QUESTION 1
i) The given function is
![f(x)=\frac{3(x-1)(x+1)}{(x-3)(x+3)}](/tpl/images/0476/7415/4a4c1.png)
The domain is
![(x-3)(x+3)\ne0](/tpl/images/0476/7415/10bcf.png)
![(x-3)\ne0,(x+3)\ne0](/tpl/images/0476/7415/e2511.png)
![x\ne3,x\ne-3](/tpl/images/0476/7415/fe142.png)
ii) To find the vertical asymptote equate the denminator to zero.
![(x-3)(x+3)\=0](/tpl/images/0476/7415/b9a3c.png)
![(x-3)=0,(x+3)=0](/tpl/images/0476/7415/ca904.png)
![x=3,x=-3](/tpl/images/0476/7415/74613.png)
iii) To find the roots equate the numerator zero.
![3(x-1)(x+1)=0](/tpl/images/0476/7415/3cc1a.png)
![3(x-1)=0,(x+1)=0](/tpl/images/0476/7415/74c4f.png)
![(x-1)=0, (x+1)=0](/tpl/images/0476/7415/8f493.png)
![x=1, x=-1](/tpl/images/0476/7415/3da3a.png)
iv) To find the y-intercept substitute
into the function;
![f(0)=\frac{3(0-1)(0+1)}{(0-3)(0+3)}](/tpl/images/0476/7415/72bec.png)
![f(0)=\frac{-3}{(-3)(3)}](/tpl/images/0476/7415/fb065.png)
![f(0)=\frac{1}{3}](/tpl/images/0476/7415/8499f.png)
The y-intercept is ![\frac{1}{3}](/tpl/images/0476/7415/5506e.png)
v) The horizontal asymptote is given by
![lim_{x\to \infty}\frac{3(x-1)(x+1)}{(x-3)(x+3)}=3](/tpl/images/0476/7415/a0cab.png)
The horizontal asymptote is y=3
vi) The rational function has no common linear factor.
This rational function has no holes.
vii) This rational function is a proper function. It has no oblique asymptote.
![domain: v.a: roots: y-int: h.a: holes: o.a: also, draw on the graph attached.](/tpl/images/0476/7415/be33b.jpg)