College Algebra Quiz: Relations And Functions

What this quiz covers

This quiz focuses on Relations And Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for College 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.

Question 1

1. The student is correct; piecewise relations are functions only when all pieces give the same value at boundary points.
2. The student identifies a valid concern; the relation should be redefined to avoid conflicting outputs at $x = 3$.
3. The student is partially correct; this creates a discontinuity but does not affect whether the relation is a function.
4. The student misunderstands piecewise notation; only the piece whose condition is satisfied applies to each input value. (correct answer)

Explanation: Understanding piecewise functions requires careful attention to the domain conditions that determine which piece applies to each input value. The key insight is that piecewise notation explicitly tells you which rule to use for each part of the domain. In this piecewise function, each piece has a specific condition: the first piece ($2x + 1$) applies only when $x < 3$, the second piece gives $f(3) = 5$, and the third piece ($x^2 - 4$) applies when $x > 3$. Since $x = 3$ does not satisfy the condition $x < 3$, you cannot use the first piece to evaluate $f(3)$. The student's calculation of $2(3) + 1 = 7$ is mathematically correct, but it's irrelevant because that piece doesn't apply when $x = 3$. Answer D correctly identifies that only the piece whose condition is satisfied applies to each input. For $x = 3$, only the second piece applies, so $f(3) = 5$ definitively. Answer A is wrong because piecewise functions don't require all pieces to agree at boundaries—the domain conditions prevent conflicts. Answer B incorrectly suggests there's actually a conflict to resolve, when the notation already resolves it by specifying conditions. Answer C acknowledges the student's misunderstanding but incorrectly validates the concern about conflicting outputs. When working with piecewise functions, always check the domain condition first before applying any piece. The conditions act like gatekeepers—they tell you exactly which rule applies, eliminating any ambiguity about which piece to use.

Question 2

1. Remove the arrow from $3$ to $7$ and keep the arrow from $3$ to $9$.
2. Remove the arrow from $3$ to $9$ and keep the arrow from $3$ to $7$. (correct answer)
3. Remove the arrow from $5$ to $7$ since $7$ already receives an arrow from $3$.
4. Add an additional arrow from $0$ to $9$ to balance the multiple mappings from $3$.

Explanation: The relation fails to be a function because element 3 in Set A maps to two different elements in Set B (both 7 and 9). To make it a function with minimum changes, we need to remove one of these mappings from element 3. Both choices A and B accomplish this with one removal, but the question asks what 'should be done' - either would work. Choice B is correct as it represents a valid solution. Choice C is wrong because having multiple inputs map to the same output is allowed in functions. Choice D would make the problem worse by creating another element that maps to multiple outputs.

Question 3

1. The student incorrectly assumes that geometric shapes can never represent functions in algebraic form.
2. The student fails to recognize that most x-values correspond to two different y-values when solving for $y$. (correct answer)
3. The student correctly identifies this as a function, but fails to note that the domain must be restricted.
4. The student misunderstands that functions require each point to map to exactly one other point, not one-to-one correspondence.

Explanation: The equation $x^2 + y^2 = 25$ represents a circle with radius 5. When we solve for y, we get $y = \pm\sqrt{25 - x^2}$, which means for most x-values in the domain $[-5, 5]$ (except x = ±5), there are two corresponding y-values. For example, when x = 0, y = ±5. This violates the definition of a function. The student's error is not recognizing this multi-valued nature. Choice A is wrong because geometric shapes can represent functions. Choice C is wrong because this is not a function even with domain restrictions on x. Choice D misrepresents what the student said and the definition of functions.

Question 4

1. This is a function because solving for $y$ gives $y = x + 2$ or $y = -(x + 2)$, providing exactly two outputs for each input.
2. This is not a function because absolute value equations always fail the vertical line test in their standard graphical form.
3. This is a function because the absolute value ensures that all outputs are positive, creating a one-to-one correspondence.
4. This is not a function because when $x = 3$, we get $|y| = 5$, which means $y = 5$ or $y = -5$, giving multiple outputs for one input. (correct answer)

