Question 1153003
 Let {{{f(x)= (3x^2+4x+1)/(3x^2+11x+6)}}}............factor

{{{f(x)= (3x^2+3x+x+1)/(3x^2+9x+3x+6) }}}  

{{{f(x)= ((3x^2+3x)+(x+1))/((3x^2+9x)+(2x+6))}}}

{{{f(x)= (3x(x+1)+(x+1))/(3x(x+3)+2(x+3))}}}

{{{f(x)= ((3x+1)(x+1))/((3x+2)(x+3))}}}



This function has:

1) A y intercept at the point: set {{{x=0}}}

{{{f(x)= ((3*0+1)(0+1))/((3*0+2)(0+3))}}}

{{{f(x)= (1*1)/(2*3)}}}

{{{f(x)= 1/6}}}
y intercept at the point: 

({{{0}}},{{{1/6}}}) ->your ordered pair



2) x intercepts at the point(s) : set {{{f(x)=0}}}

{{{ ((3x+1)(x+1))/((3x+2)(x+3))=0}}}

{{{ ((3x+1)(x+1))=0*((3x+2)(x+3))}}}
{{{ (3x+1)(x+1)=0}}}

solutions:

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

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

x intercepts at the point(s) : 

({{{-1/3}}},{{{0}}}),({{{-1}}},{{{0}}})

3) Vertical asymptotes at x => look for values that make denominator equal to zero

{{{f(x)= ((3x+1)(x+1))/((3x+2)(x+3))}}}
=> {{{(3x+2)(x+3)=0}}}, 

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

solutions: {{{x=-2/3}}}, {{{x=-3}}}

<a href="https://www.imageupload.net/image/AU8fg"><img src="https://imagehost.imageupload.net/2020/02/25/MSP11.th.gif" alt="MSP11.gif" border="0" /></a>



4) Horizontal asymptote at y =

limit of {{{(3x^2 + 4x + 1)/(3x^2 + 11 x + 6)->1}}} as {{{x}}}-> ± {{{infinity}}}


<a href="https://www.imageupload.net/image/AU8fg"><img src="https://imagehost.imageupload.net/2020/02/25/MSP11.th.gif" alt="MSP11.gif" border="0" /></a>


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