Chapter 4.6, Problem 22E

### Calculus: Early Transcendental Fun...

7th Edition
Ron Larson + 1 other
ISBN: 9781337552516

Chapter
Section

### Calculus: Early Transcendental Fun...

7th Edition
Ron Larson + 1 other
ISBN: 9781337552516
Textbook Problem

# Analyzing the Graph of a Function In Exercises 5-34, analyze and sketch a graph of the function. Label any intercepts, relative extrema, points of inflection, and asymptotes. Use a graphing utility to verify your results. y = x x 2 − 4

To determine

To graph: The function y=xx24.

Explanation

Given: The function y=xx2âˆ’4.

Graph:

Consider the provided function.

The function being a rational function has a domain as all the real numbers less that âˆ’2 and greater than 2. Thus the domain of the function would be (âˆ’âˆž,âˆ’2)âˆª(2,âˆž).

The range of the function would be (âˆ’âˆž,1)âˆª(1,âˆž).

Now find the x and y intercepts by equating f(x) and x to zero respectively to obtain:

There would be no xâ€“-intercepts or y-intercepts for the provided function.

The function has two vertical asymptotes x=âˆ’2,x=2 as the denominator is not defined at those points.

Also,

limxâ†’âˆžxx2âˆ’4=limxâ†’âˆž11âˆ’4x2=11âˆ’limxâ†’âˆž4x2=11âˆ’0=1

This implies that the function has a horizontal asymptote as y=1.

Similarly,

limxâ†’âˆ’âˆžxx2âˆ’4=limxâ†’âˆ’âˆžâˆ’11âˆ’4x2=âˆ’11âˆ’limxâ†’âˆž4x2=âˆ’11âˆ’0=âˆ’1

This happens because when xâ†’âˆ’âˆž, x2â†’âˆ’x.

This implies that the function has a horizontal asymptote as y=âˆ’1.

Now, differentiate the function with respect to x and equate it to zero to obtain the critical points.

f(x)=âˆ’4(x2âˆ’4)32<0

The first derivative will never be zero but with the denominator being positive; the derivative would always be negative

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