• March 30th 2006, 08:04 AM
mazdatuner
A question I have to answer is from the given equasion
2x^2 - 3x - 14
----------------- =F(x)
x^2 - 2x -8

I need to find the Vertical asymtope
Horozontal asymtope, slant asymtope , any holes and the x and y intercepts.

So far I have divided and come up with x^2 - x - 24
I am not sure If I am correct in doing that or even close any help would be greatly apreciated
• March 30th 2006, 09:34 AM
topsquark
Quote:

Originally Posted by mazdatuner
A question I have to answer is from the given equasion
2x^2 - 3x - 14
----------------- =F(x)
x^2 - 2x -8

I need to find the Vertical asymtope
Horozontal asymtope, slant asymtope , any holes and the x and y intercepts.

So far I have divided and come up with x^2 - x - 24
I am not sure If I am correct in doing that or even close any help would be greatly apreciated

Umm...no, that's not what you get when you divide them.

Horizontal asymptotes:
You want to take $\lim_{x \to \infty}\frac{2x^2-3x-14}{x^2-2x-8}$
Note that as x blows up, the most important terms are the x^2. So:
$\lim_{x \to \infty}\frac{2x^2-3x-14}{x^2-2x-8} \approx \frac{2x^2}{x^2} = 2$
(Or you can just plug a large number, say x=10000, and see what you get.)
So we have a horizontal asymptote at y=2 on the +x side.

Similarly:
$\lim_{x \to -\infty}\frac{2x^2-3x-14}{x^2-2x-8} \approx \frac{2x^2}{x^2} = 2$
(Squaring takes away the "-" sign, so the expressions are the same.)
So we have the same horizontal asymptote on both sides of the y-axis: y=2.

Vertical asymptotes:
It's a good idea to factor your expression here before we figure these. I'll show you why in a minute.
$F(x)=\frac{2x^2-3x-14}{x^2-2x-8}=\frac{(2x-7)(x+2)}{(x-4)(x+2)}$

Generally you find vertical asymptotes where the denominator of the expression goes to zero. This means that x = 4 is a vertical asymptote. Normally x = -2 would be another, but the x+2 in the denominator is cancelled by the x+2 in the numerator. That's why you ALWAYS want to factor these things.

Slant asymptotes:
You get a slant asymptote only when the degree of the numerator is one more than the degree of the denominator. In this case they are both quadratics, so we don't have any slant asymptotes.

Holes:
Not sure what you mean by this. Perhaps we are speaking of x = -2? This is a point in the domain that is not allowed.

x intercepts:
This is where F(x) is zero. This will happen when the numerator goes to zero (assuming that none of these x values are where we have a zero in the denominator!) The only non-cancelling term in the numerator is the 2x-7, so if we set that to zero we find an x intercept at the point (7/2,0).

y intercepts:
These are easy. Just set x = 0 in your function:
$F(0)=\frac{2*0^2-3*0-14}{0^2-2*0-8}=\frac{-14}{-8}=\frac{7}{4}$
So we have a y intercept at (0, 7/4).

-Dan