Question 1093955
{{{x/(x^2+4)}}}

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The numerator gives ONE ROOT at x=0.  


The denominator has no factor of x, so 0 will not be an empty point of the graph.  Not a discontinuity.


The denominator can accept all real values for x, and {{{x^2+4}}} will be POSITIVE for all values of x.


{{{x/(x^2+4)}}} is defined for all x values.  NO vertical asymptote.  


The sign of the function changes around x=0, so this is the only critical x value.  


(note how the function will get increasingly nearer to 0 as x goes unbound to the left or to the right... )


{{{graph(400,400,-4,4,-4,4,x/(x^2+4))}}}


{{{graph(400,400,-10,10,-2,2,x/(x^2+4))}}}