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Math Help - Rational Functions

  1. #1
    Newbie cheet0face's Avatar
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    Rational Functions

    Ok, we've been graphing rational functions for about a week now and i've been able to do most of it... until now
    Past functions have been factorable or divisble with long division.
    But the new problems are different.

    ex. y=(x^4-3x^3+x^2-3x+3) / (x^2-3x)

    another one is

    ex. y=(2x^5-x^3+2) / (x^3-1)

    They don't factor and i've tried dividing both with long division and synthetic.
    I might just be dumb and dividing wrong but I dont see how it works.
    If you could just show me how to do it I would be grateful =)

    Thanks
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  2. #2
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    Quote Originally Posted by cheet0face View Post
    Ok, we've been graphing rational functions for about a week now and i've been able to do most of it... until now
    Past functions have been factorable or divisble with long division.
    But the new problems are different.

    ex. y=(x^4-3x^3+x^2-3x+3) / (x^2-3x)

    another one is

    ex. y=(2x^5-x^3+2) / (x^3-1)

    They don't factor and i've tried dividing both with long division and synthetic.
    I might just be dumb and dividing wrong but I dont see how it works.
    If you could just show me how to do it I would be grateful =)

    Thanks
    Long division works.

    For the first one you should get

    x^2 + 1 + \frac{3}{x^2 - 3x}

    and for the second you should get

    2x^2 - 1 + \frac{2x^2 + 1}{x^3 - 1}.


    If what you end up subtracting completely cancels a term, just bring down the second and go from there. If you're trying to add/subtract something two things that are not alike, e.g. -x^3 + 2x^2 just leave it as is and go from there.
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  3. #3
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    You can use long division on both problems. You might have a remainder, but you can still do it.

    For the first problem you begin with
    \frac{x^4 - 3x^3 + x^2 - 3x + 3}{x^2 - 3x}.

    x^2 - 3x goes into x^4 x^2 times. So you subtract x^2(x^2 - 3x) from x^4 - 3x^3 + x^2 - 3x + 3, which is:

    x^4 - 3x^3 + x^2 - 3x + 3 - (x^4 - 3x^3)

    x^4 - 3x^3 + x^2 - 3x + 3 - x^4 + 3x^3

    x^2 - 3x + 3.

    Save the factor of x^2 for the quotient.

    Now, x^2 - 3x goes into x^2 once, so you subtract x^2 - 3x from x^2 - 3x + 3, which gives you 3.

    So your quotient is x^2 + 1 with a remainder of 3, or exactly x^2 + 1 + \frac{3}{x^2 - 3x}.
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  4. #4
    Newbie cheet0face's Avatar
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    Ok, so I divided the first one and got x^2+1 (we're supposed to drop the remainder), but I still don't see how that fits into graphing it?

    So would the Root be x=-1?

    We have to find the Roots/holes, the Vertical Asymptotes, the Horizontal Asymptotes, and the Y-intercept.

    On previous problems they were like so: (x+3)(x-1)(x+2) / (x+2)(x+1)
    So I could find the Roots/holes to be x=-3, -2 and 1
    VA: x=-1
    HA: y=x
    y=-3

    But I'm not getting anything close to that on these :/
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  5. #5
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    x^2 + 1 has no real roots.

    Also, a root and a "hole" are different things.

    For the equation \frac{(x+3)(x-1)(x+2)}{(x+2)(x+1)}

    the roots are -3 and 1, and the "holes" occur at x = -1 and x = -2. -2 is not a root because the function is not defined there.

    A vertical asymptote occurs wherever there is a "hole".
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  6. #6
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    Quote Originally Posted by icemanfan View Post
    x^2 + 1 has no real roots.

    Also, a root and a "hole" are different things.

    For the equation \frac{(x+3)(x-1)(x+2)}{(x+2)(x+1)}

    the roots are -3 and 1, and the "holes" occur at x = -1 and x = -2. -2 is not a root because the function is not defined there.

    A vertical asymptote occurs wherever there is a "hole".
    If you're trying to graph the original functions you can't disregard the remainder.

    For the first one
    y = x^2 + 1 + \frac{3}{x^2 - 3x}

    To find the x-intercepts, let y equal 0 and solve for x.

    To find the y-intercept, let x = 0 and simplify. Is it possible to have a y-intercept in this case?

    To find the points of discontinuity, we know that the denominator can't be 0, and so x can't be 0 or 3. Also, there are points of discontinuity where y=x^2 + 1. This is because if x^2 + 1 = x^2 + 1 + \frac{3}{x^2 - 3x}, how are you supposed to solve for x?
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  7. #7
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    We'll, she says as x goes farther away, the remainder gets closer and closer to 0 so we're just supposed to drop it.
    And what we're supposed to do is draw a graph grid and put in the roots and holes and the x axis and draw dotted lines going up and down for the asymptotes and put in the y-intercept and pretty much just connect the dots without crossing a vertical asymptote.

    And there is no y-intercept in this problem so we would have to pick an x value and plug it in to get a y value for an extra point.

    What my question is, is I get   x^2+1  but I don't know what it means. :/ Get what i'm saying?
    Like what exactly is that number?
    Is it the simplified equation or just the top part?
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  8. #8
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    Quote Originally Posted by cheet0face View Post
    We'll, she says as x goes farther away, the remainder gets closer and closer to 0 so we're just supposed to drop it.
    And what we're supposed to do is draw a graph grid and put in the roots and holes and the x axis and draw dotted lines going up and down for the asymptotes and put in the y-intercept and pretty much just connect the dots without crossing a vertical asymptote.

    And there is no y-intercept in this problem so we would have to pick an x value and plug it in to get a y value for an extra point.

    What my question is, is I get  x^2+1 but I don't know what it means. :/ Get what i'm saying?
    Like what exactly is that number?
    Is it the simplified equation or just the top part?
    This means that the entire function x^2 + 1 is an asymptote.
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