What are the asymptote(s) and hole(s), if any, of $f(x)= (x(x-3))/(x^2-9x+3)$ ?

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Answer 1 · 2018-08-10T12:32:02

No hole, Vertical asymptotes : $x~~ 0.35 and x~~ 8.65$ ,
Horizontal asymptote: $y=1$

$f(x)= (x(x-3))/(x^2-9 x+3)$ , since no factor in numerator and

denominator is being cancelled , so there is no hole.

Vertical asymptote occur when denominator is zero.

$:. x^2-9 x+3=0 ; a=1 ,b= -9 , c= 3 ; [ax^2+bx+c]$

Discriminant $D= b^2-4 a c = 81- 12 =69$

Quadratic formula: $x= (-b+-sqrt D)/(2 a)$ or

$x= (9+-sqrt 69)/2 or x = 4.5+- sqrt 69/2$

$x~~ 8.65 , x ~~ 0.35$ , therefore, vertical asymptotes

are $x= 0.35 and x= 8.65$

Horizontal asymptote: $y= (x^2-3x)/(x^2-9x+3)$ or

$y= (1-3/x)/(1-9/x+3/x^2)$

$lim(x-> oo) , y= (1-0)/(1-0+0)=1$

$:. lim(x->+-oo) ,y->1$ , hence, horizontal asymptote is

at $y=1$

graph{(x(x-3))/(x^2-9 x+3) [-40, 40, -20, 20]}[Ans]

What are the asymptote(s) and hole(s), if any, of f(x)= (x(x-3))/(x^2-9x+3)f(x)= (x(x-3))/(x^2-9x+3)?