Find the Asymptotes f(x)=x/(x^2-16)

Problem

ƒ(x)=x/(x2−16)

Solution

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

x2−16=0

(x−4)*(x+4)=0

x=4,x=−4

2. Verify the numerator at these values to ensure they are not holes.

ƒ(4)⇒4/0

ƒ*(−4)⇒(−4)/0

Since the numerator is non-zero at these points, x=4 and x=−4 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 x2−16 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/(x2−16))=0

(lim_x→−∞)(x/(x2−16))=0

This results in a horizontal asymptote at y=0

5. Check for slant asymptotes by noting that the degree of the numerator must be exactly one higher than the degree of the denominator.
   Since 1<2 there are no slant asymptotes.

Final Answer

Vertical: *x=4,x=−4; Horizontal: *y=0

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