Question

find all asymptotes for the function y=√((x^4)+(6x^2)+9)/((2x^2)-10)

Answer 1

\[y=\frac{ \sqrt{x ^{4}+6x ^{2}+9} }{ 2x ^{2}-10}\]

Answer 2

Notice that \[x ^{4} + 6x ^{2} + 9 \] is in the form of \[a ^{2} + 2ab + b ^{2}\].
So \[x ^{4} + 6x ^{2} + 9 = x^{4} + 3x^{2} + 3x^{2} + 9 = x^{2}(x^{2} + 3) + 3(x^{2} + 3) = (x^{2} + 3)^{2}\]

Answer 3

So you would get \[y = \frac{ x^{2} + 3 }{ 2x^{2} - 10 }\]. To find the asymptote, simply divide \[x^{2}\] by \[2x^{2}\] and you will get the asymptote \[y = \frac{ 1 }{ 2 }\]

Answer 4

Next, the second asymptote is by setting the denominator to 0. That is \[2x ^{2} - 10 = 0\]\[2(x ^{2} - 5) = 0\]\[x ^{2} = 5\]\[x = \pm \sqrt{5}\]

Answer 5

which asymptote is which? vertical/horizontal??

Answer 6

\[y = \frac{ 1 }{ 2 }\] is a horizontal asymptote.\[x = \pm \sqrt{5}\] are vertical asymptotes.

Answer 7

ok thanks! for the horizontal asymptote what happened to the (x^2 + 3)^2 the square on the outside? @TURITW

Answer 8

It got square rooted.

Answer 9

oh yeah i see! thanks so much!!!

Answer 10

@Loser66 @TURITW how would you write the answers as equations? is that just x=... y=...

Answer 11

The asymptotes are x = , y = ...