Find a rational function with:

a hole at x=4,

a vertical asymptote of x=2,

a slant asymptote of y=2x+1

Guest Dec 5, 2018

#1**+1 **

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

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

.Rom Dec 5, 2018