Algebra Quiz: Understanding Functions Domain And Range

20 questions · exam conditions

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Understanding Functions Domain And RangeQuestion 1 of 20

Let $g(x)=\dfrac{1}{x-5}$ . To keep $g$ a function with one output for each input, $g(x)$ must be defined for the input. For what values of $x$ is $g(x)$ defined? (Answer in words or interval notation.)

All real numbers except

x=5

All real numbers except

x=0

[5,\infty)

All real numbers

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Algebra Quiz

Algebra Quiz: Understanding Functions Domain And Range

Practice Understanding Functions Domain And Range in 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 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.

All questions

Question 1

All real numbers except $x=5$ (correct answer)
All real numbers except $x=0$
$[5,\infty)$
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

A function $f$ is defined by $f(x)=\sqrt{x-2}$ . Because a function assigns exactly one output to each input in its domain, we must restrict inputs so the output is real. What is the domain of $f$ ? (Answer in interval notation.)

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

The function $g$ is defined by $g(x)=\dfrac{5}{x-2}$ . For what values of $x$ is $g(x)$ defined? (Give your answer in words or set description.)

All real numbers $x\ge 2$
All real numbers except $x=5$
All real numbers except $x=2$ (correct answer)
All real numbers $x\le 2$

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) = 5/(x-2), we need to avoid division by zero, which happens when the denominator x - 2 = 0, so x = 2 must be excluded from the domain. Choice A is correct because it states that g(x) is defined for all real numbers except x = 2, which is exactly where the denominator becomes zero. Choice B incorrectly excludes x = 5, but plugging in x = 5 gives g(5) = 5/3, which is perfectly valid. 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 you see a fraction, always check where the denominator equals zero—those are your excluded values!

Question 4

The function $p(x)=\sqrt{x}$ assigns one output to each input in its domain. What is the range of $p$ ? (Answer in interval notation.)

$(-\infty,\infty)$
$[0,\infty)$ (correct answer)
$(-\infty,0]$
$(0,\infty)$

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

What is the domain of the function $f(x)=\sqrt{x-5}$ ? Give your answer in interval notation.

$(5,\infty)$
$(-\infty,5]$
$[5,\infty)$ (correct answer)
$(-\infty,\infty)$

Explanation: This question tests your understanding of what functions are, and how to determine their domains (possible inputs) and ranges (possible outputs). The 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). To find the domain of f(x) = √(x-5), set the inside of the square root greater than or equal to zero: x - 5 ≥ 0, so x ≥ 5. Choice C is correct because [5, ∞) includes x = 5 (where the output is 0) and all larger values, ensuring the square root is defined for real numbers. A common mistake is thinking it's (-∞, 5], but that would make x - 5 negative, which isn't allowed for real square roots—keep practicing identifying those restrictions! 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 6

The function $k$ is defined by $k(x)=\sqrt{9-x}$ . What is the domain of $k$ ? (Give your answer in interval notation.)

$( -\infty,\infty)$
$[9,\infty)$
$( -\infty,9)$
$( -\infty,9]$ (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 7

The function $p$ is defined by $p(x)=-(x-1)^2+4$ . What is the range of $p$ ? (Give your answer in interval notation.)

$[4,\infty)$
$(-\infty,4]$ (correct answer)
$(-\infty,4)$
$(-\infty,\infty)$

(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 8

The function $h$ is defined by $h(x)=x^2-9$ . What is the range of $h$ ? (Give your answer in interval notation.)

$( -\infty,\infty)$
$[-9,\infty)$ (correct answer)
$( -\infty,-9]$
$(-9,\infty)$

Explanation: This question tests your understanding of what functions are, and how to determine their domains (possible inputs) and ranges (possible outputs). The 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 h(x) = x² - 9, this is a parabola that opens upward (positive coefficient on x²) with vertex at (0, -9), found by completing the square or noting that x² has its minimum at x = 0. Choice C is correct because [-9,∞) represents all y-values greater than or equal to -9, with the square bracket showing -9 is included since h(0) = -9. Choice B uses a parenthesis at -9, incorrectly suggesting -9 isn't in the range, but it is the minimum value. 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). The function type tells you a lot!

Question 9

What are the possible output values (the range) of the function $p(x)=\sqrt{x}$ ? Give your answer in interval notation.

