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Math Help - x and y intercepts

  1. #1
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    x and y intercepts

    1) Given: F(x) =
    a. Find the x-intercept(s).
    b. Find the y-intercept(s).
    c. Find the vertical asymptote(s) (write as an equation of a line).
    d. Find the horizontal asymptote(s) (write as an equation of a line).
    e. For x>4, is there a value which F(x) cannot exceed, AND/OR a value which F(x) cannot fall below? Explain your answers.
    f. Find the maximum or minimum function value in the interval -4 x 4, and state whether it is a maximum or minimum value for that interval.
    g. Describe what happens on the graph when x approaches infinity.

    12.Given: F(x) =
    a. Find the x-intercept(s).
    b. Find the y-intercept(s).
    c. Find the vertical asymptote(s) (write as an equation of a line).
    d. Find the horizontal asymptote(s) (write as an equation of a line).
    e. For x<-3, is there a value which F(x) cannot exceed, AND/OR a value which F(x) cannot fall below? Explain your answers.
    f. Is the function increasing or decreasing in the interval -3 x 3?
    g. As x approaches 3 from the left, what happens to the function values?
    h. As x approaches 3 from the right, what happens to the function values?

    I am stumped big time. Help if you can. Thanks in advance.

    Kasey
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by flippin4u View Post
    1) Given: F(x) =
    a. Find the x-intercept(s).
    b. Find the y-intercept(s).
    c. Find the vertical asymptote(s) (write as an equation of a line).
    d. Find the horizontal asymptote(s) (write as an equation of a line).
    e. For x>4, is there a value which F(x) cannot exceed, AND/OR a value which F(x) cannot fall below? Explain your answers.
    f. Find the maximum or minimum function value in the interval -4 x 4, and state whether it is a maximum or minimum value for that interval.
    g. Describe what happens on the graph when x approaches infinity.
    f(x) = \frac{4x^2 - 1}{x^2 - 16} = \frac{(2x + 1)(2x - 1)}{(x + 4)(x - 4)}

    a) x intercepts at f(x) = 0:
    So
    f(x) = \frac{4x^2 - 1}{x^2 - 16} = \frac{(2x + 1)(2x - 1)}{(x + 4)(x - 4)} = 0

    Implies 2x + 1 = 0 or 2x - 1 = 0

    b)y intercept at x = 0. So what's f(0)?

    c) Vertical asymptotes are where the denominator is 0.

    d) Horizontal asymptotes are of the form y = c and x is very large. What is the function value when x is very large and when x is very large and negative? (Feel free to use a calculator if you can't see it without one.)

    e) Take a look at your answers to d).

    f) The easy way is to graph it and take a look.

    g) Again, see your answer to d).

    -Dan
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