Algebra Quiz: Understanding Functions Domain And Range

What this quiz covers

This quiz focuses on Understanding Functions Domain And Range, giving you a quick way to practice the rules, question types, and explanations that matter most for 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. All real numbers except $x=5$ (correct answer)
2. All real numbers except $x=0$
3. $[5,\infty)$
4. All real numbers

Explanation: This question tests your understanding of what functions are, and how to determine their domains (possible inputs) and ranges (possible outputs). The domain is the set of all possible input values (x-values) that make sense for the function: for formulas, we exclude values that would cause division by zero or square roots of negatives, and in real-world contexts, we only include values that are realistic (like you can't have -3 people or 2.5 items if they're discrete). For g(x) = 1/(x-5), we need to avoid division by zero, so the denominator x-5 cannot equal 0, which means x cannot equal 5. Choice A is correct because it states 'all real numbers except x=5', which excludes only the problematic value while allowing all others. Choice B incorrectly excludes x=0, but plugging in x=0 gives g(0)=1/(-5)=-1/5, which is perfectly defined. For finding domain from a formula: (1) Start by assuming all real numbers are okay, (2) Then look for restrictions—is there a square root (need inside ≥ 0)? A fraction (need denominator ≠ 0)? (3) Write the domain excluding or including only the values that work. Most Algebra 1 functions have domains like 'all reals' or 'x ≥ some number' or 'all reals except one value.'

Question 2

1. $[4,\infty)$
2. $(-\infty,4]$ (correct answer)
3. $(-\infty,4)$
4. $(-\infty,\infty)$

Explanation: This question tests your understanding of what functions are, and how to determine their domains (possible inputs) and ranges (possible outputs). The range is the set of all possible output values (y-values) the function can produce: for a quadratic that opens upward with vertex at $(2, -3)$, the range is all y-values greater than or equal to $-3$ because the parabola goes up from that minimum point forever. For $p(x) = -(x-1)^2 + 4$, the negative sign in front means this parabola opens downward, and the vertex form shows the vertex is at $(1, 4)$, giving a maximum value of 4. Choice B is correct because $(-\infty,4]$ represents all y-values less than or equal to 4, with the square bracket showing 4 is included since $p(1) = 4$. Choice A would suggest the range starts at 4 and goes up, but this parabola opens down from its maximum. For finding range: linear functions usually have range = all real numbers. For quadratics, find the vertex first—if it opens up, range is $y \geq$ (vertex y-value); if it opens down, range is $y \leq$ (vertex y-value). The negative coefficient flips everything!

Question 3

1. All real numbers except $x=2$ (correct answer)
2. All real numbers $x\le 2$
3. All real numbers $x\ge 2$
4. All real numbers except $x=3$

Explanation: This question tests your understanding of what functions are, and how to determine their domains (possible inputs) and ranges (possible outputs). The domain is the set of all possible input values (x-values) that make sense for the function: for formulas, we exclude values that would cause division by zero or square roots of negatives, and in real-world contexts, we only include values that are realistic (like you can't have -3 people or 2.5 items if they're discrete). For g(x) = 3/(x - 2), the function is undefined when the denominator is zero, so $x - 2 = 0$ means $x = 2$ is excluded, but all other real numbers work. Choice A is correct because it states all real numbers except $x=2$, which matches the restriction from the denominator. Choice B might seem tempting if you mistakenly thought the numerator caused an issue, but the numerator is just 3, a constant, so no problem there—always check the denominator for zeros! For finding domain from a formula: (1) Start by assuming all real numbers are okay, (2) Then look for restrictions—is there a square root (need inside $\geq 0$)? A fraction (need denominator $\neq 0$)? (3) Write the domain excluding or including only the values that work. Most Algebra 1 functions have domains like 'all reals' or '$x \geq$ some number' or 'all reals except one value.'

Question 4

1. $(-\infty,\infty)$
2. $[0,\infty)$ (correct answer)
3. $(-\infty,0]$
4. $(0,\infty)$

Explanation: This question tests your understanding of what functions are, and how to determine their domains (possible inputs) and ranges (possible outputs). The range is the set of all possible output values (y-values) the function can produce: for a quadratic that opens upward with vertex at (2, -3), the range is all y-values greater than or equal to -3 because the parabola goes up from that minimum point forever. For p(x) = √x, the square root function only produces non-negative outputs—you can never get a negative number from a principal square root. The smallest output is √0 = 0, and as x increases, √x increases without bound. Choice B is correct because [0,∞) represents all non-negative real numbers, with the square bracket showing 0 is included as the minimum possible output. Choice C incorrectly suggests negative outputs are possible, but √x is always non-negative by definition. For finding range: linear functions usually have range = all real numbers. For quadratics, find the vertex first—if it opens up, range is y ≥ (vertex y-value); if it opens down, range is y ≤ (vertex y-value). For square roots, range is usually y ≥ 0. The function type tells you a lot!

