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Find a rational function with:
a hole at x=4,
a vertical asymptote of x=2,
a slant asymptote of y=2x+1

 Dec 5, 2018
 #1
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\(\text{the hole is created by }\dfrac{x-4}{x-4}\\ \text{the vertical asymptote is created by }\dfrac{1}{x-2}\\ \text{as we end up with a single power of }x \text{ in the denominator }\\ \text{the horizontal asymptote is created by }2x^2+x \\ \text{combining all this we have}\\ f(x) = \dfrac{(x-4)(2x^2+x)}{(x-4)(x-2)} = \dfrac{x(x-4)(2x+1)}{(x-4)(x-2)}\)

 

\(f(x) = \dfrac{2x^3-7x^2-4x}{x^2-6x+8}\)

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 Dec 5, 2018

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