horzontal and vertical asymptote..

[candy101]

im not sure if this is the right place im putting this question in (if its not right sorry)

i want to know how to find the horizontal asymptote?

i know that in order to find the vertical asymptote you gotta set the bottom number to zero
and for horizontal you gotta use limit i kind of get the vertical one but im stuck on the horizontal one

such as this problem

3x^2 + 11x -7/ x^2 - 2x -8

so i will set the bottom to zero which is the vertical but what to do with the top?


please help!

AND THANKS FOR ANY HELP!! (oh n sorry if this is not in the right place but there is no pre-cal)

 

[galactus]

For the horizontal asymptote, note that the powers of the numnerator and denominator are the same. We can ignore all but the highest powers and basically get \(\displaystyle \frac{x^{2}}{x^{2}}\)

Now, see what the horizontal asymptote is?.

 

[Aladdin]

For the horizontal asymptote, note that the powers of the numnerator and denominator are the same. We can ignore all but the highest powers and basically get \(\displaystyle \frac{x^{2}}{x^{2}}\)

Now, see what the horizontal asymptote is?.

You get : \(\displaystyle 3\frac{x^{2}}{x^2}}\)

Simplify you get : horizantal asymptote equal to y=3.

 

[BigGlenntheHeavy]

\(\displaystyle What \ is \ f(x)? \ If \ f(x) \ is \ 3x^{2}+11x-\frac{7}{x^{2}-2x-8}, \ then \ there \ is \ no \ horizontal \ asymptote.\)

 

[galactus]

Sorry for the typo. I meant \(\displaystyle \frac{3x^{2}}{x^{2}}=3\)

If you ever want to check for horizontal asymptotes, follow this criteria:

\(\displaystyle f(x)=\frac{a_{n}x^{n}+........}{b_{k}x^{k}+...........}\)

If n<k, then the x-axis is the HA

If n=k, then the line \(\displaystyle y=\frac{a_{n}}{b_{k}}\), the ratio of leading coefficients is the HA. This is your case where the ratio is 3.

If n>k, then there is no HA.

 

[candy101]

i have to use limit and infinity for the horizontal asymptote..is this the same as limits and infinity?



Thanks for the reply!