Question 5

1. $( -\infty,\infty)$
2. $[9,\infty)$
3. $( -\infty,9)$
4. $( -\infty,9]$ (correct answer)

Explanation: This question tests your understanding of what functions are, and how to determine their domains (possible inputs) and ranges (possible outputs). The domain is the set of all possible input values (x-values) that make sense for the function: for formulas, we exclude values that would cause division by zero or square roots of negatives, and in real-world contexts, we only include values that are realistic (like you can't have -3 people or 2.5 items if they're discrete). For k(x) = √(9-x), we need 9 - x ≥ 0, which means 9 ≥ x, or equivalently x ≤ 9. Choice A is correct because (-∞,9] represents all real numbers less than or equal to 9, with the square bracket showing 9 is included since √0 = 0 is valid. Choice B would be the domain if we had √(x-9) instead, requiring x ≥ 9. For finding domain from a formula: (1) Start by assuming all real numbers are okay, (2) Then look for restrictions—is there a square root (need inside ≥ 0)? A fraction (need denominator ≠ 0)? (3) Write the domain excluding or including only the values that work. When the variable is subtracted inside a square root, flip your thinking!

Question 6

1. $( -\infty, -9)\cup(-9, 9)\cup(9, \infty)$
2. $( -\infty, \infty)$
3. $( -\infty, -3)\cup(-3, 3)\cup(3, \infty)$ (correct answer)
4. $( -\infty, 3)\cup(3, \infty)$

Explanation: This question tests your understanding of what functions are, and how to determine their domains (possible inputs) and ranges (possible outputs). The domain is the set of all possible input values (x-values) that make sense for the function: for formulas, we exclude values that would cause division by zero or square roots of negatives, and in real-world contexts, we only include values that are realistic (like you can't have -3 people or 2.5 items if they're discrete). For f(x) = (x+1)/(x² - 9), factor the denominator as (x-3)(x+3), so it's undefined at x=3 and x=-3 where denominator is zero. Choice A is correct because (-∞, -3) ∪ (-3, 3) ∪ (3, ∞) excludes exactly those two points. Choice C excludes only 3, but you need to catch both roots of the denominator—always factor to find all zeros! For finding domain from a formula: (1) Start by assuming all real numbers are okay, (2) Then look for restrictions—is there a square root (need inside ≥ 0)? A fraction (need denominator ≠ 0)? (3) Write the domain excluding or including only the values that work. Most Algebra 1 functions have domains like 'all reals' or 'x ≥ some number' or 'all reals except one value.'

Question 7

1. All integers (positive, negative, and zero)
2. All whole numbers $n\in\{0,1,2,3,\dots\}$ (correct answer)
3. All real numbers except $n=0$
4. All real numbers $n\ge 0$

Explanation: This question tests your understanding of what functions are, and how to determine their domains (possible inputs) and ranges (possible outputs). The domain is the set of all possible input values (x-values) that make sense for the function: for formulas, we exclude values that would cause division by zero or square roots of negatives, and in real-world contexts, we only include values that are realistic (like you can't have -3 people or 2.5 items if they're discrete). For C(n) = 5 + 12n where n is the number of tickets, n must be a non-negative integer because you can't buy a negative or fractional number of tickets in this context. Choice D is correct because it specifies all whole numbers starting from 0, which includes buying zero tickets (just the booking fee) up to any positive integer. Choice A might confuse you if you thought n could be any real number greater than or equal to zero, but tickets are discrete, so only integers make sense—always consider the real-world meaning! For finding domain from a formula: (1) Start by assuming all real numbers are okay, (2) Then look for restrictions—is there a square root (need inside ≥ 0)? A fraction (need denominator ≠ 0)? (3) Write the domain excluding or including only the values that work. Most Algebra 1 functions have domains like 'all reals' or 'x ≥ some number' or 'all reals except one value.'

