Question 1137489

make a graph of 


{{{f(x)= (3x^2+3x-6)/(x^2+x-12)}}}

{{{f(x)= (3x^2-3x+6x-6)/(x^2-3x+4x-12)}}}

{{{f(x)= ((3x^2-3x)+(6x-6))/((x^2-3x)+(4x-12))}}}

{{{f(x)= (3x(x-1)+6(x-1))/(x(x-3)+4(x-3))}}}

{{{f(x)= ((3x+6)(x-1))/((x + 4) (x - 3))}}}


asymptotes:

Horizontal asymptote: 

{{{(3 x^2 + 3 x - 6)/(x^2 + x - 12)->3}}} as{{{ x}}}-> ±{{{infinity}}}

=>horizontal asymptote is {{{y=3}}}


Vertical asymptotes:

{{{(3 x^2 + 3 x - 6)/(x^2 + x - 12)}}}-> ± {{{infinity}}} as {{{x->-4}}}
{{{(3 x^2 + 3 x - 6)/(x^2 + x - 12)}}}-> ± {{{infinity}}} as{{{ x->3}}}

=> vertical asymptotes are {{{x=-4}}} and {{{x=3}}}


domain: 

{ {{{x}}} element {{{R}}} : {{{x<>-4}}} and {{{x<>3}}} }


the {{{x}}} intercepts:

{{{(3x+6)(x-1)=0}}}

if {{{(3x+6)=0}}}=>{{{3x=-6}}}=>{{{x=-2}}}
if {{{(x-1)=0}}}=>{{{x=1}}}

the {{{x}}} intercepts:({{{ 1}}},{{{0}}}) and ({{{-2}}},{{{0}}})


the {{{y}}} intercept: 

{{{f(0)= (3*0^2+3*0-6)/(0^2+0-12)=-6/-12=1/2}}}

 ({{{0}}}, {{{1/2}}})


{{{drawing ( 600, 600, -10,10, -10, 10,
blue(line(-4,10,-4,-10)),blue(line(3,10,3,-10)),
circle(0,1/2,.12),circle(-4,0,.12),circle(3,0,.12),
graph( 600, 600, -10,10, -10, 10, (3x^2+3x-6)/(x^2+x-12),3)) }}}