Thread: Asymptotes of Rational Functions

Originally Posted by fdrhs

[SIZE="3"]Find the vertical, horizontal, and oblique asymptotes, if any, of each rational function. Do not graph.

[1] R(x) = (6x^2 + x + 12)/(3x^2 - 5x - 2)

You need to synthetically divide.
To get,

The horizontal asymptotes are the solutions to,


And the vertical asymptotes are .
Because the second part,
for large .

There are no slant asymptotes.

How did you get y = 2 as the vertical asymptote?

Also, I don't understand the second part.

Originally Posted by fdrhs

Find the vertical, horizontal, and oblique asymptotes, if any, of each rational function. Do not graph.

[1] R(x) = (6x^2 + x + 12)/(3x^2 - 5x - 2)...

Hello,

1. To me the vertical asymptotes are perpendicular to the x-axis. You get them if you calculate the zeros of the denominator:





The vertical asymptotes are and


2. To get the horizontal or oblique asymptotes do long division:

Code:

 (6x^2 +   x + 12)  (3x^2 - 5x - 2) = 2 
-(6x^2 - 10x -  4)
----------------
         11x + 16    <---- remainder

Therefore the equation of your function becomes:



Now calculate the limit if x approaches positive or negative infinity. You see that the fraction approaches zero in both cases. Thus the asymptote is horizantal and has the equation: y = 2

EB

PS.: Even though you are not allowed to graph the function it isn't such a bad idea to control the calculations by a sketch. But don't show it around!

Originally Posted by fdrhs

Find the vertical, horizontal, and oblique asymptotes, if any, of each rational function. Do not graph.
...
[2] F(x) = (x - 1)/(x - x^3)

Hello,

first factor the denominator:



Thus you get 2 vertical asymptotes at x = 0 and x = -1 and a hole at x = 1 that means the graph is punctuated(?).

Because the degree of the nummerator is smaller than the degree of the denominator you get the x-axis as the only horizontal asymptote: y = 0.

EB