Ask your own question, for FREE!
Mathematics 28 Online
OpenStudy (anonymous):

Identify the oblique asymptote of f(x) = quantity x squared minus 4 x plus 8 over quantity x plus 2. HELPPP

OpenStudy (johnweldon1993):

The oblique asymptote will come from the long division \[\large \frac{x^2 - 4x + 8}{x + 2}\] |dw:1407715655355:dw|

OpenStudy (anonymous):

THANKS SO MUCH

OpenStudy (johnweldon1993):

No problem :)

OpenStudy (anonymous):

CAN YOU HELP ME WITH A NOTHER

OpenStudy (anonymous):

@johnweldon1993

OpenStudy (johnweldon1993):

Yeah sure :)

OpenStudy (anonymous):

awesomeee 1 sec

OpenStudy (anonymous):

Identify the horizontal asymptote of f(x) = quantity x squared plus 5 x minus 3 over quantity 4 x minus 1.

OpenStudy (johnweldon1993):

Horizontal asymptotes happen when we see what the function equals when we let x go to infinity

OpenStudy (johnweldon1993):

so \[\large \frac{x^2 + 5x - 3}{4x - 1}\] If we plug in infinity here for 'x' we would essentially have \[\large \frac{\infty^2}{\infty} = \infty\]

OpenStudy (johnweldon1993):

Now we dont say this has a horizontal asymptote of infinity...we just say it doesn't have one

OpenStudy (anonymous):

oh. okay great thanks so much, i really appreciate it

OpenStudy (johnweldon1993):

Anytime!

OpenStudy (anonymous):

so for oblique asymptotes we do long division?

OpenStudy (johnweldon1993):

Right Oblique = long division Horizontal = plug in infinity for 'x' and simplify Vertical = See what makes the denominator = 0

OpenStudy (anonymous):

okay i sorta get it now thanks thanks loads

OpenStudy (johnweldon1993):

If you have any more questions feel free to message me :)

OpenStudy (anonymous):

okay

OpenStudy (anonymous):

@johnweldon1993

OpenStudy (anonymous):

What is the discontinuity of the function f(x) = the quantity of x squared minus 4 x minus 12, all over x plus 2?

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!