i) The given function is
[tex]f(x)=\frac{x^2+4x-4}{x^2-2x-8}[/tex]
We can rewrite in factored form to obtain;
[tex]f(x)=\frac{x^2+4x-4}{x^2-2x-8}[/tex]
[tex]f(x)=\frac{(x+2\sqrt{2}+2)(x-2\sqrt{2}+2)}{(x-4)(x+2)}[/tex]
The domain is
[tex](x-4)(x+2)\ne0[/tex]
[tex](x-4)\ne0,(x+2)\ne0[/tex]
[tex]x\ne4,x\ne-2[/tex]
ii) To find the vertical asymptotes equate the denominator to zero.
[tex](x-4)(x+2)=0[/tex]
[tex](x-4)=0,(x+2)=0[/tex]
[tex]x=4,x=-2[/tex]
iii) To find the roots, equate the numerator to zero.
[tex](x+2\sqrt{2}+2)(x-2\sqrt{2}+2)=0}[/tex]
[tex](x+2\sqrt{2}+2)=0,(x-2\sqrt{2}+2)=0}[/tex]
[tex](x=-2\sqrt{2}-2,x=2\sqrt{2}-2)}[/tex]
iv) To find the y-intercept, substitute [tex]x=0[/tex] into the equation.
[tex]f(0)=\frac{0^2+4(0)-4}{0^2-2(0)-8}[/tex]
We simplify to obtain;
[tex]f(0)=\frac{-4}{-8}[/tex]
[tex]f(0)=\frac{1}{2}[/tex]
v) The horizontal asymptote is
[tex]lim_{\to \infty}\frac{x^2+4x-4}{x^2-2x-8}=1[/tex]
The equation of the horizontal asymptote is y=1
vi) The function does not have a variable factor that is common to both the numerator and the denominator.
The function has no holes in it.
vii) The given function is a proper rational function.
Proper rational functions do not have oblique asymptotes.
