Problem 30
Question
Determine whether the relation defines \(y\) to be a function of \(x .\) If it does not, find two ordered pairs where more than one value of \(y\) corresponds to a single value of \(x\). See Example 2. $$ \\{(-2,4),(-3,8),(-3,12),(-4,16)\\} $$
Step-by-Step Solution
Verified Answer
The relation is not a function because the x-value -3 has two different y-values, 8 and 12.
1Step 1: Understand the Definition of a Function
A relation is a function if for every input value (x) there is exactly one output value (y). This means there can't be two ordered pairs with the same x-value but different y-values.
2Step 2: Identify x-value Repetitions
Examine the given pairs: \((-2,4),(-3,8),(-3,12),(-4,16)\). Look for any x-values that are repeated with different y-values.
3Step 3: Analyze the x-values
Observe that the x-value \(-3\) is repeated: in the pairs \((-3,8)\) and \((-3,12)\). These pairs show that two different y-values correspond to the same x-value, meaning the relation does not define y as a function of x.
4Step 4: State the Conclusion
Since the x-value \(-3\) corresponds to two different y-values (8 and 12), the relation is not a function.
Key Concepts
Relation vs FunctionOrdered Pairsx-values and y-valuesRepeated x-values
Relation vs Function
In algebra, understanding the difference between a relation and a function is vital. A relation is simply a set of ordered pairs. It tells us how two elements are related but does not impose restrictions on their correspondence. On the other hand, a function is a special kind of relation where each input value (often represented as \(x\)) has a unique output value (commonly denoted as \(y\)).
This means, in a function:
This means, in a function:
- For every \(x\), there is exactly one \(y\).
- No \(x\) in the set can pair with multiple \(y\) values.
Ordered Pairs
Ordered pairs are a fundamental concept that help describe relationships between two elements. An ordered pair is written in the form \((x, y)\). The first element \(x\) is known as the input or independent variable, while the second element \(y\) is referred to as the output or dependent variable.
These pairs provide the foundation for defining relations and identifying functions. In practical terms, ordered pairs help us map out the connection between two variables, allowing for clear visualization and analysis of possible patterns or rules governing their relationship.
These pairs provide the foundation for defining relations and identifying functions. In practical terms, ordered pairs help us map out the connection between two variables, allowing for clear visualization and analysis of possible patterns or rules governing their relationship.
x-values and y-values
Within the context of ordered pairs, the terms \(x\)-values and \(y\)-values are integral. The \(x\)-value represents the independent or input variable, while the \(y\)-value signifies the dependent or output variable.
- The \(x\)-values are the first elements of each ordered pair.
- The \(y\)-values are the second elements.
Repeated x-values
One of the key tasks in determining if a relation is a function is to look for repeated \(x\)-values within the set of ordered pairs. In a function, no \(x\) value can repeat unless the corresponding \(y\) value is the same every time.
Consider the ordered pairs: \((-2,4), (-3,8), (-3,12), (-4,16)\). Here, we observe that the \(x\) value \(-3\) is repeated. But critically, it pairs with two different \(y\) values: 8 and 12. This repetition with differing \(y\) values immediately tells us the relation is not a function. By checking for repeated \(x\)-values, we efficiently ascertain the functional quality of a relation, making it a crucial step in algebraic analysis.
Consider the ordered pairs: \((-2,4), (-3,8), (-3,12), (-4,16)\). Here, we observe that the \(x\) value \(-3\) is repeated. But critically, it pairs with two different \(y\) values: 8 and 12. This repetition with differing \(y\) values immediately tells us the relation is not a function. By checking for repeated \(x\)-values, we efficiently ascertain the functional quality of a relation, making it a crucial step in algebraic analysis.
Other exercises in this chapter
Problem 29
Solve each equation. $$ \frac{1}{5}(x+6)=\frac{5}{8}(2 x-1)+2 $$
View solution Problem 30
Find the domain of each rational function. Express your answer in words and using interval notation. See Example 2. $$f(x)=\frac{x^{2}+36}{x^{2}-36}$$
View solution Problem 30
Factor each polynomial. $$ 18 p^{3} q^{2}-5 t^{5} $$
View solution Problem 30
Factor completely. \(16 n^{4}-1\)
View solution