Find the Asymptotes f(x)=x/(x^2-9)

Problem

ƒ(x)=x/(x^2−9)

Solution

1. Identify vertical asymptotes by finding the values of x that make the denominator zero while the numerator is non-zero.

x^2−9=0

(x−3)*(x+3)=0

x=3,x=−3

2. Verify vertical asymptotes by checking if the numerator x is non-zero at these points.

ƒ(3)⇒3/0

ƒ*(−3)⇒(−3)/0

Since both result in a non-zero constant divided by zero, x=3 and x=−3 are vertical asymptotes.

3. Identify horizontal asymptotes by comparing the degrees of the numerator and the denominator.
   The degree of the numerator x is 1
   The degree of the denominator x^2−9 is 2

4. Apply the rule for horizontal asymptotes where the degree of the denominator is greater than the degree of the numerator.

(lim_x→∞)(x/(x^2−9))=0

(lim_x→−∞)(x/(x^2−9))=0

This indicates a horizontal asymptote at y=0

5. Check for oblique asymptotes by noting that they only occur when the degree of the numerator is exactly one higher than the degree of the denominator.
   Since 1<2 there are no oblique asymptotes.

Final Answer

Vertical: *x=3,x=−3; Horizontal: *y=0

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