Finding the equation of a graph with asymptotes (help please)

jenni2010

1. The problem statement, all variables and given/known data

This is the question:

Shown in the figure below is the graph of a rational function with vertical asymptotes x=2, x=6, and horizontal asymptote y= -2 . (All x-intercepts of the graph of f are also shown, and a point on the graph is indicated.) The equation for f(x) has one of the five forms shown below. Choose the appropriate form for f(x) , and then write the equation.

I can't get the graph on but point are at (-4,0); (0,2); (3,0).



2. Relevant equations

The answers choices are:

A) f(x)= a / x-b

B) f(x)= a(x-b) / x-c

C) f(x)= a / (x-b)(x-c)

D) f(x)= a(x-b) / (x-c)(x-d)

E) f(x)= a(x-b)(x-c) / (x-d)(x-e)



3. The attempt at a solution

What I first did was try and get rid of equations that couldn't possible work. So I knew I had vertical asymptotes at x=2,6. I had two asymptotes so I knew that options A and B couldn't work. I then got rid of option C because dividing a by x to get my horizontal asymptote would not give me y= -2 it would give me y=0. I then got rid of option E by factoring the top part through. Since having ax^2 as my leading coificent would give me a diagonal asymptote I got rid of it. So I got D as my answer but then I have to find the equation. So I got (x-2)(x-6) for the bottom because those would give me vertical asymptotes at x=2,6 But I don't know how to find the top. I got -2(x-12) divided by (x-2)(x-6) because sticking 0 in for x would give me 2. So I had the point (0,2) but I can't get any of the others to match up. I'm not sure how to figure it out.

CompuChip

You are right, that the other points don't match up. To put it in a mathematical way: the set of equations for a, b, c... you get from plugging in the points is overdetermined and does not have a solution. Therefore D is wrong.

I think your argument for getting rid of F is not entirely correct. You are saying, that your leading behaviour is ~ x^2. However, you also have something ~ x^2 in the denominator.
Did you learn to take limits yet? Can you properly calculate
[tex]\lim_{x \to \infty} \frac{a (x - b)(x - c) }{ (x - d)(x - c) }[/tex]

Also note that you will need all the points to determine the constants. To take your (wrong) example of option D:
if you plug in x = 0 then you get
f(0) = - a b / 12 = 3.
Although you get a relation a = -36 / b, this alone will not allow you to determine a and b, you will need at least one other equation.

HallsofIvy

Does option F really have "(x- c)" in both numerator and denominator? And what happened to option E?

CompuChip

Ah right, I assumed that was a typo and it should be (x - e) in the denominator.

lol that I didn't even notice.

jenni2010

No I really don't know how to do limits. We've touched on it a little bit last year. But I am in an online pre-calculus course and the program doesn't really explain things in terms I can understand. The program has never shown me how to do this so I have no idea how to do it. I'm completly confused and lost. I'm slow at learning math so most of the time I need it in the simplest terms possible.