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Find an equation of a function with a vertical asymptote at x=2 and a hole at (-3,4).
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Answered by
oobleck
vertical asymptote at x=2
y = a/(x-2)
a hole at x = -3
y = a(x+3) / (x-2)(x+3)
to fill in the hole at (-3,4) we need
a/(-3-2) = 4
a = -20
so, we wind up with
y = -20(x+3) / (x+3)(x-2) = -20(x+3)/(x^2+x-6)
y = a/(x-2)
a hole at x = -3
y = a(x+3) / (x-2)(x+3)
to fill in the hole at (-3,4) we need
a/(-3-2) = 4
a = -20
so, we wind up with
y = -20(x+3) / (x+3)(x-2) = -20(x+3)/(x^2+x-6)
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