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Math Help - New member needing some help.

  1. #1
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    Question New member needing some help.

    Hello all. I am a new member to this site, although I have been browsing for quite some time now.

    I have a couple questions that I just can't seem to solve for tonight's homework assignment.

    I was hoping that I might be able to get some help from the great members of this site...


    1. Construct a rational function whose vertical asymptotes are located at x = -2 and x=k, where k > 0.



    2. Construct a rational function whose vertical asymptote is located at
    x = 3 and a horizontal asymptote at y = 1/2.



    Any help would be greatly appreciated, and thanks in advance to anyone who gives it a shot!
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  2. #2
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    Quote Originally Posted by MathNoob11 View Post

    1. Construct a rational function whose vertical asymptotes are located at x = -2 and x=k, where k > 0.
    One possibilty is

     y=\frac{1}{(x+2)(x-k)} as x \neq -2,k
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  3. #3
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    Quote Originally Posted by MathNoob11 View Post


    2. Construct a rational function whose vertical asymptote is located at
    x = 3 and a horizontal asymptote at y = 1/2.

    y = \frac{a}{x-3}+\frac{1}{2} where a \in \mathbb{R}
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  4. #4
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    Thank you for the quick response.

    I am also having trouble with the following problem.

    1. Sketch the function y(x) = 2-x / (3+x)(x-1)


    a. x-axis intercept
    b. y-axis intercept
    c. horizontal asymptote(s)
    d.vertical asymptote(s)

    Would anyone be willing to offer their assistence with this problem?

    Again thank you in advance!
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  5. #5
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    Quote Originally Posted by MathNoob11 View Post

    a. x-axis intercept
    Make y(x)=0 and solve for x

    Quote Originally Posted by MathNoob11 View Post


    b. y-axis intercept
    Make x=0 and solve for y(x)

    Quote Originally Posted by MathNoob11 View Post

    c. horizontal asymptote(s)
    d.vertical asymptote(s)
    My previous 2 posts have covered this.
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  6. #6
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    I was able to calculate the y-axis intercept by doing the following.

    y(x)= 2 - x / (3 + x)(x-1)

    y(x) = 2 - 0 / (3 +0)(0-1)

    = y(x) = 2 / (3)(-1)

    =y(x) = 2/-3

    y(x) = -2/3

    So the y-intercept is (0,-2/3) or (0, -.667) right??


    But I am still having trouble with the other steps to the problem.

    Im having trouble foiling out once I set the denominator equal to 0.

    Can you offer some assistence please?
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  7. #7
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    (The way you wrote the function is vague. The way it was written I wasn't sure if your function is
    \frac{2 - x}{(3 + x)(x - 1)} = 0 or
    2 - \frac{x}{(3 + x)(x - 1)} = 0.
    I'm going to assume it's the former. Please either learn LaTex or use parentheses.)

    To find the x-intercept, set the function equal to zero:
    \frac{2 - x}{(3 + x)(x - 1)} = 0.
    But since this is a fraction, you just set the numerator equal to zero and solve for x:
    2 - x = 0.

    To find the vertical asymptotes set the denominator equal to zero:
    (3 + x)(x - 1) = 0.
    . By the way, you don't need to FOIL! Just set each factor equal to zero.

    To find the horizontal asymptotes, find the degrees of the polynomials in the numerator and denominator (I guess you will have to FOIL after all) and compare them.


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