anonymous
  • anonymous
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.
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
f(x) = x^2/4 - x^2
anonymous
  • anonymous
is it \[\frac{x^2}{4-x^2}\]?
anonymous
  • anonymous
yes

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anonymous
  • anonymous
ok so for the vertical asymptotes, set the denominator equal to zero and solve for \(x\)
anonymous
  • anonymous
ok
anonymous
  • anonymous
you get \[4-x^2=0\] or \[(2-x)(2+x)=0\] and the solutions are \(x=-2\) or \(x=2\) those are your vertical asymptotes (there are two of them)
anonymous
  • anonymous
ok
anonymous
  • anonymous
for 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
  • anonymous
okay
anonymous
  • anonymous
the leading coefficient of \(x^2\) is 1 and the leading coefficient of \(4-x^2\) is \(-1\) therefore the horizontal asymptote is \(y=-1\)
anonymous
  • anonymous
that 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
anonymous
  • anonymous
okay thx
anonymous
  • anonymous
yw

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