f(x)=\frac{ (-2x+x^2)(3x^2-4) }{ (x^2-2x-3)(-x^2+8) }$$

suwhitney · Answer

I already got the vertical asymptotes.. But i have no clue how to find the horizontal ones, please help.

anonymous · Answer

From what I remember in precal, for horizontal, if the degrees in the denominator and numerator the same, in this case, degree of 2, then divide.$$\frac{ x ^{2} }{ x ^{2} }=1$$Horizontal asymptote at 1

anonymous · Answer

After checking that rule above only works for a simpler case like this$$\frac{ x ^{2}-4 }{ x ^{2}-1 }$$

suwhitney · Answer

Oh okay, Thank you though

anonymous · Answer

Yes, we were on the right track. If you multiply out and move brackets, the dominant degree top and bottom would be$$-\frac{ 3x ^{4} }{ x ^{4} }$$There is a horizontal asymptote at -3

suwhitney · Answer

Yes that's the answer he has! :) Please help me understand it a little more..What do you mean when you say multiply out?

anonymous · Answer

You don't have a true sense or a true look as the question is posed. You have to restate it like this$$\frac{ 3x ^{4}-6x ^{3}-4x ^{2}+8x }{ -x ^{4}+12x ^{2}-16x-24 }$$

suwhitney · Answer

Why do you have to restate it like that?

anonymous · Answer

Because you have to find the 'dominant' term which happens to be x^4. As the question is posed you have to terms in brackets that have to be multiplied together. You might be tempted to read x^2 as the dominant term.

suwhitney · Answer

I got it!! :) yayyyy, i have been so confused on this for days. Thank you, thank you, thank you :P