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

  1. Feb 20, 2009 #1
    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.
    Last edited: Feb 21, 2009
  2. jcsd
  3. Feb 21, 2009 #2


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    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.
  4. Feb 21, 2009 #3


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    Does option F really have "(x- c)" in both numerator and denominator? And what happened to option E?
  5. Feb 21, 2009 #4


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

    lol that I didn't even notice. :biggrin:
  6. Feb 21, 2009 #5
    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.
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