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

Question

1032 views around the world

You can reuse this answer
Creative Commons License

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]