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!!
