   Chapter 10.5, Problem 16E ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042

#### Solutions

Chapter
Section ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042
Textbook Problem

# For each function in Problems 11-18, find any horizontal and vertical asymptotes, and use information from the first derivative to sketch the graph. y = ( x + 2 x − 3 ) 2

To determine

To calculate: The horizontal and vertical for the provided function y=(x+2x3)2 and sketch its graph.

Explanation

Given Information:

The provided function is y=(x+2x3)2.

Formula used:

A vertical asymptote of a rational function h(x)=f(x)g(x) is x=a where g(a)=0 and f(a)0.

A horizontal asymptote of a rational function h(x)=f(x)g(x) is,

Step 1. A line y=0 if the degree of the numerator is less than the degree of the denominator.

Step 2. The line y=anbn ratio of the leading coefficients if the degree of the numerator is equal to the degree of the denominator.

Step 3. Does not exist the degree of the numerator is greater than the degree of the denominator.

To find relative maxima and minima of a function,

1. Set the first derivative of the function to zero f'(x)=0, to find the critical values of the function.

2. Substitute the critical values into f(x) and calculate the critical points.

3. Calculate the sign of the function to the left and right of the critical values.

If f(x)<0 and f(x)>0 on the left and right side respectively of a critical value, then the point is relative minimum.

If f(x)>0 and f(x)<0 on the left and right side respectively of a critical value, then the point is relative maximum.

Calculation:

Consider the provided function,

y=(x+2x3)2

Recall that a vertical asymptote of a rational function h(x)=f(x)g(x) is x=a where, g(a)=0 and f(a)0.

Set the denominator x of the function equal to zero.

(x3)3=0x3=0x3+3=0+3x=3

Substitute 3 for x in the numerator.

x+2=3+2=50

Thus, the vertical asymptote is the line x=3.

Degree of the numerator is 2 and denominator is 2 of the function y=(x+2x3)2.

Since, the degree of the numerator is same as the denominator.

Thus, the horizontal asymptote is ratio of the leading coefficients.

y=11=1

Now, calculate the first derivative of the function y=(x+2x3)2.

y'=2(x+2x3)×1×(x3)1×(x+2)(x3)2=2(x+2x3)×x3x2(x3)2=2(x+2x3)×5(x3)2=10(x+2)(x3)3

Now, to obtain the critical values, set y'=0 as,

10(x+2)(x3)3=010(x+2)=0

Thus, x+2=0

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

#### Solve the equations in Exercises 126. (x3+1)x+1(x3+1)2x+1=0

Finite Mathematics and Applied Calculus (MindTap Course List)

#### In Exercises 107-120, factor each expression completely. 112. x2 2x 15

Applied Calculus for the Managerial, Life, and Social Sciences: A Brief Approach

#### True or False: converges conditionally.

Study Guide for Stewart's Multivariable Calculus, 8th

#### True or False: f(a)=lima0f(a+h)f(a)h.

Study Guide for Stewart's Single Variable Calculus: Early Transcendentals, 8th 