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

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Answer 1 · 2017-11-22T12:37:21

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

none removable discontinuities ("holes")

$f_((x))={(3x)^2}/{x^2-x-6}+3=$

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

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

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

$= 3{4x^2-x-6}/{(x-3)(x+2)}$

$x_(u_(1,2))={3+-sqrt(9+4*12*18)}/24={3+-3sqrt(97)}/24={1+-sqrt(97)}/8$

$x_(u_(1,2))~~1.356 , -1.106 != 3,-2$

$=>$

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

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

$=>$

horizontal asymptote is $y=12$

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