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Chapter 2: Problem 13

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

Short Answer

Expert verified
Yes, the equation \(x^{2}+y-16=0\) defines y as a function of x.

Step by step solution

01

Write the given equation

The given equation is \(x^{2}+y-16=0\).
02

Solve the equation for y

By rearranging the terms of our equation, we will isolate y on one side of the equation. So, \(y=16-x^{2}\)
03

Examine the obtained equation for y

After rearranging, we got \(y=16-x^{2}\). An x-value will correspond to only one y-value. Hence, it meets the definition of a function.

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Key Concepts

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

Understanding Algebraic Equations
Algebraic equations are foundational in mathematics and help us describe relationships using letters and numbers. When we see an equation like \(x^2 + y - 16 = 0\), it tells us how two variables, \(x\) and \(y\), relate to each other.

In algebraic terms, an equation is a statement that asserts the equality between two expressions. Here are some key aspects to remember:
  • An equation can have one or more variables.
  • It can be linear, quadratic, or of higher degree, depending on the highest power of the variable. For example, \(x^2\) indicates a quadratic equation.
  • Solving an equation typically involves finding the value(s) of the variable(s) that make the equation true.
Equations are the language of algebra, allowing us to describe patterns, solve problems, and understand mathematical relationships.
Defining Functions
A function in mathematics sets up a relationship between input and output values. It assigns exactly one output to each input.

Given the equation \(y = 16 - x^2\), we want to check if it defines \(y\) as a function of \(x\).
  • For a relationship to be a function, every input must have a unique output.
  • If you can express \(y\) in terms of \(x\) such that for each \(x\), there is only one \(y\), it confirms \(y\) is a function of \(x\).
  • The equation \(y = 16 - x^2\) does this, so \(y\) is indeed a function of \(x\).
Functions are powerful tools in algebra and calculus. They let us model real-world phenomena mathematically and perform various operations like differentiation and integration.
Solving Equations
Solving an equation means finding the value of the variable that makes the equation true. Here鈥檚 how we do it step by step based on our original exercise:
  • Start by writing down the equation: \(x^2 + y - 16 = 0\).
  • Next, rearrange the equation to solve for \(y\). In this case, isolate \(y\) on one side: \(y = 16 - x^2\).
  • Check if the structure of the equation follows the criteria of a function. Here, for every \(x\), there's only one \(y\), confirming it is a function.
When solving equations, always aim to isolate the variable of interest by using algebraic operations like addition, subtraction, multiplication, or division.

Solving equations is critical in everything from engineering to everyday problem-solving, enabling us to find unknown values and make informed decisions.

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Most popular questions from this chapter

Begin by graphing the cube root function, \(f(x)-\sqrt[3]{x} .\) Then use transformations of this graph to graph the given function. $$ g(x)-\sqrt[3]{-x-2} $$

Begin by graphing the standard cubic function, \(f(x)-x^{3} .\) Then use transformations of this graph to graph the given function. $$ h(x)=-x^{3} $$

Use a graphing utility to graph the function. Use the graph to determine whether the function has an inverse that is a function (that is, whether the function is one-to-one). $$ f(x)=|x-2| $$

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 \((x-4)+(y+6)=25\) is a circle with radius 5 centered at \((4,-6)\)

Write a piecewise function that models each cellphone billing plan. Then graph the function. \(\$ 50\) per month buys 400 minutes. Additional time costs \(\$ 0.30\) per minute.
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