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Posted by on Tuesday, August 4, 2009 at 3:28pm.

Let f(x) = (x)/(x^2-4)


(a) State the y-intercept.

(b) State the vertical asymptote(s).

(c) State the horizontal asymptote.

  • Algebra - , Tuesday, August 4, 2009 at 4:53pm

    First you would factorize the denominator to visualize what's going on.
    f(x) = x/(x²-4) = x/((x+2)(x-2)
    A vertical asymptote is a value of x whereby as the graph nears this value, the y-value approaches positive or negative infinity.
    At what value(s) does the given graph approach infinity?
    A horizontal asymptote is the y-value approaches as x approaches ± ∞.
    What value does f(x)=x/(x²-4) take when x → ± ∞?
    You will need the l'Hôpital's rule since both the numerator and denominator become infinity when evaluating f(∞).
    Using l'Hôpital's rule, we calculate the derivative at both the numerator and denominator to evaluate the resulting expression, thus
    Lim x→&infin 1/(2x) = 0
    So the horizontal asymptotes are at y=0.

    You can view a sketch of the graph at:
    http://i263.photobucket.com/albums/ii157/mathmate/Jessica-1.png

    This kind of problem takes a lot of practice. I hope you will work out a few more similar problems to give yourself the facility to tackle the graph sketching questions.

  • Algebra - , Tuesday, August 4, 2009 at 5:13pm

    Yes I will need the practice without a doubt! Now you say the

    The horizontal asymptote is y = 0

    What about the

    the y-intercept and the

    vertical asymptote(s)?

  • Algebra - , Tuesday, August 4, 2009 at 6:13pm

    The y-intercept to a function f(x) is the constant term, i.e. the term that does not contain any variables.
    For example, in
    f(x) = x²+4x-3
    the y-intercept is -3.
    In your particular case, there is no constant term, so the y-intercept is zero, as illustrated in the figure.
    Inspect the denominator of the function:
    f(x) = x/(x²-4) = x/((x+2)(x-2))
    and figure out what values of x would make the denominator zero. These are the values where the vertical asymptotes are located. You can confirm you answer from the figure, or you can post again for confirmation.

  • Algebra - , Tuesday, August 4, 2009 at 7:46pm

    I got 0 for the y-intercept why -3??? and

    The vertical asymptotes are x = -2,
    x = 2 ????

  • Algebra - , Tuesday, August 4, 2009 at 8:20pm

    You have correctly identified the vertical asymptotes, namely x=2 and x=-2.

    You may have taken the -3 out of context, which was:

    For example, in
    f(x) = x²+4x-3
    the y-intercept is -3.

  • Algebra - , Tuesday, August 4, 2009 at 8:21pm

    0 for the y-intercept is also correct, since there is no constant term in the function expression.

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