Question 910314
as far as i can tell, this is what's happening.


vertical asymptote will occur when x^2 + 4 = 0


solve for x^2 + 4 = 0 and you get x = plus or minus sqrt(-4) which is not a real number so there is no vertical asymptote.


since the degree of the numerator is less than the degree of the denominator, the asymptote will be at y = 0


y intercept is the value of y when x is equal to 0.


when x = 0, the value of y becomes -22/4 which is equal to -5.5


x intercept is the value of x when y is equal to 0.


set the equation equal to 0 and solve for x.


(x-22) / (x^2 + 4) = 0
multiply both sides of this equation by (x^2+4) to get x - 22 = 0
add 22 to both sides of this equation to get x = 22.


you should have:


no vertical asymptote.
horizontal asymptote at y = 0
y intercept at y = -5.5
x intercept at x = 22


fyi - the graph of the equation can cross the horizontal asymptote, but the graph of the equation will never cross the vertical asymptote.


the graph of your equation is shown below.
it confirms the solutions provided above.


<img src = "http://theo.x10hosting.com/2014/100701.jpg" alt="$$$" </>