Problem 12 - Precalculus Essentials

Question

Determine whether each equation defines y as a function of \(x .\) $$x+y=25$$

Step-by-Step Solution

Answer

Yes, the equation \(y = 25 - x\) does define \(y\) as a function of \(x\). For every \(x\), there is exactly one corresponding \(y\) value.

Step 1: Rewrite in terms of y

Rewrite the given equation \(x + y = 25\) in terms of \(y\). Start by transposing \(x\) to the other side of the equation. So, the equation becomes: \(y = 25 - x\)

Step 2: Determine if y is a function of x

Consider if there are any values of \(x\) which would give more than one corresponding \(y\). In this equation, for any given \(x\), there exists exactly one \(y\) value. Therefore, \(y\) is a function of \(x\).

Key Concepts

Equations and Functions

When studying precalculus, it's important to recognize the relationship between equations and functions. An equation is a mathematical statement that asserts the equality of two expressions, whereas a function is a special type of equation that describes a relationship where each input corresponds to exactly one output.

In the context of equations and functions, a function can often be recognized if it passes the 'vertical line test.' This test checks whether a graph of the equation intersects a vertical line at no more than one point. If it does, then every input (or x-value) is associated with a unique output (or y-value), qualifying it as a function.

The exercise provided x + y = 25is an equation that can be analyzed to determine if it defines a function. By solving for one variable in terms of the other, we seek to express this relationship in function form, such as y = f(x). The rewritten equation y = 25 - xnow clearly shows that for every x-value, there is only one y-value, thus satisfying the criteria of a function. This makes it easier to visualize and understand the concept of functions in the realm of algebraic equations.

Defining Functions Algebraically

Defining functions algebraically involves expressing one variable explicitly in terms of another. This practice allows us to see the direct relationship between the variables, making it clear how the value of one variable affects the other.

In Step 1 of our exercise, a key move was made — expressing y in terms of x. This was achieved by isolating y on one side of the equation, resulting in y = 25 - x. Now, y is defined algebraically as a function of x, denoted as f(x) = 25 - x.

Identifying Independent and Dependent Variables

Here, x is the independent variable, while y is the dependent variable, indicating y's value depends on the chosen value of x. This formulation allows for the clear interpretation of the function, providing a path to plot the equation on a graph or to easily calculate y-values for different x-values.

Having a function defined algebraically is essential in precalculus and beyond, as it forms the basis for understanding more complex functions and algebraic structures.

Transposing Equations

Transposing equations involves moving terms from one side of an equation to the other to isolate a specific variable—useful for both solving equations and defining functions. While transposing, it's critical to maintain the balance of the equation by performing the same operation on both sides.

For example, in our exercise, we are presented with the equation x + y = 25. To define y as a function of x, we transpose x to the other side to isolate y, which is done by subtracting x from both sides. We get y = 25 - x. It's essential to apply operations consistently to preserve the equality of the equation.

Preserving Equality

Always remember that whatever you do to one side of the equation, you must do to the other. This ensures that the two sides remain in balance. This is a fundamental rule in algebra that prevents altering the solution or the behavior of the original equation.

Transposing isn't just a mechanical process; it's part of understanding and manipulating the structure of equations to serve our purpose, particularly when defining functions algebraically.