Vertical/Horizontal Asymptotes

[tachu101]

Homework Statement

Find the Vertical and horizontal asymptotes of this function

f(x)=(4x^2+25)/(x^2+9)

The Attempt at a Solution

I think that the horizontal asym. is y=4
But the Vertical if (x^2+9)=0 then x is an imaginary number. So is the asym. (3i) or does the asym not exist?

 

[Dick]

You are right on horizontal. For the second one, no, does not exist. Vertical asymptotes apply to real functions. For complex, the corresponding concept is called a 'pole'.

 

[dynamicsolo]

But the Vertical if (x^2+9)=0 then x is an imaginary number.

For (real-valued) rational functions of polynomials, you will have a vertical asymptote where the denominator is zero, unless the numerator is also zero there. (If that happens, you have to look for cancelling terms or look at the limit of the ratio as x approaches the value where both "top" and "bottom" are zero.)

In the case of your function, the denominator can never be zero (for x real), so the rational function is always defined. So it will have no vertical asymptotes.