$( -\infty, \infty)$
$( -\infty, 0]$
$[0, \infty)$ (correct answer)
$(0, \infty)$

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, inputs are x ≥ 0, and outputs start at 0 (when x=0) and go to infinity as x increases, never negative. Choice C is correct because [0, ∞) includes 0 and all positive reals, matching the non-negative outputs of square roots. Choice B suggests negative outputs, but square roots can't produce negatives in real numbers—think about how √x is always ≥ 0! 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 10

For what values of $x$ is the function $g(x)=\dfrac{3}{x-2}$ defined? Give your answer in interval notation.

$(-\infty,2]\cup[2,\infty)$
$(-\infty,2)\cup(2,\infty)$ (correct answer)
$(-\infty,\infty)$
$[2,\infty)$

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 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 (-∞, 2) ∪ (2, ∞) includes everything except x=2, where the function would involve division by zero. If you chose B thinking it's all reals, that's understandable, but remember to always check for denominator zeros—it's a key restriction! 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 11

What is the domain of the function $h(x)=\sqrt{9-x}$ ? Give your answer in interval notation.

$( -\infty, 9)$
$( -\infty, 9]$ (correct answer)
$( -\infty, \infty)$
$[9, \infty)$

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 12

Let $r(x)=\dfrac{x+2}{x-1}$ . What is the domain of $r$ ? (Answer in interval notation.)

$(-\infty,1)\cup(1,\infty)$ (correct answer)
$(-\infty,\infty)$
$(-\infty,1]\cup(1,\infty)$
$(-\infty,-1)\cup(-1,\infty)$

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 r(x) = (x+2)/(x-1), we need to avoid division by zero, so the denominator x-1 cannot equal 0, which means x cannot equal 1. The domain is all real numbers except x = 1. Choice A is correct because (-∞,1)∪(1,∞) represents all real numbers less than 1 (using a parenthesis to exclude 1) union with all real numbers greater than 1 (again excluding 1 with a parenthesis). Choice D incorrectly excludes x = -1, but r(-1) = (-1+2)/(-1-1) = 1/(-2) = -1/2, 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 13

What is the range of the function $f(x)=x^2-4$ ? Give your answer in interval notation.

$( -4, \infty)$
$[-4, \infty)$ (correct answer)
$( -\infty, \infty)$
$( -\infty, -4]$

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 14

For what values of $x$ is the function $g(x)=\dfrac{3}{x-2}$ defined? Write your answer in words or using interval notation.

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

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 15

The function $f$ is defined by $f(x)=\sqrt{x-4}$ . What is the domain of $f$ ? (Give your answer in interval notation.)

$( -\infty,\infty)$
$( -\infty,4]$
$[4,\infty)$ (correct answer)
$(4,\infty)$

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-4), we need the expression inside the square root to be non-negative, so x - 4 ≥ 0, which means x ≥ 4. Choice B is correct because [4,∞) includes all real numbers greater than or equal to 4, with the square bracket showing that 4 itself is included since √0 = 0 is valid. Choice A would allow negative values under the square root, which isn't allowed in real numbers. 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 16

Consider the piecewise function $f(x) = \begin{cases} 2x + 1 & \text{if } x < 0 \\ x^2 & \text{if } x \geq 0 \end{cases}$ . What is $f(-2) + f(2)$ ?

1 (correct answer)
3
7
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 17

Function $h$ is defined by the equation $h(x) = \sqrt{x + 3}$ . Which statement about the domain of $h$ is correct?

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

If $g(x) = 3x - 2$ and $g(a) = 10$ , what is the value of $g(a + 1)$ ?

11
13 (correct answer)
31
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 19

Consider the relation $R = \{(2,5), (3,7), (2,9), (4,1)\}$ . If we remove exactly one ordered pair to make this relation a function, what is the range of the resulting function?

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

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 20

The height of a toy rocket above the ground is modeled by $H(t)=-2t^2+8t+1$ , where $t$ is time in seconds. The model is only valid from launch until it hits the ground at $t=4$ . What is the domain of $H$ in this context?

$(0,4)$
$( -\infty, \infty)$
$[0,4]$ (correct answer)
$[4, \infty)$

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 the rocket height H(t) = -2t² + 8t + 1, the context limits t from launch at t=0 to when it hits the ground at t=4, so only those times are meaningful. Choice C is correct because [0,4] includes t=0 (launch) and t=4 (landing), capturing the valid interval. Choice A ignores the context and assumes all real t, but negative time or after landing doesn't make sense here—always factor in the real-world scenario! 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.'