Explanation: When determining whether a relation represents a function, you need to check if each input value (x) corresponds to exactly one output value (y). The key test is whether any single x-value produces multiple y-values. Let's analyze $|y| = x + 2$ by testing a specific value. When $x = 3$, we get $|y| = 3 + 2 = 5$. Since the absolute value of $y$ equals 5, this means $y$ could be either 5 or -5 (because $|5| = 5$ and $|-5| = 5$). Having two different y-values for the same x-value violates the definition of a function. This confirms answer D is correct. Answer A makes a critical error in reasoning. While it correctly identifies that solving gives $y = x + 2$ or $y = -(x + 2)$, it then incorrectly claims this creates a function because there are "exactly two outputs." Having two outputs for one input is precisely what makes this NOT a function. Answer B incorrectly generalizes about absolute value equations. While this particular equation isn't a function, not all absolute value equations fail to be functions—it depends on which variable is inside the absolute value bars. Answer C misunderstands the role of absolute values here. The equation doesn't ensure positive outputs for y; rather, $|y|$ represents the absolute value of y, which can itself be positive or negative. Remember: A relation is a function if and only if each input produces exactly one output. When in doubt, test specific values to see if you get multiple y-values for a single x-value.

Question 5

1. This is a function because both rules $y = x + 4$ and $y = x + 5$ are linear functions with well-defined outputs.
2. This is a function because all elements in set $A$ are mapped to elements within the specified set $B$.
3. This is not a function because each element in set $A$ is related to two elements in set $B$ according to the given rule. (correct answer)
4. This is not a function because the rule creates an inconsistent mapping where some elements in $B$ receive no input from $A$.

Explanation: When analyzing whether a relation is a function, you need to check if each input (element in the domain) maps to exactly one output (element in the codomain). A relation fails to be a function if any single input produces multiple outputs. Let's examine this relation systematically. For each element in set $A$, we apply both conditions $y = x + 4$ and $y = x + 5$:

• When $x = 1$: $y = 5$ or $y = 6$
• When $x = 2$: $y = 6$ or $y = 7$
• When $x = 3$: $y = 7$ or $y = 8$
• When $x = 4$: $y = 8$ or $y = 9$

Question 6

1. The modified relation is still a function because the original quadratic function remains valid for all other x-values.
2. The modified relation becomes a better representation of the quadratic because it includes additional data points.
3. The modified relation is no longer a function because $x = 2$ now corresponds to two different y-values. (correct answer)
4. The modified relation is still a function because adding points to a discrete set cannot violate the function definition.

Explanation: When you encounter questions about relations and functions, focus on the fundamental definition: a relation is a function if and only if each input (x-value) corresponds to exactly one output (y-value). This is the vertical line test in graphical terms, or the "one-to-one correspondence" rule for ordered pairs. Let's trace through this problem systematically. The original quadratic $y = x^2 - 4x + 3$ generates ordered pairs for $x$ from $-1$ to $5$. When $x = 2$, we get $y = (2)^2 - 4(2) + 3 = 4 - 8 + 3 = -1$, so the original relation includes the point $(2, -1)$. Adding the point $(2, 10)$ means the input $x = 2$ now maps to two different outputs: $-1$ and $10$. This violates the definition of a function. Answer A incorrectly assumes that the validity of the original quadratic equation somehow preserves the function property, but functions are defined by their complete set of ordered pairs, not their underlying equations. Answer B misses the point entirely—this isn't about data quality but about mathematical definitions. Answer D contains a fundamental misconception: adding points to any set can absolutely violate the function definition if those points create multiple y-values for the same x-value. Study tip: Always check for repeated x-values when analyzing whether a relation is a function. If any x-value appears with different y-values, it's not a function—regardless of how the relation was created or what it represents.

Question 7

1. The relation is not a function because the input value 2 corresponds to two different output values, 3 and 7. (correct answer)
2. The relation is not a function because the output value 5 corresponds to two different input values, 4 and 8.
3. The relation is not a function because there are an odd number of ordered pairs in the set.
4. The relation is not a function because the ordered pairs do not follow a consistent mathematical pattern.

Explanation: A relation is a function if and only if each input (x-value) corresponds to exactly one output (y-value). In this relation, the input value 2 appears in two ordered pairs: (2, 3) and (2, 7), meaning it corresponds to two different outputs. This violates the definition of a function. Choice B is incorrect because having multiple inputs correspond to the same output is allowed in functions. Choices C and D are incorrect because neither the number of pairs nor following a pattern affects whether a relation is a function.