Question 8

1. $( -\infty, 9)$
2. $( -\infty, 9]$ (correct answer)
3. $( -\infty, \infty)$
4. $[9, \infty)$

Explanation: This question tests your understanding of what functions are, and how to determine their domains (possible inputs) and ranges (possible outputs). The domain is the set of all possible input values (x-values) that make sense for the function: for formulas, we exclude values that would cause division by zero or square roots of negatives, and in real-world contexts, we only include values that are realistic (like you can't have -3 people or 2.5 items if they're discrete). For h(x) = √(9 - x), the inside must be non-negative, so 9 - x ≥ 0, which solves to x ≤ 9. Choice B is correct because (-∞, 9] includes all x up to and including 9, where h(9) = √0 = 0 is defined. Choice C excludes 9 with a parenthesis, but that's not needed since x=9 works—double-check if the boundary value is allowed! For finding domain from a formula: (1) Start by assuming all real numbers are okay, (2) Then look for restrictions—is there a square root (need inside ≥ 0)? A fraction (need denominator ≠ 0)? (3) Write the domain excluding or including only the values that work. Most Algebra 1 functions have domains like 'all reals' or 'x ≥ some number' or 'all reals except one value.'

Question 9

1. $(-\infty,2]$
2. $[2,\infty)$ (correct answer)
3. $(2,\infty)$
4. All real numbers

Explanation: This question tests your understanding of what functions are, and how to determine their domains (possible inputs) and ranges (possible outputs). The domain is the set of all possible input values (x-values) that make sense for the function: for formulas, we exclude values that would cause division by zero or square roots of negatives, and in real-world contexts, we only include values that are realistic (like you can't have -3 people or 2.5 items if they're discrete). For f(x) = √(x-2), we need the expression under the square root to be non-negative, so x-2 ≥ 0, which means x ≥ 2. Choice B is correct because [2,∞) represents all real numbers greater than or equal to 2, using a square bracket at 2 to show it's included. Choice A would give negative values under the square root, making the output imaginary rather than real. For finding domain from a formula: (1) Start by assuming all real numbers are okay, (2) Then look for restrictions—is there a square root (need inside ≥ 0)? A fraction (need denominator ≠ 0)? (3) Write the domain excluding or including only the values that work. Most Algebra 1 functions have domains like 'all reals' or 'x ≥ some number' or 'all reals except one value.'

Question 10

1. $( -4, \infty)$
2. $[-4, \infty)$ (correct answer)
3. $( -\infty, \infty)$
4. $( -\infty, -4]$

Explanation: This question tests your understanding of what functions are, and how to determine their domains (possible inputs) and ranges (possible outputs). The range is the set of all possible output values (y-values) the function can produce: for a quadratic that opens upward with vertex at (2, -3), the range is all y-values greater than or equal to -3 because the parabola goes up from that minimum point forever. For f(x) = x² - 4, this is a parabola opening upward with vertex at (0, -4), so the minimum y is -4, and it goes to infinity as x moves away from 0. Choice C is correct because [-4, ∞) captures all y ≥ -4, including -4 when x=0. Choice B is incorrect because it suggests y ≤ -4, but that's for a downward-opening parabola—remember to check if the coefficient of x² is positive (up) or negative (down)! For finding range: linear functions usually have range = all real numbers. For quadratics, find the vertex first—if it opens up, range is y ≥ (vertex y-value); if it opens down, range is y ≤ (vertex y-value). For square roots, range is usually y ≥ 0. The function type tells you a lot!

Question 11

1. Yes; each $x$ gives exactly one $y$.
2. Yes; it passes the vertical line test.
3. No; some $x$ values correspond to two different $y$ values. (correct answer)
4. No; it has no real solutions.

Explanation: This question tests your understanding of what functions are, and how to determine their domains (possible inputs) and ranges (possible outputs). A function is a special relationship where each input gives exactly one output—like a vending machine where pressing one button (input) gives you one specific item (output). If one input could give multiple outputs, it's not a function. The vertical line test checks this on graphs: if any vertical line crosses the graph more than once, it's not a function. For the relation x = y², solving for y gives y = ±√x, meaning for any positive x-value, there are two possible y-values (one positive, one negative). For example, when x = 4, y could be 2 or -2. Choice C is correct because it identifies that some x values correspond to two different y values, violating the function definition. Choice A incorrectly claims each x gives exactly one y, missing the ± issue. The vertical line test is your quick function-checker: imagine sliding a vertical line across the graph from left to right. If it ever crosses the graph at two or more points at the same time, you've found a spot where one x gives multiple y's—not a function! Circles fail this test, parabolas (y = x²) pass it!

