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:
$\displaystyle 3x^2-5x-2=0 \Longleftrightarrow x=\frac{-(-5) \pm \sqrt{(-5)^2-4 \cdot 3 \cdot (-2)}}{2 \cdot 3}$
$\displaystyle x=\frac{5 \pm 7}{6} \Longleftrightarrow x = 2\ \vee \ x=-\frac{1}{3}$
The vertical asymptotes are $\displaystyle x = 2$ and $\displaystyle x=-\frac{1}{3}$
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:
$\displaystyle R(x) = \frac{6x^2 + x + 12}{3x^2 - 5x - 2}=2 + \frac{11x+16}{3x^2 - 5x - 2}$
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!