We are asked to graph the function ` y=(x^2-9)/(2x^2+1) ` :

Note that the denominator is always positive so there are no vertical asymptotes. (The domain is all real numbers.)

The numerator factors as (x+3)(x-3), so the x-intercepts are at -3 and 3. For x<-3 the numerator is positive so...

## Unlock

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

We are asked to graph the function ` y=(x^2-9)/(2x^2+1) ` :

Note that the denominator is always positive so there are no vertical asymptotes. (The domain is all real numbers.)

The numerator factors as (x+3)(x-3), so the x-intercepts are at -3 and 3. For x<-3 the numerator is positive so the function is positive; for -3<x<3 the numerator is negative, and for x>3 the numerator is positive.

Since the degree of the numerator is the same as the degree of the denominator, there is a horizontal asymptote at y=1/2.

If you have calculus, the first derivative is `(38x)/((2x^2+1)^2) ` . The function is decreasing on x<0 and increasing on x>0 and has a global minimum at (0,-9).

The graph: