What are the asymptotes and removable discontinuities, if any, of f(x)= (3x)^2/(x^2-x-6)+3 f(x)=(3x)2x2−x−6+3?

1 Answer
Tamir E.
Nov 22, 2017

vertical asymptotes are x=3x=3 and x=-2x=−2
horizontal asymptote is y=12y=12

none removable discontinuities ("holes")

Explanation:

f_((x))={(3x)^2}/{x^2-x-6}+3=f(x)=(3x)2x2−x−6+3=

= {9x^2}/{(x-3)(x+2)}+3==9x2(x−3)(x+2)+3=

= {9x^2+3x^2-3x-18}/{(x-3)(x+2)}==9x2+3x2−3x−18(x−3)(x+2)=

= {12x^2-3x-18}/{(x-3)(x+2)}==12x2−3x−18(x−3)(x+2)=

= 3{4x^2-x-6}/{(x-3)(x+2)}=34x2−x−6(x−3)(x+2)

x_(u_(1,2))={3+-sqrt(9+4*12*18)}/24={3+-3sqrt(97)}/24={1+-sqrt(97)}/8xu1,2=3±√9+4⋅12⋅1824=3±3√9724=1±√978

x_(u_(1,2))~~1.356 , -1.106 != 3,-2xu1,2≈1.356,−1.106≠3,−2

=>⇒

vertical asymptotes are x=3x=3 and x=-2x=−2

lim_{x rarr +-oo}f_((x))=12

=>

horizontal asymptote is y=12

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