Question

although most graphs have at one horizontal asymptote, it is possible for a graph to have more than one. Which of the following functions have graphs with more than one horizontal asymptote? (state them and give reasons, please?)

a) f(x)=abs(x^3+1)/(8-x^3)
b) g(x)= abs(x-1)/(x^2-4)
c)h(x)= x/√(x^2-4)

Answer 1

\[\frac{\left| x^3 + 1 \right|}{-x^3 + 8}\] Has two horizontal asymptotes. One at \[y = -1\] and one at \[y = 1\]

Answer 2

I know how to explain how to get the asymptote y = -1.. When the numerator and denominator are of the same degree.. You look at the leading coefficients of the highest degree term and that is your horizontal asymptote. So looking at the x^3 the leading coefficients are 1/-1 Hence a horizontal asymptote is at y = -1.. However i do not know how to explain y = 1 without using limits.

Answer 3

well I'm learning limits now, so please help me with the other two, you are VERY helpful

Answer 4

I dont believe the other two have multiple asymptotes. Let me check.

Answer 5

i got y=0 for b and
y=-1 , and y=1 for c is that right?

Answer 6

\[\frac{x}{\sqrt{x^2-4}}\] Does have two horizontal asymptotes. One at \[y = 1\] and \[y = -1\]And this site helps in explaining.

Answer 7

http://home.scarlet.be/~ping1339/asym.htm#Horizontal-asymptote

Answer 8

YESSSSSS! thanks you soooo much. and for part b did you get 0? please say yes.

Answer 9

Let me check!

Answer 10

B has a asymptote at y=0 you are correct!

Answer 11

yesssss!!!!!! your way of explaining worked better for me than the way my teacher explained it

Answer 12

Just evaluate the limit of the function as x approaches infinity and that is how you find horizontal asymptotes. And if you ever need to find vertical asymptotes, just set the denominator equal to 0 :D.

Answer 13

yeah vertical ones i know that way, but horizontal ones messed me up because he didn't say it right