Mathematics
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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
11 years ago
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OpenStudy (anonymous):
@ParthKohli
11 years ago
OpenStudy (anonymous):
@cj49
11 years ago
OpenStudy (anonymous):
@TheGaMer007
11 years ago
OpenStudy (thegamer007):
ill try
11 years ago
OpenStudy (anonymous):
thanks so much!!!
11 years ago
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OpenStudy (thegamer007):
oh man
11 years ago
OpenStudy (anonymous):
lol its hard
11 years ago
OpenStudy (thegamer007):
too hard
11 years ago
OpenStudy (thegamer007):
dang
11 years ago
OpenStudy (anonymous):
well thanks for trying
11 years ago
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OpenStudy (anonymous):
@aryandecoolest
11 years ago
OpenStudy (thegamer007):
so sry
11 years ago
OpenStudy (thegamer007):
wat grade u in?
11 years ago
OpenStudy (anonymous):
11 lol hbu
11 years ago
OpenStudy (thegamer007):
WOW im in 8th
11 years ago
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OpenStudy (anonymous):
haha yea...
11 years ago
OpenStudy (anonymous):
@mathmath333
11 years ago
OpenStudy (thegamer007):
well ill try to find the answer for ya
11 years ago
OpenStudy (anonymous):
thanks so much!! @TheGaMer007
11 years ago
OpenStudy (thegamer007):
np :)
11 years ago
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OpenStudy (mathmath333):
have u got the solution
11 years ago
OpenStudy (anonymous):
yea its f(x)=3x^2-9/x^2-4
11 years ago
OpenStudy (thegamer007):
u found it
11 years ago
OpenStudy (mathmath333):
\(\large \begin{align} \color{black}{ f(x)=\dfrac{3x^2-9}{x^2-4}\hspace{.33em}\\~\\
}\end{align}\)
11 years ago
OpenStudy (anonymous):
yes
11 years ago
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OpenStudy (thegamer007):
cool
11 years ago
OpenStudy (anonymous):
so the top is squared root 3?? @mathmath333
11 years ago
OpenStudy (anonymous):
the bottom x=2,x=-2?
11 years ago
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 \(\large x\)
11 years ago
OpenStudy (anonymous):
its 0?
11 years ago
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OpenStudy (anonymous):
wait +-3
11 years ago
OpenStudy (mathmath333):
u have to solve this
\(\large \begin{align} \color{black}{
x^2-4=0 \hspace{.33em}\\~\\
}\end{align}\)
use formula
\(\large \begin{align} \color{black}{a^2-b^2=(a+b)(a-b) \hspace{.33em}\\~\\
}\end{align}\)
11 years ago
OpenStudy (mathmath333):
\(\large \begin{align} \color{black}{
x^2-4=0 \hspace{.33em}\\~\\
x^2-2^2=0 \hspace{.33em}\\~\\
}\end{align}\)
11 years ago
OpenStudy (anonymous):
x-2=0 right @mathmath333
11 years ago
OpenStudy (mathmath333):
\(\large \begin{align} \color{black}{
x^2-4=0 \hspace{.33em}\\~\\
x^2-2^2=0 \hspace{.33em}\\~\\
(x-2)(x+2)=0 \hspace{.33em}\\~\\
x=2~~\normalsize \text{or} \hspace{.33em}\\~\\
x=-2
}\end{align}\)
11 years ago
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OpenStudy (anonymous):
so its y=+-2
11 years ago
OpenStudy (mathmath333):
see above if it is
x or y that i found \(\pm 2\)
11 years ago
OpenStudy (anonymous):
lol sorry its x
11 years ago
OpenStudy (mathmath333):
still need help ?
11 years ago
OpenStudy (mathmath333):
for options
11 years ago
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OpenStudy (anonymous):
11 years ago
OpenStudy (anonymous):
@mathmath333
11 years ago
OpenStudy (anonymous):
should i go with e or d?
11 years ago
OpenStudy (mathmath333):
what do u think ?
11 years ago
OpenStudy (anonymous):
i think e because d is y not x
11 years ago
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OpenStudy (mathmath333):
\(\Huge \checkmark\)
11 years ago
OpenStudy (anonymous):
thanks
11 years ago