Question

State the horizontal asymptote of the rational function.
f(x) = x^2 + 8 -x/ x - 2

Answer 1

degree of the numerator (2) is larger than the degree of the denominator (1), so none

Answer 2

oh maybe i am reading it incorrectly

Answer 3

\[f(x) =\frac{ x^2 + 8 -x}{ x - 2}\] is that it?

Answer 4

Yes.

Answer 5

ok then answer i gave above is correct

Answer 6

okay what about this one? State the horizontal asymptote of the rational function.
f(x) = 5x+1/9x-2

Answer 7

thank you for your help by the way. satellite73

Answer 8

Think about what happens when you let x get really large, say, 1000 for starters. x^2 will be 1,000,000, and x will be 1000. So already, x^2 completely dominates x. So, when you have the numerator growing a lot bigger than the denominator, there's not going to be a horizontal asymptote: the function will just keep on increasing, without stopping.

That's a good way to think it through, but here are the "rules":
If the degree of the numerator is greater than the degree of the denominator, then there's no asymptote.
If the degree of the numerator is less than the degree of the denominator, then the asymptote is y=0.
If the degrees are equal, then the asymptote is the ratio of the leading coefficients.

So, can you see how to answer your next question with that info?

Answer 9

degrees are the same (both 1) so it is the ratio of the leading coefficients
\[y=\frac{5}{9}\]
yw

Answer 10

last one i promise! State the vertical asymptote of the rational function.
f(x) = (x-3)(x+4)/ x^2-1

Answer 11

this one is a trick

Answer 12

oh vertical, not horizontal]
set the denominator equal to zero and solve for \(x\)

Answer 13

Vertical asymptotes happen when the denominator is 0 and the numerator isn't. In this case, the denominator is 0 when x^2 = 1, so x = 1 or -1.

Answer 14

\[x^2-1=0\implies x^2=1\] so \(x=1\) or \(x=-1\)

Answer 15

thank you thank you thank you sooo much !