/*! 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 15 Determine whether each equation ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 15

Determine whether each equation defines \(y\) as a function of \(x .\) $$ x^{2}+y^{2}=16 $$

Short Answer

Expert verified
No, the given equation does not define y as a function of x. For an equation to define \(y\) as a function of \(x\), each \(x\) value must correspond to exactly one \(y\) value. In the case of the given equation, \(x^{2}+y^{2}=16\), each \(x\) value corresponds to two \(y\) values (one positive and one negative), thus, it isn't a function.

Step by step solution

01

Standard form of a circle equation

Recognize that the given equation \(x^{2}+y^{2}=16\) is a standard form of a circle equation. The radius \(r\) can be determined by the square root of the constant, in this case 16. Thus, \(r=4\).
02

Solve for y

In order to see if y is a function of x, solve the equation for \(y\). The given equation is \(x^{2}+y^{2}=16\). This can be rearranged to solve for \(y\) by the operation \(y^{2}=16-x^{2}\). Taking the square root of both sides to solve for \(y\) results in \(y = \pm\sqrt{16-x^{2}}\).
03

Determine if y is a function of x

The presence of a \(+\) and \(-\) in the solution indicates that for a given \(x\), there are two possible values of \(y\): one positive and one negative. This means that multiple output values exist for a single input value which defies the definition of a function. Thus, it can be concluded that the given equation does not define \(y\) as a function of \(x\).

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.

Circle Equation
A circle equation in the standard form is represented as \(x^2 + y^2 = r^2\) where \(r\) is the radius of the circle. In this form, both \(x\) and \(y\) are squared, indicating the symmetry about the center of the circle, which is positioned at the origin \((0,0)\) in this case. This equation represents all the points \((x, y)\) that are exactly \(r\) units away from the center. It鈥檚 important because it provides a simple yet complete description of the circle in a Cartesian plane. By examining and manipulating this equation, you can infer many characteristics of the circle, such as radius, position, and even transformations when the equation changes.
Radius Calculation
Calculating the radius from the standard circle equation is straightforward. In an equation form as \(x^2 + y^2 = r^2\), the radius \(r\) can be found by taking the square root of the constant on the other side of the equation. For example, given the equation \(x^2 + y^2 = 16\), you can determine that \(r = \sqrt{16} = 4\).
  • The radius is always a positive number, as it represents distance.
  • This distance is from the center of the circle, which affects the size of the circle but not its shape.
Remember, knowing the radius is crucial for drawing, analyzing, and utilizing the circle in various mathematical applications.
Solving Equations
Solving equations, especially those that involve squares, involves finding unknown variables while maintaining the equation's balance. For a given circle equation like \(x^2 + y^2 = 16\), solving for \(y\) involves isolating the \(y^2\) term first, resulting in \(y^2 = 16 - x^2\). To further simplify, take the square root of both sides:- This produces \(y = \pm\sqrt{16 - x^2}\), indicating two possible values: positive and negative.When solving, it's also essential to consider all solutions. Both \(+\sqrt{}\) and \(-\sqrt{}\) solutions must be considered, as they represent the upper and lower parts of the circle, respectively.
Mathematical Definitions
In mathematics, understanding definitions is pivotal to grasping broader concepts. For example, a function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output. This definition starkly contrasts with a circle's equation like \(x^2 + y^2 = 16\), which does not fit the criteria for being a function.
  • A function must have only one output for each input; however, a circle's equation may give two outputs (positive and negative) for a single input \(x\).
  • Understanding these fundamental definitions helps in determining and analyzing various mathematical phenomena.
With these definitions in mind, we can quickly identify whether given equations describe functions or not, such as concluding that the circle equation does not define \(y\) as a function of \(x\).

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

give the center and radius of the circle described by the equation and graph each equation. Use the graph to identify the relation's domain and range. $$ x^{2}+y^{2}=49 $$

complete the square and write the equation in standard form. Then give the center and radius of each circle and graph the equation. $$ x^{2}+y^{2}+6 x+2 y+6=0 $$

Simplify: \(2(x+h)^{2}+3(x+h)+5-\left(2 x^{2}+3 x+5\right)\)

use a graphing utility to graph each circle whose equation is given. $$ x^{2}+10 x+y^{2}-4 y-20=0 $$

Does \(f(x)\) mean \(f\) times \(x\) when referring to a function \(f ?\) If not, what does \(f(x)\) mean? Provide an example with your explanation.
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

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.