Michael H.

asked • 02/12/15# I have to sketch a graph for the rational equation with all asymptotes included where are the asymptotes and what points work for drawing the graoh

F(x)=X^2-4/x^2-3x+2

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## 2 Answers By Expert Tutors

Roy P. answered • 02/12/15

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Astronomer and System Administrator

Start by remembering that you'll get a vertical asymptote where you get an undefined ...or infinite result. In rational functions, this generally happens when the denominator goes to 0 for some value .... you can't divide by 0, even with polynomials!

I'd do this by factoring the denominator into x

^{2}-3x+2 = (x-1)(x-2) Check that this works.Now, if either quantity in parentheses there goes to 0, the whole denominator is 0, and we need to know the values for which this happens to identify the asymptotes. Since If (x-1) = 0 makes the bottom all 0 just as (x-2) = 0 does, we can solve these two simple equations for x to get x = 1 and x = 2. Check that x=1 or 2 makes the denominator go to 0. There are the x values where your asymptote will be.

BUT WAIT. The numerator can be factored, too. Into (x-2)(x+2). So the function actually looks like:

f(x) = [(x-2)(x+2)]/[(x-1)(x-2)] So the (x-2) terms cancel and so does that asymptote. You have a vertical asymptote at x = 1 only.

f(x) is now ... [(x+2)]/[(x-1)]

Now, the horizontal asymptote is identified by the ratio of the coefficients on the leading x-terms. HEAVY, I KNOW.

So, you have a horizontal asymptote if and only if the highest power in the numerator is the same as the highest power in the denominator. So, here you have x up top and x down below, so you'll have a horizontal asymptote. Both the top and bottom x have coefficients of 1.

x = 1*x, right? The ratio of 1/1 is 1. So you'll have a horizontal asymptote at y = 1.

To graph, draw in the asymptotes, first, then pick some points in the regions away from the asymptotes on the x-axis. Plug and chug!

Good luck!

(http://www.freemathhelp.com/finding-horizontal-asymptotes.html)

PS: My answer is correct. idk what these other guys were doing.

Michael H.

Ur answer is correct thanks for the help, the other guy answered a different question I since then edited, my final problem is that I accidentally edited the equation incorrectly the denominator should b x^2+3x+2...my brain is fried this is due tomorrow n with all the help I've received I can't use it cuz it doesn't apply thanks again anyways

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02/12/15

Roy P.

That's an easy fix. Check to see that the correct denominator factors into (x+1)(x+2). Follow what I've written above with this change.

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02/13/15

Mark M. answered • 02/12/15

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Mathematics Teacher - NCLB Highly Qualified

The numerator does not affect the graph.

A vertical asymptote is at x = 4.

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Michael H.

02/12/15