   Chapter 3.5, Problem 51E

Chapter
Section
Textbook Problem

# 49–54 Use the guidelines of this section to sketch the curve. In guideline D find an equation of the slant asymptote. y = x 3 + 4 x 2

To determine

To sketch:

The curve of y and use guidelines D to find an equation of the slant asymptote

Explanation

1) Concept:

i) A domain is the set of x values that satisfy the function.

ii) To find x-intercept, put y=0, and to find y-intercept, put x=0 in the given function.

iii) Symmetry: To find symmetry, replace x by –x and check the behaviour of function. Thus, if f-x=fx, then it is an even function, so it has y-axis symmetry. If f-x=-fx, then it is an odd function, so it has x-axis symmetry. And if f-x-fxfx, then it has no symmetry.

iv) An asymptote is a tangent at infinity. To find horizontal, vertical, and slant asymptote, follow the rules.

v) A function is increasing if f'x>0  and decreasing if f'x<0 in that particular interval.

vi) The number f(c) is a local maximum value of f  if fcf(x) when x is near c and is a local minimum value of f if fc f(x) when x is near c.

vii) If f''x>0, the function is concave up and if f''x<0, the function is concave down in that particular interval. And if f''x=0, give the values of inflection points.

2) Given:

fx=y=x3+4x2

3) Calculations:

Here, first find the domain of the given function and the x & y intercepts. Next, check the symmetry, asymptotes, intervals of increase and decrease, local maximum and minimum values, concavity, and points of inflection. Using these, sketch the curve

A. Domain

Domain is the point where the denominator is not zero. Therefore, the domain is (-,0)(0,)

B. Intercepts-

For y intercept, plug x=0  in the given function and solve it

y-intercept is nothing but value of y at x=0. But 0 is not in the domain, therefore there is no

y-intercept

For x intercept, plug y=0 in the original function and solve it

0=x3+4x2

Solve x

x3+4=0

x=-43

x Intercept are -43, 0

C. Symmetry

For symmetry, replace each x by -x therefore,

f-x= -x3+4-x2

f-x=-x3+4x2

That means there is no symmetry

D. Asymptote:

Horizontal asymptotes: the degree of numerator is greater than the degree of denominator, so there is no horizontal asymptote.

limn±x3+4x2=±

So, there is no horizontal asymptote.

Vertical asymptotes: the points where f is not defined.

Here function is not define at x=0 so Vertical asymptote is x=0.

To find slant asymptote-

Use long division method rule,

So,fx=x+4x2

By guidelines for slant asymptote, slant asymptote is y=x

E. Intervals of increase or decrease

To find the intervals of increase or decrease, find the derivative of the given function

f'x=1-8x3

Equating this derivative with 01-8x3=0

Solve for x

x=2

Combining the critical points with the domain, the function has three intervals,

- , 0, 0, 2, (2, )

Now, take a test point from each of the above intervals, and check whether the function is increasing or decreasing in that interval

### 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

#### Expand each expression in Exercises 122. (2x3)2

Finite Mathematics and Applied Calculus (MindTap Course List)

#### In Exercises 49-62, find the indicated limit, if it exists. 62. limx24x22x2+x3

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

#### Calculate y'. 31. y = x tan1(4x)

Single Variable Calculus: Early Transcendentals

#### For , f′(x) =

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