Question

" title="f(x) = \frac{3(x-1)(x+1)}{(x-3)(x+3)}" alt="f(x) = \frac{3(x-1)(x+1)}{(x-3)(x+3)}" align="absmiddle" class="latex-formula">
Domain:
V.A:
Roots:
Y-int:
H.A:
Holes:
O.A:

Also, draw on the graph attached.

Answer 1

QUESTION 1

i) The given function is

$f(x)=\frac{3(x-1)(x+1)}{(x-3)(x+3)}$

The domain is

$(x-3)(x+3)\ne0$

$(x-3)\ne0,(x+3)\ne0$

$x\ne3,x\ne-3$

ii) To find the vertical asymptote equate the denminator to zero.

$(x-3)(x+3)\=0$

$(x-3)=0,(x+3)=0$

$x=3,x=-3$

iii) To find the roots equate the numerator zero.

$3(x-1)(x+1)=0$

$3(x-1)=0,(x+1)=0$

$(x-1)=0, (x+1)=0$

$x=1, x=-1$

iv) To find the y-intercept substitute $x=0$ into the function;

$f(0)=\frac{3(0-1)(0+1)}{(0-3)(0+3)}$

$f(0)=\frac{-3}{(-3)(3)}$

$f(0)=\frac{1}{3}$

The y-intercept is $\frac{1}{3}$

v) The horizontal asymptote is given by

$lim_{x\to \infty}\frac{3(x-1)(x+1)}{(x-3)(x+3)}=3$

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.