Question 12

1. All real numbers
2. All real numbers except $m=0$
3. All real numbers $m \geq 0$ (correct answer)
4. All real numbers $m \leq 0$

Explanation: This question tests your understanding of what functions are, and how to determine their domains (possible inputs) and ranges (possible outputs). The domain is the set of all possible input values (x-values) that make sense for the function: for formulas, we exclude values that would cause division by zero or square roots of negatives, and in real-world contexts, we only include values that are realistic (like you can't have -3 people or 2.5 items if they're discrete). In the context of a ride-share, m represents miles traveled, which cannot be negative—you can't travel -5 miles! The domain must start at $m = 0$ (no travel) and extend to positive values. Choice C is correct because "all real numbers $m \geq 0$" captures this real-world constraint that miles must be non-negative. Choice A would allow negative miles, which doesn't make sense in this context. For real-world problems, always think about what the variable represents: Can it be negative? Can it be a fraction? Are there upper limits? Context matters as much as the formula!

Question 13

1. Yes, because every output value is different.
2. Yes, because each output corresponds to exactly one input.
3. No, because the input $x=2$ has two different outputs. (correct answer)
4. No, because the output $y=8$ occurs only once.

Explanation: This question tests your understanding of what functions are, and how to determine their domains (possible inputs) and ranges (possible outputs). A function is a special relationship where each input gives exactly one output—like a vending machine where pressing one button (input) gives you one specific item (output). If one input could give multiple outputs, it's not a function. The vertical line test checks this on graphs: if any vertical line crosses the graph more than once, it's not a function. In this table, the input x=2 is listed twice with different outputs 5 and 7, so it doesn't assign exactly one output to each input. Choice C is correct because it identifies that x=2 has two outputs, violating the function rule. Choice A is a common mix-up, but functions care about unique outputs per input, not unique outputs overall—multiple inputs can have the same output, that's fine! The vertical line test is your quick function-checker: imagine sliding a vertical line across the graph from left to right. If it ever crosses the graph at two or more points at the same time, you've found a spot where one x gives multiple y's—not a function! Circles fail this test, parabolas (y = x²) pass it!

Question 14

1. $(-\infty, -2) \cup (-2, \infty)$ (correct answer)
2. $(-\infty, \infty)$
3. $(-\infty, 2) \cup (2, \infty)$
4. $[-2, \infty)$

Explanation: This question tests your understanding of what functions are, and how to determine their domains (possible inputs) and ranges (possible outputs). The domain is the set of all possible input values (x-values) that make sense for the function: for formulas, we exclude values that would cause division by zero or square roots of negatives, and in real-world contexts, we only include values that are realistic (like you can't have -3 people or 2.5 items if they're discrete). For q(x) = (x+1)/(x+2), the denominator can't be zero, so x ≠ -2, excluding just that point from all reals. Choice A is correct because (-∞, -2) ∪ (-2, ∞) properly leaves out only x=-2. Choice C excludes x=2 instead, but the denominator is x+2, not x-2—gently double-check where the denominator is zero. For finding domain from a formula: (1) Start by assuming all real numbers are okay, (2) Then look for restrictions—is there a square root (need inside ≥ 0)? A fraction (need denominator ≠ 0)? (3) Write the domain excluding or including only the values that work. Most Algebra 1 functions have domains like 'all reals' or 'x ≥ some number' or 'all reals except one value.'

Question 15

1. 1 (correct answer)
2. 3
3. 7
4. 9

Explanation: Since -2 < 0, use the first piece: f(-2) = 2(-2) + 1 = -4 + 1 = -3. Since 2 ≥ 0, use the second piece: f(2) = 2² = 4. Therefore f(-2) + f(2) = -3 + 4 = 1. Choice B might result from calculation errors. Choice C might come from using wrong pieces. Choice D uses f(2) + f(2) + 1 incorrectly.

Question 16

1. $\{2, 3, 4\}$
2. $\{5, 7, 9, 1\}$
3. $\{5, 7, 1\}$ or $\{9, 7, 1\}$ (correct answer)
4. $\{2, 3, 4, 5, 7, 9, 1\}$

Explanation: The relation fails to be a function because input 2 maps to both 5 and 9. We must remove either (2,5) or (2,9). If we remove (2,5), the range is {9,7,1}. If we remove (2,9), the range is {5,7,1}. Choice A gives the domain. Choice B gives the original range before removal. Choice D combines domain and range incorrectly.

