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.

Algebra ->  Graphs -> 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.       Log On


   

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(39621) About Me  (Show Source):
You can put this solution on YOUR website!
x%2F%28x%5E2%2B4%29
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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%5E2%2B4 will be POSITIVE for all values of x.

x%2F%28x%5E2%2B4%29 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%28400%2C400%2C-4%2C4%2C-4%2C4%2Cx%2F%28x%5E2%2B4%29%29

graph%28400%2C400%2C-10%2C10%2C-2%2C2%2Cx%2F%28x%5E2%2B4%29%29