SOLUTION: How do you graph (x/(x^2+4))? I don't know if the denominator has any vertical asymptotes or removable discontinuities since x would be imaginary.

Question 1093955: How do you graph (x/(x^2+4))? I don't know if the denominator has any vertical asymptotes or removable discontinuities since x would be imaginary.
Answer by josgarithmetic(39623) (Show Source): You can put this solution on YOUR website!

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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 will be POSITIVE for all values of x.

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... )

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