Determining Functions - Pre Algebra

1. 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?
   Answer: Student B is correct because repeated x-values with different y-values violate the function definition 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.
2. Consider the piecewise relation defined as: $f(x) = \begin{cases} 2x + 1 & \text{if } x < 0 \\ 3 & \text{if } x = 0 \\ -x + 4 & \text{if } x > 0 \end{cases}$ Which statement best explains why this relation is or is not a function?
   Answer: It is a function because each value of x in the domain corresponds to exactly one value of y 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).
3. 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?
   Answer: $b = 5$ and $a$ and $c$ can be any real numbers including $a = c$ 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.
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?
   Answer: For most positive values of x, there are two possible y-values that satisfy the equation 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).

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