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Mathematics 12 Online
OpenStudy (anonymous):

Find the equations of both the vertical and horizontal asymptotes of the function: f(x)=(3x-1)/(x+4)

OpenStudy (anonymous):

would the vertical be x=-4 and the horizontal would be x=1

OpenStudy (anonymous):

the horizontal would be at y=3, and the vertical at x=-4

OpenStudy (anonymous):

ok, i see how you got the -4 but why is the horizontal 3

OpenStudy (anonymous):

there powers of x are the same, that means that the horizontal asympotote is the ratio of the coefficietnts of the x. In other words 3/1

OpenStudy (anonymous):

ok

myininaya (myininaya):

\[f(x)=\frac{3x-1}{x+4}*\frac{\frac{1}{x}}{\frac{1}{x}}=\frac{\frac{3x}{x}-\frac{1}{x}}{\frac{x}{x}+\frac{4}{x}}=\frac{3-\frac{1}{x}}{1+\frac{4}{x}}\] what happens as x get really really big?

OpenStudy (anonymous):

i'll say that 1/x and 4/x go to zero, is that right?

myininaya (myininaya):

yes!

OpenStudy (anonymous):

yes!!!

myininaya (myininaya):

and 3 and 1 are constants so they stay the same no matter what changes in x happens

OpenStudy (anonymous):

boy, i had to fire up the old noggin for that one

OpenStudy (anonymous):

so if the powers of x arent the same for exp: g(x)=(x+7)/x^2 -4), then it would 1/2 horizontal:1 and vertical:2

OpenStudy (zarkon):

or \[\frac{3x-1}{x+4}=\frac{3x+12-12-1}{x+4}=\frac{3(x+4)-13}{x+4}=\frac{3(x+4)}{x+4}-\frac{13}{x+4}=3-\frac{13}{x+4}\] letting \[x\to\infty\] we get 3

myininaya (myininaya):

in other words: \[\lim_{x \rightarrow \infty}f(x)=\frac{3-0}{1+0}=3/1=3\]

myininaya (myininaya):

thats another one we could do long division! :)

OpenStudy (zarkon):

it is the long way though...your way is faster

OpenStudy (anonymous):

i was wrong right?

myininaya (myininaya):

lagrange was right

OpenStudy (anonymous):

in that second case, we see that the denominator has a higher power than the numerator, so the horizontal asymptote is at y=0

myininaya (myininaya):

oh i was fixing to say what i didn't realize we started talking about another function

OpenStudy (anonymous):

sorry, i meant was i right about the 2nd one

myininaya (myininaya):

no the horizontal asy is wrong and you are missing a veritcal asy

myininaya (myininaya):

what lagrange said about the horizontal asy is correct

myininaya (myininaya):

when is x^2-4, 0?

OpenStudy (anonymous):

do this: x^2-4=0

myininaya (myininaya):

so you said x was 2 is there another x that makes x^2-4=0?

OpenStudy (anonymous):

i will take a wild guess: -2?

myininaya (myininaya):

judges?

OpenStudy (zarkon):

yes ;)

myininaya (myininaya):

you win a brand new tv

OpenStudy (zarkon):

sweet...is it 3D

myininaya (myininaya):

sure

OpenStudy (zarkon):

nice

myininaya (myininaya):

so you have 2 vertical asy one is x=2 another is x=-2

myininaya (myininaya):

ok i'm gonna play some video games now

myininaya (myininaya):

you guys have fun i might look in every once in awhile

OpenStudy (zarkon):

later...have fun!

OpenStudy (anonymous):

catch you later

OpenStudy (anonymous):

ok, later, thanks for the help.

myininaya (myininaya):

lol

OpenStudy (zarkon):

lol

OpenStudy (aravindg):

hi anyone knows chemistry???

OpenStudy (aravindg):

u knw it???

OpenStudy (aravindg):

k i will post it....plz explain too

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