/*! 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 69 Make a table of values, and sket... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 69

Make a table of values, and sketch the graph of the equation. Find the x- and y-intercepts, and test for symmetry. (a) \(y=\sqrt{4-x^{2}}\) (b) \(x=y^{3}+2 y\)

Short Answer

Expert verified
Equation (a) is a semi-circle with x-intercepts at -2, 2 and y-intercept at 2; it's symmetric about the y-axis. Equation (b) has y-intercept 0 without symmetry or x-intercepts.

Step by step solution

01

Understanding Equation (a)

The equation is given as \(y = \sqrt{4 - x^2}\). This is a semi-circle function as the square root restricts the graph to be non-negative, implying it represents the upper half of a circle centered at the origin \((0,0)\) with a radius of 2. First, we'll identify the domain of \(x\), which must satisfy \(4 - x^2 \geq 0\), leading to the domain \([-2, 2]\).
02

Making a Table of Values for Equation (a)

Let's choose values of \(x\) within the domain \([-2, 2]\) and compute corresponding \(y\) values:- For \(x = -2\), \(y = \sqrt{4 - (-2)^2} = 0\).- For \(x = -1\), \(y = \sqrt{4 - (-1)^2} = \sqrt{3}\).- For \(x = 0\), \(y = \sqrt{4 - 0^2} = 2\).- For \(x = 1\), \(y = \sqrt{4 - 1^2} = \sqrt{3}\).- For \(x = 2\), \(y = \sqrt{4 - 2^2} = 0\).
03

Finding Intercepts for Equation (a)

The y-intercept is at \(x = 0\), giving \(y = 2\). The x-intercepts occur when \(y = 0\), so solving \(\sqrt{4 - x^2} = 0\), gives \(x = \pm 2\). Thus, the x-intercepts are \((2, 0)\) and \((-2, 0)\).
04

Checking Symmetry for Equation (a)

The equation \(y = \sqrt{4-x^2}\) is symmetrical about the y-axis because substituting \(-x\) for \(x\) results in the same equation. Hence, it exhibits even symmetry.
05

Understanding Equation (b)

The equation is given as \(x = y^3 + 2y\). This curve is defined implicitly. First, determine what x and y values are feasible.
06

Making a Table of Values for Equation (b)

Choose various values for \(y\) to find corresponding \(x\) values:- For \(y = -2\), \(x = (-2)^3 + 2(-2) = -12\).- For \(y = -1\), \(x = (-1)^3 + 2(-1) = -3\).- For \(y = 0\), \(x = 0^3 + 2(0) = 0\).- For \(y = 1\), \(x = 1^3 + 2(1) = 3\).- For \(y = 2\), \(x = 2^3 + 2(2) = 12\).
07

Finding Intercepts for Equation (b)

The y-intercept is where \(x = 0\), giving \(y = 0\). There are no x-intercepts because there is no \(y\) that satisfies \(x = 0\) when \(x\) itself is set to 0 other than \(y = 0\).
08

Checking Symmetry for Equation (b)

Substitute \(-y\) for \(y\) in \(x = y^3 + 2y\) resulting in \(x = -y^3 - 2y\). This is neither symmetric across the y-axis nor the x-axis. By changing \(x\) to \(-x\) or \(y\) to \(-y\), the original equation form is not preserved, indicating there is no symmetry.

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- and y-intercepts
The x- and y-intercepts are points where a graph crosses the x-axis and y-axis, respectively. These intercepts give us valuable information about a graph's shape and position.

For the equation \(y = \sqrt{4 - x^2}\), let's identify the intercepts:
  • The **y-intercept** occurs where \(x = 0\). By substituting \(x = 0\) in the equation, we find \(y = \sqrt{4 - 0^2} = 2\). This means the graph crosses the y-axis at the point \((0, 2)\).
  • The **x-intercepts** occur where \(y = 0\). Solving \(\sqrt{4 - x^2} = 0\) gives us \(x^2 = 4\), so \(x = \pm 2\). This indicates the graph crosses the x-axis at the points \((2, 0)\) and \((-2, 0)\).
For equation \(x = y^3 + 2y\), the intercepts are found similarly:
  • The **y-intercept** is found by setting \(x = 0\), giving us \(y = 0\). Thus, the graph crosses the y-axis at \((0, 0)\).
  • There are no additional **x-intercepts** for \(x = y^3 + 2y\) at values of \(x = 0\), as the only solution is \(y = 0\). Hence, it does not cross the x-axis elsewhere.
symmetry in graphs
Symmetry in graphs tells us whether a graph is mirrored across a certain line. This can help simplify graphing by reducing the need to plot every point.

