/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 85 Explain the relationship between... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 85

Explain the relationship between the degree of a polynomial function and the number of turning points on its graph.

Short Answer

Expert verified
The relationship is that a polynomial's degree, if \(n\), can have at most \(n-1\) turning points in its graph. For example, if the degree is 2, at most one turning point can exist.

Step by step solution

01

Understanding Degree of Polynomial

The degree of a polynomial is the highest power of the variable in the polynomial. For instance, if a polynomial is 3x^4 - 2x^3 + x^2 - 1, then its degree is 4 because the highest power of x is 4.
02

Understanding Turning Points

A turning point in a graph is a point where the graph changes from increasing to decreasing (rising to falling) or from decreasing to increasing (falling to rising). For instance, in the graph of a quadratic function (which is a polynomial of degree 2), the vertex is a turning point because the graph changes from increasing on one side of the vertex to decreasing on the other.
03

Relating Degree and Turning Points

The relationship between the degree of a polynomial and the number of turning points is: if a polynomial is of degree \(n\), then it will have at most \(n-1\) turning points. This means a polynomial of degree 2 (a quadratic function) can have at most 1 turning point, a polynomial of degree 3 can have at most 2 turning points, and so on.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Write the equation of a rational function \(f(x)=\frac{p(x)}{q(x)}\) having the indicated properties, in which the degrees of p and q are as small as possible. More than one correct function may be possible. Graph your function using a graphing utility to verify that it has the required properties. \(f\) has a vertical asymptote given by \(x=1,\) a slant asymptote whose equation is \(y=x, y\) -intercept at \(2,\) and \(x\)-intercepts at -1 and 2.

Use a graphing utility to graph \(y=\frac{1}{x}, y=\frac{1}{x^{3}},\) and \(\frac{1}{x^{5}}\) in the same viewing rectangle. For odd values of \(n,\) how does changing \(n\) affect the graph of \(y=\frac{1}{x^{n}} ?\)

Determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. The graph of a rational function can have three vertical asymptotes.

Use a graphing utility to graph \(y=\frac{1}{x^{2}}, y=\frac{1}{x^{4}},\) and \(y=\frac{1}{x^{6}}\) in the same viewing rectangle. For even values of \(n,\) how does changing \(n\) affect the graph of \(y=\frac{1}{x^{n}} ?\)

Determine whether each statement makes sense or does not make sense, and explain your reasoning. When solving \(f(x)>0,\) where \(f\) is a polynomial function, I only pay attention to the sign of \(f\) at each test value and not the actual function value.
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Logic and Functions

Read Explanation

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.