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Mathematics 21 Online
OpenStudy (anonymous):

what is the horizontal asymptote of the function f(x)=3x^2-9/x^2-4 a. y=3 b. x=+-3 c. x=9/4 d y=+-2 e. none

OpenStudy (anonymous):

@ParthKohli

OpenStudy (anonymous):

@cj49

OpenStudy (anonymous):

@TheGaMer007

OpenStudy (thegamer007):

ill try

OpenStudy (anonymous):

thanks so much!!!

OpenStudy (thegamer007):

oh man

OpenStudy (anonymous):

lol its hard

OpenStudy (thegamer007):

too hard

OpenStudy (thegamer007):

dang

OpenStudy (anonymous):

well thanks for trying

OpenStudy (anonymous):

@aryandecoolest

OpenStudy (thegamer007):

so sry

OpenStudy (thegamer007):

wat grade u in?

OpenStudy (anonymous):

11 lol hbu

OpenStudy (thegamer007):

WOW im in 8th

OpenStudy (anonymous):

haha yea...

OpenStudy (anonymous):

@mathmath333

OpenStudy (thegamer007):

well ill try to find the answer for ya

OpenStudy (anonymous):

thanks so much!! @TheGaMer007

OpenStudy (thegamer007):

np :)

OpenStudy (mathmath333):

have u got the solution

OpenStudy (anonymous):

yea its f(x)=3x^2-9/x^2-4

OpenStudy (thegamer007):

u found it

OpenStudy (mathmath333):

\(\large \begin{align} \color{black}{ f(x)=\dfrac{3x^2-9}{x^2-4}\hspace{.33em}\\~\\ }\end{align}\)

OpenStudy (anonymous):

yes

OpenStudy (thegamer007):

cool

OpenStudy (anonymous):

so the top is squared root 3?? @mathmath333

OpenStudy (anonymous):

the bottom x=2,x=-2?

OpenStudy (mathmath333):

the asymptote is found when \(\large f(x) \) is undefined that is it is \(\large \infty\) so \(\large \begin{align} \color{black}{ f(x)=\infty\hspace{.33em}\\~\\ \dfrac{3x^2-9}{x^2-4}=\infty \hspace{.33em}\\~\\ \dfrac{3x^2-9}{x^2-4}=\dfrac{1}{0} \hspace{.33em}\\~\\ \dfrac{x^2-4}{3x^2-9}=\dfrac{0}{1} \hspace{.33em}\\~\\ x^2-4=0 \hspace{.33em}\\~\\ }\end{align}\) solve further to get the values of \(\large x\)

OpenStudy (anonymous):

its 0?

OpenStudy (anonymous):

wait +-3

OpenStudy (mathmath333):

u have to solve this \(\large \begin{align} \color{black}{ x^2-4=0 \hspace{.33em}\\~\\ }\end{align}\) use formula \(\large \begin{align} \color{black}{a^2-b^2=(a+b)(a-b) \hspace{.33em}\\~\\ }\end{align}\)

OpenStudy (mathmath333):

\(\large \begin{align} \color{black}{ x^2-4=0 \hspace{.33em}\\~\\ x^2-2^2=0 \hspace{.33em}\\~\\ }\end{align}\)

OpenStudy (anonymous):

x-2=0 right @mathmath333

OpenStudy (mathmath333):

\(\large \begin{align} \color{black}{ x^2-4=0 \hspace{.33em}\\~\\ x^2-2^2=0 \hspace{.33em}\\~\\ (x-2)(x+2)=0 \hspace{.33em}\\~\\ x=2~~\normalsize \text{or} \hspace{.33em}\\~\\ x=-2 }\end{align}\)

OpenStudy (anonymous):

so its y=+-2

OpenStudy (mathmath333):

see above if it is x or y that i found \(\pm 2\)

OpenStudy (anonymous):

lol sorry its x

OpenStudy (mathmath333):

still need help ?

OpenStudy (mathmath333):

for options

OpenStudy (anonymous):

OpenStudy (anonymous):

@mathmath333

OpenStudy (anonymous):

should i go with e or d?

OpenStudy (mathmath333):

what do u think ?

OpenStudy (anonymous):

i think e because d is y not x

OpenStudy (mathmath333):

\(\Huge \checkmark\)

OpenStudy (anonymous):

thanks

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