Question 201553
Vertical asymptotes of a rational function, if any, will be for values of x that make the denominator zero.<br>
For your function we will try to find what values of x make {{{x^2+4}}} negative. But if one understands how squaring works we can see that this particular denominator can never be zero because:<ol><li>When you square zero you get zero. When you square any other Real number you get some positive number. So {{{x^2}}} is either zero or a positive number no matter what x is.</li><li>If you add {{{x^2}}}, which is zero or positive, to 4 you will always get a positive number, <b>never a zero</b>!</li></ol>Since the denominator can never be zero, there are no vertical asymptotes for this function.<br>
If you denominator had been {{{x^2-4}}} instead, we could not say that the denominator could never be zero because subtracting 4 from a zero or positive number <i>could</i> result in a zero. So we would have to solve the equation:
{{{x^2-4=0}}}
We could factor this into:
{{{(x+2)*(x-2)=0}}}
In order for a product to be zero one of the factors must ne zero so:
{{{x+2=0}}} or {{{x-2=0}}}
Solving each of these we get
{{{x = -2}}} or {{{x=2}}}
These would be the vertical asymptotes for a (simplified) rational function with {{{x^2-4}}} as the denominator (like {{{q(x)=5/(x^2-4)}}}).