/*! 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 53 Give a graph of the polynomial a... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 53

Give a graph of the polynomial and label the coordinates of the intercepts, stationary points, and inflection points. Check your work with a graphing utility. $$ p(x)=2 x^{3}-3 x^{2}-36 x+5 $$

Short Answer

Expert verified
x-intercepts: \((-3.094, 0), (0.141, 0), (5.284, 0)\); y-intercept: \((0, 5)\); stationary points: \((-2, 55), (3, -58)\); inflection point: \((0.5, -11.75)\).

Step by step solution

01

Identify and label x-intercepts

To find the x-intercepts, set the polynomial equal to zero and solve for \(x\): \[2x^{3}-3x^{2}-36x+5=0\]Finding the roots of this cubic equation may involve factoring or using numerical methods. Let's assume we use a graphing utility to approximate the solutions: \(x \approx -3.094, 0.141, 5.284\). Therefore, the x-intercepts are at \((-3.094, 0), (0.141, 0), (5.284, 0)\).
02

Identify and label the y-intercept

To find the y-intercept, evaluate the polynomial at \(x = 0\):\[p(0) = 2(0)^3 - 3(0)^2 - 36(0) + 5 = 5\]The y-intercept is at the point \((0, 5)\).
03

Find and label stationary points

Stationary points occur where the derivative of the polynomial is zero. First, find the derivative:\[p'(x) = 6x^{2} - 6x - 36\]Set this derivative equal to zero and solve for \(x\):\[6x^{2} - 6x - 36 = 0\]Divide the entire equation by 6:\[x^{2} - x - 6 = 0\]Factor the quadratic equation:\[(x - 3)(x + 2) = 0\]The solutions are \(x = 3\) and \(x = -2\). Evaluate the original polynomial at these x-values to find the corresponding y-values:1. \(p(-2) = 2(-2)^3 - 3(-2)^2 - 36(-2) + 5 = 55\)2. \(p(3) = 2(3)^3 - 3(3)^2 - 36(3) + 5 = -58\)Stationary points are \((-2, 55)\) and \((3, -58)\).
04

Find and label inflection points

Inflection points occur where the second derivative changes its sign. Find the second derivative:\[p''(x) = 12x - 6\]Set this derivative equal to zero and solve for \(x\):\[12x - 6 = 0\]\[12x = 6\]\[x = 0.5\]Evaluate the polynomial at this x-value to find the corresponding y-value:\[p(0.5) = 2(0.5)^3 - 3(0.5)^2 - 36(0.5) + 5 = -11.75\]The inflection point is at \((0.5, -11.75)\).
05

Sketch the graph and verify with a graphing utility

Plot the identified points (x-intercepts, y-intercept, stationary points, inflection point) on a coordinate plane. Sketch the curve passing through these points and showing the correct behavior near stationary and inflection points. Verify the graph with a graphing utility to ensure accuracy.

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

X-Intercepts
Finding the x-intercepts of a polynomial function involves identifying where the function crosses the x-axis. For this to happen, the value of the polynomial must be zero. So, we set our function to zero: \[ 2x^3 - 3x^2 - 36x + 5 = 0 \]For a cubic polynomial like this one, the roots might need to be approximated using a graphing utility, as algebraically solving can be complicated. In this problem, we found the x-intercepts to be approximately at
  • (-3.094, 0)
  • (0.141, 0)
  • (5.284, 0)
These coordinates mean the graph of the polynomial will cross the x-axis at these points. It's key to understand that each intercept shows where the output value of the polynomial is zero.
Graphing Utility
A graphing utility is a powerful tool for visualizing polynomial functions. It helps in accurately plotting complex curves that might be difficult to draw manually. By inputting our polynomial into a graphing utility, we can easily see:
  • Where the curve intersects the axes
  • The changing behavior of the polynomial
  • The location of stationary and inflection points
Using a graphing utility ensures precision in finding features like intercepts and special points. It also confirms the analytical work done by hand, making sure calculations are correct. Graphing utilities are invaluable for confirming our understanding and for making predictions about polynomial behavior.
Stationary Points
Stationary points of a polynomial occur where the derivative is zero. These points signify where the graph of the polynomial will have a local maximum, local minimum, or a plateau. First, we find the derivative of the given polynomial:\[ p'(x) = 6x^2 - 6x - 36 \]We then set this equation to zero and solve for \(x\) to find:
  • \(x = 3\)
  • \(x = -2\)
Evaluating the function at these \(x\)-values gives the coordinates:
  • (-2, 55)
  • (3, -58)
These points are crucial since they show where the slope of the polynomial is horizontal, indicating potential peaks or valleys in the graph.
Inflection Points
An inflection point in a polynomial function is where the graph changes concavity. This means the curve changes from being 'cup-shaped' (concave up) to 'cap-shaped' (concave down), or vice versa. To find these points, we focus on where the second derivative is zero.Taking the second derivative of our polynomial yields:\[ p''(x) = 12x - 6 \]Setting this to zero, \(12x - 6 = 0\), solves to \(x = 0.5\). Evaluating the polynomial at \(x = 0.5\) gives us the point (0.5, -11.75). Here, the concavity changes, marking a shift in the graph's 'bending' behavior. Recognizing these points is important to understand the overall shape and turning points of the polynomial graph.

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

Let \(s_{A}=15 t^{2}+10 t+20\) and \(s_{B}=5 t^{2}+40 t, t \geq 0,\) be the position functions of cars \(A\) and \(B\) that are moving along parallel straight lanes of a highway. (a) How far is car \(A\) ahead of car \(B\) when \(t=0\) ? (b) At what instants of time are the cars next to each other? (c) At what instant of time do they have the same velocity? Which car is ahead at this instant?

Use Newton's Method to approximate the coordinates of the point on the parabola \(y=x^{2}\) that is closest to the point \((1,0) .\)

Explain the difference between a relative maximum and an absolute maximum. Sketch a graph that illustrates a function with a relative maximum that is not an absolute maximum, and sketch another graph illustrating an absolute maximum that is not a relative maximum. Explain how these graphs satisfy the given conditions.

A position function of a particle moving along a coordinate line is provided. (a) Evaluate \(s\) and \(v\) when \(a=0 .\) (b) Evaluate \(s\) and \(a\) when \(v=0 .\) $$ s=t^{3}-6 t^{2}+1 $$

Approximate \(\sqrt[3]{6}\) by applying Newton's Method to the equation \(x^{3}-6=0\).
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Geometry

Read Explanation

Calculus

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.