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A family of functions is given by r(x) = 1/[a-(x-b)^2] Assume a > 0, b > 0. Find the vertical asymptotes of r(x)
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set the denominator equal to zero and solve for \(x\)
clear or no?
that make sense!
Vertical asymptotes will appear when division by 0 occurs, so that is why you want to find where the denominator is 0.
I can't seem to solve for the denominator to equal 0..
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\[ a-(b-x)^2 = a - (b^2-2bx+x^2) = (-1)x^2 +(2b)x+(a-b^2) \]
it looks like i can somehow use the quadratic formula right..? the answer I am supposed to get is both positive and negative
Yes, the solution must be in terms of a and b.
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