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anonymous

  • 5 years ago

I am working with Vertical and Horizontal Asymptotes. My problem is f(x)=3x/x^2-x-2 I am sure there will be no HA due to the degree of the numerator being larger than the degree of the denominator. As for the VA, do you factor the equation out, or will that mess up the answer?

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  1. anonymous
    • 5 years ago
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    For the vertical asymptote, you set the denominator equal to 0 and solve for values of x. So x^2-x-2=0 factored: (x-2)(x+1) = 0 x = -1, 2 Vertical asymptotes at -1, 2

  2. anonymous
    • 5 years ago
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    Ahhhhh! okay!, thank you!

  3. anonymous
    • 5 years ago
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    You're welcome

  4. anonymous
    • 5 years ago
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    If there is no HA, that could mean as the domain approaches -infinity and +infinity, the range is acting likewise. A HA restricts the range. Do you have a graphing calculator? ti-83 etc?

  5. anonymous
    • 5 years ago
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    no, we dont use them in class. I know my normal graph would be two lines with a curve around the intersection of the asymptotes, but not crossing. But if there is no HA to indicate a limitation in the graph, would it just be two straight lines going to infinity near to the intersection point, but not crossing?

  6. anonymous
    • 5 years ago
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    You can test say, x=-100 and x =100 to see what the end behavior looks like. If it's a big positive, then it's probably approaching positive infinity. large negative -> negative infinity close to 0 -> approaching 0

  7. anonymous
    • 5 years ago
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    The only way I can think of to indicate a range of answers would be to shade in the graph on the side around the intersection of the asymptotes. I get the same answer for each, since there is a square, the answers would lie in a positive domain. OHHHHH! I get it. Thank you again!

  8. anonymous
    • 5 years ago
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    You're welcome

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