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onegirl
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.
is it \[\frac{x^2}{4-x^2}\]?
ok so for the vertical asymptotes, set the denominator equal to zero and solve for \(x\)
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)
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
the leading coefficient of \(x^2\) is 1 and the leading coefficient of \(4-x^2\) is \(-1\) therefore the horizontal asymptote is \(y=-1\)
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