Question 844729
For {{{x=-4}}} , {{{y=(x+2)(x+4)/((x+4)(x+1))}}} does not exist, because the denominator is zero.
However, for any {{{x<>-4}}} ,
{{{y=(x+2)(x+4)/((x+4)(x+1))=(x+2)/(x+1)}}} which is continuous at {{{x=-4}}} .
So, at {{{x=-4}}} we have a hole in the graph.
The function {{{y=(x+2)/(x+1)}}} is what results from "plugging" that hole.
For {{{x=-1}}} , {{{y=(x+2)(x+4)/((x+4)(x+1))}}} does not exist, and there is no equivalent continuous function. (No plugging possible).
At {{{x=-1}}}, the functions {{{y=(x+2)(x+4)/((x+4)(x+1))}}} and {{{y=(x+2)/(x+1)}}} have a vertical asymptote.
The function changes sign at that point.
{{{y=(x+2)/(x+1)}}} is positive for {{{x>-1}}} , where {{{x+1>0}}} and {{{x+2>0}}} .
It is negative for {{{-2<x<-1}}} , where {{{x+1<0}}} and {{{x+2>0}}} .
{{{graph(300,300,-6,4,-10,10,(x+2)/(x+1),200(x+1))}}}