i) The given function is
[tex]f(x)=\frac{2x-1}{x^2-x-6}[/tex]
We factor to obtain
[tex]f(x)=\frac{2x-1}{(x-3)(x+2)}[/tex]
The domain is
[tex](x-3)(x+2)\ne0[/tex]
[tex](x-3)\ne0,(x+2)\ne0[/tex]
[tex]x\ne3,x\ne-2[/tex]
ii) The vertical asymptotes are
[tex](x-3)(x+2)=0[/tex]
[tex](x-3)=0,(x+2)=0[/tex]
[tex]x=3,x=-2[/tex]
iii) To find the root, we equate the numerator to zero.
[tex]2x-1=0[/tex]
[tex]x=\frac{1}{2}[/tex]
iv) To find the y-intercept, put x=0 into the function.
[tex]f(0)=\frac{2(0)-1}{(0)^2-(0)-6}[/tex]
[tex]f(0)=\frac{-1}{-6}[/tex]
[tex]f(0)=\frac{1}{6}[/tex]
vi) To find the horizontal asymptote, we take limit to infinity.
This implies that;
[tex]lim_{x\to \infty}\frac{2x-1}{x^2-x-6}=0[/tex]
The horizontal asymptote is y=0.
vii) The numerator and the denominator do not have common factors that are at least linear.
Therefore the function has no holes in it.
