Question

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

Answer 1

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

Explanation:

`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]