QUESTION 1
i) The given function is
[tex]f(x)=\frac{3(x-1)(x+1)}{(x-3)(x+3)}[/tex]
The domain is
[tex](x-3)(x+3)\ne0[/tex]
[tex](x-3)\ne0,(x+3)\ne0[/tex]
[tex]x\ne3,x\ne-3[/tex]
ii) To find the vertical asymptote equate the denminator to zero.
[tex](x-3)(x+3)\=0[/tex]
[tex](x-3)=0,(x+3)=0[/tex]
[tex]x=3,x=-3[/tex]
iii) To find the roots equate the numerator zero.
[tex]3(x-1)(x+1)=0[/tex]
[tex]3(x-1)=0,(x+1)=0[/tex]
[tex](x-1)=0, (x+1)=0[/tex]
[tex]x=1, x=-1[/tex]
iv) To find the y-intercept substitute [tex]x=0[/tex] into the function;
[tex]f(0)=\frac{3(0-1)(0+1)}{(0-3)(0+3)}[/tex]
[tex]f(0)=\frac{-3}{(-3)(3)}[/tex]
[tex]f(0)=\frac{1}{3}[/tex]
The y-intercept is [tex]\frac{1}{3}[/tex]
v) The horizontal asymptote is given by
[tex]lim_{x\to \infty}\frac{3(x-1)(x+1)}{(x-3)(x+3)}=3[/tex]
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.
