Pre-Algebra Quiz: Determining Functions

4 questions · exam conditions

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Determining FunctionsQuestion 1 of 4

A relation is defined by the set of ordered pairs $\{(a,3), (2,b), (c,3), (2,5)\}$ where $a$ , $b$ , and $c$ are real numbers. For this relation to be a function, which condition must be satisfied?

a \neq c

and

b

can be any real number except 5

b = 5

and

a

and

c

can be any real numbers with

a \neq c

a \neq 2

c \neq 2

, and

b

can be any real number

b = 5

and

a

and

c

can be any real numbers including

a = c

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Pre-Algebra Quiz

Pre-Algebra Quiz: Determining Functions

Practice Determining Functions in Pre-Algebra with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Determining Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for Pre-Algebra.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A relation is defined by the set of ordered pairs $\{(a,3), (2,b), (c,3), (2,5)\}$ where $a$ , $b$ , and $c$ are real numbers. For this relation to be a function, which condition must be satisfied?

$a \neq c$ and $b$ can be any real number except 5
$b = 5$ and $a$ and $c$ can be any real numbers with $a \neq c$
$a \neq 2$ , $c \neq 2$ , and $b$ can be any real number
$b = 5$ and $a$ and $c$ can be any real numbers including $a = c$ (correct answer)

Explanation: For a relation to be a function, each x-value must correspond to exactly one y-value. Looking at the ordered pairs, we have (2,b) and (2,5). Since both have the same x-value (2), they must have the same y-value for the relation to be a function. Therefore, b = 5. The pairs (a,3) and (c,3) both have y-value 3, but a and c can be equal or different since having the same y-value for different x-values is allowed in functions. Choice A incorrectly requires a ≠ c and restricts b. Choice B unnecessarily requires a ≠ c. Choice C incorrectly focuses on preventing a and c from equaling 2, which isn't the issue.

Question 2

Consider the piecewise relation defined as:

2x + 1 & \text{if } x < 0 \\ 3 & \text{if } x = 0 \\ -x + 4 & \text{if } x > 0 \end{cases}

Which statement best explains why this relation is or is not a function?

It is not a function because different formulas are used for different values of x
It is not a function because at x = 0, the value changes from one formula to another
It is a function because each value of x in the domain corresponds to exactly one value of y (correct answer)
It is a function only if the pieces connect continuously at the boundary points

Explanation: This piecewise relation is a function because for every possible input value of x, there is exactly one corresponding output value. For x < 0, use 2x + 1; for x = 0, use 3; for x > 0, use -x + 4. No x-value is assigned to multiple y-values. Choice A is incorrect because using different formulas for different intervals doesn't prevent something from being a function. Choice B is wrong because the change in formula doesn't create multiple outputs for any single input. Choice D is incorrect because continuity is not required for a relation to be a function (though it may be required for other properties).

Question 3

Three students are analyzing whether the relation $\{(-2,4), (0,1), (3,4), (5,1), (-2,7)\}$ is a function. Their reasoning is shown below:

Student A: "It's not a function because 4 and 1 each appear twice as y-values." Student B: "It's not a function because -2 appears twice as an x-value with different y-values." Student C: "It's a function because all the ordered pairs are written correctly."

Which student's reasoning is correct?

Student A is correct because repeated y-values indicate the relation is not a function
Student B is correct because repeated x-values with different y-values violate the function definition (correct answer)
Student C is correct because properly formatted ordered pairs always represent functions
All three students are incorrect because this relation actually is a function

Explanation: Student B correctly identifies that a function cannot have the same x-value paired with different y-values. In the given set, x = -2 is paired with both y = 4 and y = 7, which violates the definition of a function (each input must have exactly one output). Student A is incorrect because repeated y-values are allowed in functions (many-to-one relationships are permitted). Student C is incorrect because the format of ordered pairs doesn't determine whether a relation is a function. Choice D is wrong because the relation is indeed not a function due to the repeated x-value with different y-values.

Question 4

A relation is described by the rule: "For each input x, the output y is determined by $y^2 = x + 3$ ". For which reason is this relation not a function?

The equation contains a squared term, which automatically disqualifies it from being a function
For most positive values of x, there are two possible y-values that satisfy the equation (correct answer)
The equation cannot be solved for negative values of x when x < -3
The relation is undefined at x = -3, creating a gap in the domain

Explanation: The equation

y^2 = x + 3

can be rewritten as

y = ±\sqrt{x + 3}

. For any value of x > -3, there are two possible y-values: one positive and one negative. For example, when x = 1, we get

y^2 = 4

, so y = 2 or y = -2. This violates the definition of a function because a single input (x-value) produces multiple outputs (y-values). Choice A is incorrect because squared terms don't automatically disqualify relations from being functions. Choice C is incorrect because the relation is defined for x ≥ -3. Choice D is incorrect because the relation is defined at x = -3 (where y = 0).