Help with horizontal asymtopes!?

Question

How do I find the horizontal asymtopes of
H(x) = (x^3-1)/(x^2-9) ?

I thought that there wouldn't be any because 3>2?? But it says in the back of my book that the horizontal asymptope is x=y, intersected at (1/9, 1/9)

Help please!! Thank you

Answer 1

y = x is not a horizontal asymptote. It's an oblique asymptote.

Since limH(x)/x = 1, when x approaches to infinity, the oblique asymptote is y = limH(x) = x.

Plug in y = x for H(x),
x = H(x) = (x^3-1)/(x^2-9)
x^3 - 9x = x^3 - 1
Solve for x,
x = 1/9
y = x = 1/9

Therefore, the coordinates of intersection between the oblique asymptote and H(x) is (1/9, 1/9)

Answer 2

Sounds to me like an oblique asymotote, not a horizontal asymtote. Could be a typo in your book, check with your teacher. This happens sometimes, especially if your book is in the first edition.
You're right, since the top is of higher degree than the bottom, there won't be a horizontal asymptote, because the horizontal asymptote indicates that the function's behavior as x becomes arbitrarily large, y "settles down" to 0 or some other finite number. However, if the numerator is of higher degree than the denominator, the function may have oblique asymptotes, which can be found by dividing the numerator by the denominator with polynomial or synthetic division. This will, in this case, yield a function which is the sum or difference of a linear and a rational function. You can think of the linear part as the oblique asymptote, and the rational part as the deviation of the function from the asymptote.
The intersected part means that the graph of the function actually crosses the line x=y at that point.
For more information see
http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut40_ratgraph.htm

Answer 3

For

H(x) = (x^3 - 1) / (x^2 - 9)

What you have to do is long division on this problem. What you should get is something of the form

H(x) = [f(x) / g(x)] + r(x)

Your asymptote is going to be y = r(x).

Without showing you the details, the result of the long division of (x^3 - 1) / (x^2 - 9) will be

H(x) = (9x - 1)/(x^2 - 9) + x

Therefore, as we said earlier regarding the function separate from the fraction, our r(x) is x. Therefore, y = x is our asymptote.

Note that this isn't a horizontal asymptote; in order for it be a horizontal asymptote, r(x) must be some constant, like the following example below:

****
Let's take a second (more common) example:

y = (2x + 1)/(x - 1)

Using long division will give us the result

y = 2/(x - 1) + 2

Which would result in our horizontal asymptote being y = 2.

Answer 4

there's no horizontal asymtopes !
lim h(x) when x -------> +or- infini is x
let' s see: lim h(x) whenx---> +or- infini is:
'' = x^3/x^2=x^2*x/x^2=x
this is an oblique asymptote like A.O having
a straight y=x+(9x-1)/x^2-9 finding by y=Q+R/D
y=x+(9x-1)/x^2-9
set x=0 and y=0 to find the point intersected by this straight.
for x=0 , y=-1/-9 or y=1/9
for y=0 , x+(9x-1)/x^2-9=0 or x=1/9
the point intersected is shown by K(1/9,1/9)

Answer 5

H(x) = (x^3-1)/(x^2-9)

First: look at the degrees (exponents) with the "x" variables >

x^3/x^2

Sec: the numerator's degree (3) is larger than the numerator's degree (2) >

Third: the horizontal assymptote is - there is no horizontal assymptote

Answer 6

y = x is an oblique asymptote. It sounds like someone imprecisely used the term "horizontal asymptote" to simply mean an asymptote that is not vertical.