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Math Help - asymptotes

  1. #1
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    asymptotes

    find the vertical asymptote for

    a) f(x)=\frac{1}{x^2-25}

    domain for

    a) f(x)=\frac{2x+3}{3x-4}

    range for

    a) f(x)=\frac{1}{x^2-9}

    horizontal asymptote for

    a) f(x)=\frac{4x-7}{x+8}
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by william View Post
    find the vertical asymptote for

    a) f(x)=\frac{1}{x^2-25}
    Hi william,

    For which values of x is f(x) undefined? Those will be your vertical asymptotes.

    Quote Originally Posted by william View Post
    domain for

    a) f(x)=\frac{2x+3}{3x-4}
    The denominator cannot = zero. What value would cause the denominator to = zero? Everything except that value would constitute the domain.
    Quote Originally Posted by william View Post

    horizontal asymptote for

    a) f(x)=\frac{4x-7}{x+8}

    The location of the horizontal asymptote is determined by looking at the degrees of the numerator (n) and denominator (m). If the degrees are the same, the horizontal asymptote is the ratio of the leading coefficients.

    y=\frac{4}{1}=4
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  3. #3
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    Another way of thinking about the horizontal asymptote. For very very large x (either positive or negative), each power of x is far larger than lower powers (example if x= 1000000, x^2 is 1000000 times as large and, x^3 is 1000000 times as large as that, etc. That is, for very large x, you can ignore all but the highest powers. For very large x, \frac{4x- 7}{x+ 8} if very very close to \frac{4x}{x}= 4.
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  4. #4
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    Quote Originally Posted by masters View Post
    Hi william,

    For which values of x is f(x) undefined? Those will be your vertical asymptotes.



    The denominator cannot = zero. What value would cause the denominator to = zero? Everything except that value would constitute the domain.
    ok thanks. for a) i get \frac{1}{x^2-25}=\frac{1}{(x-5)(x+5)}= x=5 and x=-5 (vertical asymptotes)

    i still don't get b and c
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  5. #5
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    Quote Originally Posted by william View Post
    ok thanks. for a) i get \frac{1}{x^2-25}=\frac{1}{(x-5)(x+5)}= x=5 and x=-5 (vertical asymptotes)

    i still don't get b and c
    What do you mean, b and c? All of your problems are labeled "a"! Do you mean the domain and range questions?

    For the domain, there are three ways to restrict it. Masters already mentioned one of them:
    The denominator cannot = zero. What value would cause the denominator to = zero? Everything except that value would constitute the domain.
    Now tell us the domain of f(x)=\frac{2x+3}{3x-4} is.


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