How do you find the horizontal asymptotes of these two problems?

Question

a) f(x) = (3x)/(x-2)

b) f(x)= (x-5)/(x^2+6x+5)

In problem "a" I tried solving for x, but when I graphed it the answer didn't seem to be right (according to the graphing calc.).

In problem "b", you can't solve for x because the of the x^2 (I think?), so I tried the other method, which involved dividing all the numbers by the highest power of x. The result left me somewhat confused. Can anyone explain this please? x_x

Answer 1

a) Multiply top and bottom by 1/x:
3/(1-2/x)
as x grows larger and larger, 2/x becomes 0 so that fraction becomes 3/(1-0) or 3.

b) same deal but this time, divide top and bottom by x^2.
(1/x-5/x^2)/(1+6/x+5/x^2)
as x grows larger, the top becomes zero and the bottom becomes 1. 0/1 is 0.

Answer 2

To find the horizontal asymptotes of a function you must solve for the limit of the function as it approaches +/- infinity. To do this divide each term by the est power in the denomenetor, ex (x)/(x^2+3x the largest power in the denomenetor is x^2)So...

a)Lim = (3x)/(x-2)
Lim = (3x/x)/(x/x-2/x)
Lim = (3)/(1-2/x)
as x approaches +/- infinity 2/x approaches 0 so...
Lim = (3)/(1-0)
Lim = 3/1=3
and so the horizontal asymptote is at y = 3

b) Lim = (x - 5)/(x^2 + 6x + 5)
Lim = (x/x^2 - 5/x^2)/ (x^2/x^2 + 6x/x^2 + 5/x^2)
Lim = (1/x -5/x^2) / (1 + 6/x + 5/x^2)
Terms begin to approach 0 as the limit approachs +/-infinity so...
Lim = (0 - 0) / (1 + 0+ 0)
Lim = (0/1) = 0
and so the horizontal asymptote is at y=0

Answer 3

The way you find a horizontal assymptote is by approaching the value of x to infinity. We also know that any number divided by infinity is 0. (Just divide a number by a bigger number and you'll get a very small number. keep increasing the denominator and eventually you'll go to 0)

a) y= 3x / x-2 divide everything by x
y = 3 / 1-(2/x) we approach x to infinity and 2/x becomes 0.
y=3

b) Do the same stuff as in 'a' and it'll approach 0.

Answer 4

Derek:

The answers provided are indeed correct. Another way to think about this problem is as follows:

a) f(x)= 3x/(x-2) = 3 + 6/(x-2)

as x approaches plus infinity, f(x) approaches 3
as x approaches minus infinity, f(x) approaches 3

b) f(x) = x-5 / (x^2+6x+5) = (5/2) / (x+5) + (-3/2) / (x+1)
Note that as x goes to infinity, f(x) goes to zero in both cases.

Hope this explanation helps you visualize these two problems.

Answer 5

My teacher told me a really easy way to find horizontal asymptotes. Just divide the numerator by the denominator.

in a) y = 3
in b) y = 0 (If it doesn't divide in at all, it's y = 0).

Answer 6

For the first one the asymptote is y=3.

For the second one you get y=0 because the degree below is higher

Answer 7

a. f(x) = (3x)/(x-2)

First: look at the degrees on the variables "x" on the top and bottom: x/x and they have the same degree = 1.

Second: when both variables have the same degree, you take the coefficients: 3/1 and divide if they're whole numbers.

H.A. or y = 3