Question 17

1. The domain is all real numbers because square roots can be taken of any number
2. The domain is $x > -3$ because the expression under the square root must be strictly positive
3. The domain is $x \geq 3$ because we cannot have negative values inside the square root function
4. The domain is $x \geq -3$ because the expression under the square root must be non-negative (correct answer)

Explanation: When you encounter a function involving a square root, you need to determine what values make the function undefined. The key principle is that square roots of negative numbers are not defined in the real number system. For $h(x) = \sqrt{x + 3}$, the expression under the square root is $x + 3$. This expression must be non-negative (greater than or equal to zero) for the function to produce real outputs. Setting up the inequality: $x + 3 \geq 0$. Solving for $x$: $x \geq -3$. This means the domain includes $x = -3$ and all values greater than $-3$. Let's check why each wrong answer misses the mark. Choice A incorrectly claims square roots can be taken of any number. While this is true for complex numbers, in Algebra 1 we work with real numbers only, where square roots of negative numbers are undefined. Choice B states the domain is $x > -3$, incorrectly excluding $x = -3$. However, when $x = -3$, we get $\sqrt{-3 + 3} = \sqrt{0} = 0$, which is perfectly valid. Choice C gives $x \geq 3$, which stems from confusing the sign inside the square root—the expression is $x + 3$, not $x - 3$. Choice D correctly identifies that the domain is $x \geq -3$ because the radicand (expression under the square root) must be non-negative. Study tip: For square root functions, always set the expression inside the radical ≥ 0 and solve. Remember that zero is allowed under square roots, so use "greater than or equal to" in your final domain.

Question 18

1. $mn = 1$ because this is the constant term of the quadratic function
2. $m - n = 4$ because this represents the distance between the two equal function values
3. $m + n = 4$ because the parabola is symmetric about its axis of symmetry (correct answer)
4. $m + n = -4$ because this is the negative of the coefficient of the linear term

Explanation: When you see that $p(m) = p(n)$ where $m \neq n$ for a quadratic function, you're dealing with the symmetry property of parabolas. Since two different x-values produce the same y-value, these points must be equidistant from the parabola's axis of symmetry. For any quadratic $ax^2 + bx + c$, the axis of symmetry occurs at $x = -\frac{b}{2a}$. In $p(x) = x^2 - 4x + 1$, we have $a = 1$ and $b = -4$, so the axis of symmetry is at $x = -\frac{(-4)}{2(1)} = 2$. Since $m$ and $n$ are symmetric about $x = 2$, they're equidistant from this line. If the axis of symmetry is at $x = 2$, then $\frac{m + n}{2} = 2$, which means $m + n = 4$. Answer A incorrectly assumes the relationship involves the constant term. While $mn$ might equal 1 for this specific function, this isn't a general property of symmetric points on quadratics. Answer B confuses the sum with the difference. The difference $m - n$ tells you the distance between the points, but it's not necessarily 4. Answer D applies the wrong sign to the linear coefficient. The relationship $m + n = -\frac{b}{a}$ gives us $m + n = -\frac{(-4)}{1} = 4$, not $-4$. Remember: when two different inputs give the same output for a quadratic, those inputs always sum to twice the x-coordinate of the vertex. Look for the axis of symmetry formula $x = -\frac{b}{2a}$.

Question 19

1. 11
2. 13 (correct answer)
3. 31
4. 28

Explanation: First find a: if g(a) = 10, then 3a - 2 = 10, so 3a = 12, thus a = 4. Then g(a + 1) = g(5) = 3(5) - 2 = 13. Choice A incorrectly adds 1 to g(a). Choice C uses g(3a) instead of g(a+1). Choice D uses g(a) + 3a incorrectly.

Question 20

1. The value 7 is in the domain of $f$ and appears exactly twice in the range of $f$
2. The range of $f$ contains exactly three elements and the domain contains exactly three elements
3. The function $f$ is not a valid function because two different inputs produce the same output
4. The value 7 is in the range of $f$ and the function is not one-to-one over the given points (correct answer)

Explanation: When you encounter a problem giving you specific function values, you're being tested on fundamental concepts about functions, domains, ranges, and the relationships between inputs and outputs. Let's analyze what the given information tells us. We know $f(2) = 7$, $f(5) = 7$, and $f(8) = 12$. This means that when we input 2 or 5, we get the same output of 7, while inputting 8 gives us 12. This immediately tells us two important things: the value 7 is definitely in the range of $f$ (since it's an output), and the function is not one-to-one over these points because two different inputs (2 and 5) produce the same output (7). This makes choice D correct. Now let's examine why the other choices are wrong. Choice A incorrectly states that 7 is in the domain - but 7 is an output value, so it's in the range, not the domain. The domain consists of the input values: 2, 5, and 8. Choice B makes an unjustified assumption about the complete function. We only know information about three specific points, but the function could have many more elements in both its domain and range. Choice C reflects a common misconception - functions are absolutely allowed to have multiple inputs produce the same output. What's forbidden is one input producing multiple outputs. Remember: functions can be many-to-one (multiple inputs giving the same output) but never one-to-many. When analyzing function problems, carefully distinguish between domain (inputs) and range (outputs), and don't assume the given information represents the entire function unless explicitly stated.