i) The given function is
![f(x)=\frac{x^2+4x-4}{x^2-2x-8}](/tpl/images/0477/4460/df142.png)
We can rewrite in factored form to obtain;
![f(x)=\frac{x^2+4x-4}{x^2-2x-8}](/tpl/images/0477/4460/df142.png)
![f(x)=\frac{(x+2\sqrt{2}+2)(x-2\sqrt{2}+2)}{(x-4)(x+2)}](/tpl/images/0477/4460/e96d4.png)
The domain is
![(x-4)(x+2)\ne0](/tpl/images/0477/4460/d7506.png)
![(x-4)\ne0,(x+2)\ne0](/tpl/images/0477/4460/56171.png)
![x\ne4,x\ne-2](/tpl/images/0477/4460/ac4fc.png)
ii) To find the vertical asymptotes equate the denominator to zero.
![(x-4)(x+2)=0](/tpl/images/0477/4460/aebcc.png)
![(x-4)=0,(x+2)=0](/tpl/images/0477/4460/41728.png)
![x=4,x=-2](/tpl/images/0477/4460/a467f.png)
iii) To find the roots, equate the numerator to zero.
![(x+2\sqrt{2}+2)(x-2\sqrt{2}+2)=0}](/tpl/images/0477/4460/38589.png)
![(x+2\sqrt{2}+2)=0,(x-2\sqrt{2}+2)=0}](/tpl/images/0477/4460/fc545.png)
![(x=-2\sqrt{2}-2,x=2\sqrt{2}-2)}](/tpl/images/0477/4460/8538c.png)
iv) To find the y-intercept, substitute
into the equation.
![f(0)=\frac{0^2+4(0)-4}{0^2-2(0)-8}](/tpl/images/0477/4460/c6936.png)
We simplify to obtain;
![f(0)=\frac{-4}{-8}](/tpl/images/0477/4460/2e9ee.png)
![f(0)=\frac{1}{2}](/tpl/images/0477/4460/154e0.png)
v) The horizontal asymptote is
![lim_{\to \infty}\frac{x^2+4x-4}{x^2-2x-8}=1](/tpl/images/0477/4460/831e6.png)
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.
![domain: v.a: roots: y-int: h.a: holes: o.a: also, draw on the graph attached.](/tpl/images/0477/4460/a1090.jpg)