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natali 33 [55]
3 years ago
6

" 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.

Mathematics
1 answer:
xxTIMURxx [149]3 years ago
8 0

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.

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