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Rational functions(just need explanation)

  1. Oct 3, 2005 #1
    okay i need help on rational functions im not sure if that what they are called but it all should be the same.What i need to find is the

    vertical asymtotes
    horizontal asymtotes
    and the slant asymtotes

    can someone please tell me how to find each on of those and the factors that go along with them the teacher had to teach us by herself because it isnt in our book and he had to review it for sats or somthing thanks alot.
  2. jcsd
  3. Oct 4, 2005 #2
    It's a pretty long process to find all of those.

    First, you have to factor the numerator and denominator.

    To find zeros aka x-intercepts: they are the zeros of the numerator.
    Ex. Numerator: (x+3)(x-2) zeros are -3 and 2.

    Find the y intercept by plugging in the value of the function as x=0.
    Basically, find f(0). (Plug in 0 for x)

    To find vertical asymptotes, find the zeros of the denominator. You can see why right? This makes the resulting fraction undefined.
    (x+3)(x-2) v-asymptotes are -4 and 8.

    I'll show how to find the other stuff later. For now i have to go.
  4. Oct 4, 2005 #3


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    Homework Helper

    That's not an easy question, luv, especially in a medium like this. Some general information:

    It's useful to look at what I call the "overall order" of the function. For instance, suppose the numerator is cubic and the denominator is quadratic. In that case, the overall order is 3-2 = 1, so overall the function is linear. If the numerator is fifth order and the denominator is second, then the overall order is 5 - 2 = 3, so overall the thing is cubic.

    Now: Horizontal asymptotes occur when the function approaches (but does not reach) horizontal lines when x gets very large. This will generally occur when the overall order is 0 or less - i.e. when the order of the numerator is less than or equal to the order of the denominator. You can find the asymptote by plugging in large values for x and seeing what the result is. (Note: as with all of the hints I'm going to give you here, this one isn't infallible. For instance, there's really no way to know how "large" a value of x to use to find the asymptote. It's just that I despair of trying to give you a general way to find the location of a horizontal asymptote through a message board like this.)

    Slant asymptotes are like horizontal asymptotes in the sense that they're lines that the function approaches as x increases (positive or negative) without bound. However, instead of being horizontal lines, they're slanted. Hence the name. This occurs generally when the function has overall order of 1, aka linear. The quickest way to explain a technique for finding them would be to tell you to multiply out both the numerator and the denominator and then carry out the division. If the overall order is actually 1, you should end up with a result like

    y = mx + b + R/d(x)

    in which case the slant asymptote would be y = mx + b

    In both of these cases, you're concerned with the behaviour of the function as x increases without bound. In the case of zeros, holes, and vertical asymptotes, you are concerned with the local behaviour of the function. As the previous poster indicated, you are looking for places where the numerator and the denominator are equal to zero. If just the numerator is zero, then you have a zero in the function. If just the denominator is zero, then you have a vertical asymptote. If they are both zero at the same time, then you have a hole.

    Again - exceptions occur, but if what you're doing is preparing for the SAT's (which I gather), then this should stand you in good stead.
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