Find the equations of both the vertical and horizontal asymptotes of the function: f(x)=(3x-1)/(x+4)
would the vertical be x=-4 and the horizontal would be x=1
the horizontal would be at y=3, and the vertical at x=-4
ok, i see how you got the -4 but why is the horizontal 3
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
ok
\[f(x)=\frac{3x-1}{x+4}*\frac{\frac{1}{x}}{\frac{1}{x}}=\frac{\frac{3x}{x}-\frac{1}{x}}{\frac{x}{x}+\frac{4}{x}}=\frac{3-\frac{1}{x}}{1+\frac{4}{x}}\] what happens as x get really really big?
i'll say that 1/x and 4/x go to zero, is that right?
yes!
yes!!!
and 3 and 1 are constants so they stay the same no matter what changes in x happens
boy, i had to fire up the old noggin for that one
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
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
in other words: \[\lim_{x \rightarrow \infty}f(x)=\frac{3-0}{1+0}=3/1=3\]
thats another one we could do long division! :)
it is the long way though...your way is faster
i was wrong right?
lagrange was right
in that second case, we see that the denominator has a higher power than the numerator, so the horizontal asymptote is at y=0
oh i was fixing to say what i didn't realize we started talking about another function
sorry, i meant was i right about the 2nd one
no the horizontal asy is wrong and you are missing a veritcal asy
what lagrange said about the horizontal asy is correct
when is x^2-4, 0?
do this: x^2-4=0
so you said x was 2 is there another x that makes x^2-4=0?
i will take a wild guess: -2?
judges?
yes ;)
you win a brand new tv
sweet...is it 3D
sure
nice
so you have 2 vertical asy one is x=2 another is x=-2
ok i'm gonna play some video games now
you guys have fun i might look in every once in awhile
later...have fun!
catch you later
ok, later, thanks for the help.
lol
lol
hi anyone knows chemistry???
u knw it???
k i will post it....plz explain too
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