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anonymous
 3 years ago
Determine all horizontal, slant, and vertical asymptotes. For each vertical asymptote, determine whether f(x) > infinity sign of f(x) > negative infinity sign on either side of the asymptote.
anonymous
 3 years ago
Determine all horizontal, slant, and vertical asymptotes. For each vertical asymptote, determine whether f(x) > infinity sign of f(x) > negative infinity sign on either side of the asymptote.

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0is it \[\frac{x^2}{4x^2}\]?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok so for the vertical asymptotes, set the denominator equal to zero and solve for \(x\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you get \[4x^2=0\] or \[(2x)(2+x)=0\] and the solutions are \(x=2\) or \(x=2\) those are your vertical asymptotes (there are two of them)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0for the horizontal asymptote, note that the numerator and denominator have the same degree (both are degree 2) so it is the ratio of the leading coefficients

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the leading coefficient of \(x^2\) is 1 and the leading coefficient of \(4x^2\) is \(1\) therefore the horizontal asymptote is \(y=1\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that is in the case where the degrees are the same. there is no slant asymptote for there to be a slant asymptote , the degree of the numerator would have to be one more than the degree of the denominator
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