Write a rational function f(x) such that f has vertical asymptotes at x = 3 and x = -1, no horizontal asymptote, and end behavior that can be modeled by y = 2x.

Respuesta :

Answer:

[tex]f(x)=\frac{2x^3-4x^2-6x}{x^2-2x-3}[/tex]

Step-by-step explanation:

Roots of a denominator in a rational function gives to us the vertical asymptotes. Hence we can take the denominator as

[tex](x-3)(x+1)=x^2-2x-3[/tex]

if we want that the end behavior as y=2x we can choose a polynomial whose factors cancel out with the denominator. Thus

[tex]2x(x-3)(x+1)=2x^3-4x^2-6x[/tex]

Hence, the function is

[tex]f(x)=\frac{2x^3-4x^2-6x}{x^2-2x-3}[/tex]

Hope this helps!!