With the function \(y = \sqrt{4-x^2}\), we have even symmetry because the equation remains unchanged when substituting \(-x\) for \(x\). This means:
  • The graph is symmetric about the **y-axis**, as the positive and negative x-values yield identical y-values.
For the function \(x = y^3 + 2y\), symmetry is more complex:
  • Substituting \(-y\) for \(y\) does not lead the equation back to itself, and similarly with \(x\).
  • This indicates there is **no symmetry** about either axis or the origin for this graph.
Understanding symmetry makes plotting points efficient by revealing patterns and helping redraw fewer sections of the curve or line.
table of values
Creating a table of values is a fundamental step in graphing an equation. It involves selecting values for one variable and calculating the corresponding values for the other variable. This step lays the groundwork for accurately plotting a graph.

For equation \(y = \sqrt{4 - x^2}\):
  • Choose values of \(x\) from its domain \([-2, 2]\).
  • Calculate the corresponding \(y\) values using the equation:
    • For \(x = -2\), \(y = 0\).
    • For \(x = -1\), \(y = \sqrt{3}\).
    • For \(x = 0\), \(y = 2\).
    • For \(x = 1\), \(y = \sqrt{3}\).
    • For \(x = 2\), \(y = 0\).
For equation \(x = y^3 + 2y\):
  • Choose various values for \(y\).
  • Compute the corresponding \(x\) values:
    • For \(y = -2\), \(x = -12\).
    • For \(y = -1\), \(x = -3\).
    • For \(y = 0\), \(x = 0\).
    • For \(y = 1\), \(x = 3\).
    • For \(y = 2\), \(x = 12\).
A well-prepared table helps visualize the relation between x and y-values, making graphing more straightforward and precise.
semi-circle function
The semi-circle function is a specific kind of function involving a square root which produces a circular graph but only in half, typically allowing only non-negative values.

The function \(y = \sqrt{4 - x^2}\) is an example of a semi-circle function as it represents the upper half of a circle. Here鈥檚 how you can identify such a function:
  • This semi-circle is centered at the point \((0,0)\), with a radius of 2.
  • The domain is restricted because the square root is defined only for non-negative values, hence, the values of \(x\) must satisfy \(4 - x^2 \geq 0\).
  • This creates a graph that extends from \(x = -2\) to \(x = 2\).
  • Notice that the radius determines the range of the function, here from 0 to 2.
The characteristics of a semi-circle function make them immediately recognizable. Understanding these properties allows such functions to be easily graphed and interpreted.

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

The Lens Equation If \(F\) is the focal length of a convex lens and an object is placed at a distance \(x\) from the lens, then its image will be at a distance \(y\) from the lens, where \(F, x,\) and \(y\) are related by the lens equation $$\frac{1}{F}=\frac{1}{x}+\frac{1}{y}$$ Suppose that a lens has a focal length of \(4.8 \mathrm{cm}\) and that the image of an object is \(4 \mathrm{cm}\) closer to the lens than the object itself. How far from the lens is the object?

Simplify the expression. (a) \(\sqrt{9 a^{3}}+\sqrt{a}\) (b) \(\sqrt{16 x}+\sqrt{x^{5}}\)

Simplify the expression. (a) \(\frac{\left(8 s^{3} t^{3}\right)^{2 / 3}}{\left(s^{4} t^{-8}\right)^{1 / 4}}\) (b) \(\frac{\left(32 x^{5} y^{-3 / 2}\right)^{2 / 5}}{\left(x^{5 / 3} y^{2 / 3}\right)^{3 / 5}}\)

Depth of a Well One method for determining the depth of a well is to drop a stone into it and then measure the time it takes until the splash is heard. If \(d\) is the depth of the well (in feet) and \(t_{1}\) the time (in seconds) it takes for the stone to fall, then \(d=16 t_{1}^{2},\) so \(t_{1}=\sqrt{d} / 4 .\) Now if \(t_{2}\) is the time it takes for the sound to travel back up, then \(d=1090 t_{2}\) because the speed of sound is 1090 ft's. So \(t_{2}=d / 1090\) Thus the total time elapsed between dropping the stone and hearing the splash is $$t_{1}+t_{2}=\frac{\sqrt{d}}{4}+\frac{d}{1090}$$ How deep is the well if this total time is 3 s? PICTURE CANT COPY

Without using a calculator, determine which number is larger in each pair. (a) \(2^{1 / 2}\) or \(2^{1 / 3}\) (b) \(\left(\frac{1}{2}\right)^{1 / 2}\) or \(\left(\frac{1}{2}\right)^{1 / 3}\) (c) \(7^{1 / 4}\) or \(4^{1 / 3}\) (d) \(\sqrt[3]{5}\) or \(\sqrt{3}\)
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Statistics

Read Explanation

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Geometry

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.