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Math Help - Determin an equation from a rational fucntion graph

  1. #1
    Junior Member
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    Exclamation Determin an equation from a rational fucntion graph

    Please Help!!

    Basically the graph is showing:

    vertical asymptote (V)on x=3
    and horizontal asymptote(H) on x=2
    and a point in the graph (1.5,0)

    So this is how I got an answer
    y=ax+b
    ------
    cx+d
    since H=a/c a=2c
    V=-d/c d=-3c
    to find b, i did some substitution

    y=2cx+b
    -------
    cx-3c

    substitute a point from the graph

    0=2c(1.5)+b
    -----------
    c(1.5)-3c
    0=3c+b
    -----------
    -1.5c
    therefore
    b=-4.5 c

    I sub in b into the general equation

    y=2cx-4.5c
    ----------
    cx-3c
    y= c(2x-4.5)
    ---------- c cancel out
    c(x-3)

    final answer is
    y=2x-4.5
    ------
    x-3

    but my book's answer is

    y=2x-3
    ---------
    x-3


    if it helps im using nelson advanced functions book 2008
    pg 154 question 7

    please help!! and if anyone can point me to a website that has better explaining in terms of rational functions in the form


    y=ax+b
    ------
    cx+d
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  2. #2
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    Quote Originally Posted by jepal View Post
    Basically the graph is showing:

    vertical asymptote (V)on x=3
    and horizontal asymptote(H) on x=2
    and a point in the graph (1.5,0)
    If the vertical asymptote is at x = 3, then x - 3 is a factor of the denominator, and no other such linear (neatly solveable in the whole numbers) factor exists there.

    If the function has an horizontal asymptote that is off the x-axis, then the degree of the numerator is the same as the degree of the denominator.

    The simplest conclusion would then be that the rational function is of the form f(x) = (2x + b)/(x - 3), where "b" is some constant.

    We are given that f(1.5) = 0, so (3 + b)/(1.5 - 3) = 0. A fraction is zero only when its numerator is zero, so 3 + b = 0.

    What then is the value of "b"?
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  3. #3
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    Quote Originally Posted by stapel View Post
    If the vertical asymptote is at x = 3, then x - 3 is a factor of the denominator, and no other such linear (neatly solveable in the whole numbers) factor exists there.

    If the function has an horizontal asymptote that is off the x-axis, then the degree of the numerator is the same as the degree of the denominator.

    The simplest conclusion would then be that the rational function is of the form f(x) = (2x + b)/(x - 3), where "b" is some constant.

    We are given that f(1.5) = 0, so (3 + b)/(1.5 - 3) = 0. A fraction is zero only when its numerator is zero, so 3 + b = 0.

    What then is the value of "b"?
    ok thanks for the reply but im still a bit confuse. In terms of f(x)=(2x+b), where did you get the 2 from or how did u go about determing the two. Im assuming its the horizontal asymptote. Will this be case all the time, if the horizontal asymptote is off the x-axis.
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  4. #4
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    Quote Originally Posted by jepal View Post
    In terms of f(x)=(2x+b), where did you get the 2 from or how did u go about determing the two.
    What is the horizontal asymptote?

    How is this value obtained?

    Given a (simplest) denominator of "x - 3", what must the leading coefficient of the numerator be?

    Quote Originally Posted by jepal View Post
    Will this be case all the time, if the horizontal asymptote is off the x-axis.
    No. Review the rules for asymptotes; in particular, what happens when the degree of the denominator is greater than the degree of the numerator.
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