Q:

[tex]f(x) = \frac{3(x-1)(x+1)}{(x-3)(x+3)}[/tex]Domain:V.A:Roots:Y-int:H.A:Holes:O.A:Also, draw on the graph attached.

Accepted Solution

A:
QUESTION 1i) 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=3vi) 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.