Question 928592
{{{3x/( x^2+5x+4)}}}



Horizontal asymtptote for unbounded to the left and the right, of y=0, because degree of denominator is one more than degree of numerator.



Factor and look at critical points values.
{{{(3x)/((x+1)(x+4))}}}
Critical values are -4, -1, 0.
Zero at x=0;
Vertical asymptotes at x=-4 and x=-1.



{{{x<-4}}}:
{{{3(neg)/((neg)(neg))}}}
{{{3*neg/(pos)}}}
{{{negative}}}.


{{{-4<x<-1}}}:
{{{3*neg/((neg)(pos))}}}
{{{positive}}}



{{{-1<x<0}}}:
{{{3*neg/((pos)(pos))}}}
{{{negative}}}


{{{0<x}}}:
{{{positive}}}.


Think about how the intervals should look, and then compare to the graph you would draw.......




{{{graph(300,300, -8,5,-10,10,3x/( x^2+5x+4))}}}