SAT Domain And Range Practice Questions | Download PDF

Correct answer: B) All real numbers

A linear function has no denominator, square root, or other restriction. Any real value of x can be used, so the domain is all real numbers.

SAT trap: Do not assume every function has a visible restricted domain. Linear functions usually have domain all real numbers.

Correct answer: D) All real numbers

A polynomial function is defined for every real x-value. There is no denominator or even root that creates a restriction.

SAT trap: A quadratic may have a limited range, but its domain is still all real numbers unless a real-world context limits it.

Correct answer: B) 6

The denominator cannot be 0. Set x – 6 = 0, so x = 6 is excluded.

SAT trap: For rational expressions, set the denominator equal to zero, not the numerator.

Correct answer: D) x ≠ -2

The denominator x + 2 cannot equal 0. Therefore x cannot equal -2.

SAT trap: The excluded value has the opposite sign of the number shown in the denominator.

Correct answer: B) x ≠ -3 and x ≠ 3

Factor the denominator: x² – 9 = (x – 3)(x + 3). The denominator is zero at x = 3 and x = -3, so both are excluded.

SAT trap: Even if a factor cancels later, the original denominator still creates a domain restriction.

Correct answer: D) x ≥ 4

The expression inside an even root must be nonnegative. Solve x – 4 ≥ 0 to get x ≥ 4.

SAT trap: For square roots, include the boundary when the radicand can equal zero.

Correct answer: B) x ≥ -5

Set the radicand greater than or equal to 0: 2x + 10 ≥ 0. Then 2x ≥ -10, so x ≥ -5.

SAT trap: Do not flip an inequality sign unless you multiply or divide by a negative number.

Correct answer: D) x ≤ 4

Require 12 – 3x ≥ 0. Then -3x ≥ -12, and dividing by -3 flips the sign, giving x ≤ 4.

SAT trap: When solving a square-root domain problem, sign flips still matter.

Correct answer: B) x > 1

The expression inside the square root must be positive because the square root is in the denominator. x – 1 > 0, so x > 1.

SAT trap: A square root in the denominator cannot equal zero, so use > instead of ≥.

Correct answer: D) x ≠ 2

The denominator factors as (x – 2)². It equals zero at x = 2, so x = 2 must be excluded.

SAT trap: A repeated factor still creates a restriction.

Correct answer: B) x ≥ -4 and x ≠ 3

The square root requires x + 4 ≥ 0, so x ≥ -4. The denominator requires x – 3 ≠ 0, so x ≠ 3. Both conditions must be true.

SAT trap: For combined expressions, handle every restriction, not just the first one you see.

Correct answer: D) x ≠ -3 and x ≠ 4

Factor the denominator: x² – x – 12 = (x – 4)(x + 3). Exclude x = 4 and x = -3.

SAT trap: Factor carefully. The signs in the factors determine the excluded values.

Correct answer: B) -3 ≤ x ≤ 3

The radicand must be nonnegative: 9 – x² ≥ 0. This means x² ≤ 9, so -3 ≤ x ≤ 3.

SAT trap: When x is squared, solve both sides of the interval, not just x ≤ 3.

Correct answer: D) x ≥ -2 and x ≠ 14

The square root requires x + 2 ≥ 0, so x ≥ -2. The denominator cannot be zero, so √(x + 2) – 4 ≠ 0. That gives √(x + 2) ≠ 4, so x + 2 ≠ 16 and x ≠ 14.

SAT trap: Check whether a square-root expression appears inside a denominator.

Correct answer: B) 5

The square root requires x ≥ 2. The denominator excludes x = 5 and x = -5. Among the choices, 5 is not in the domain.

SAT trap: A value can satisfy the square-root condition but still fail because of the denominator.

Correct answer: D) y ≥ 0

A square is never negative. The smallest value of x² is 0, so the range is y ≥ 0.

SAT trap: Do not confuse domain and range. The domain is all real numbers, but the range is restricted.

Correct answer: B) y ≥ 3

The minimum value of x² is 0. Adding 3 shifts the graph up, so the minimum value is 3.

SAT trap: A vertical shift changes the range but not the domain.

Correct answer: D) y ≥ -5

The vertex is (2, -5), and the parabola opens upward. The minimum value is -5, so the range is y ≥ -5.

SAT trap: The range uses the y-value of the vertex, not the x-value.

Correct answer: B) y ≤ 4

The parabola opens downward because the coefficient is negative. The vertex is (-1, 4), so the maximum value is 4 and the range is y ≤ 4.

SAT trap: A negative coefficient makes the vertex a maximum, not a minimum.

Correct answer: D) y ≥ 1

The coefficient 2 is positive, so the parabola opens upward. The vertex y-value is 1, so the range is y ≥ 1.

SAT trap: The coefficient changes width, but the vertex y-value still gives the boundary of the range.

Correct answer: B) y ≤ -7

The coefficient is negative, so the graph opens downward. The vertex is (-2, -7), making -7 the maximum value.

SAT trap: A maximum can be negative. Focus on the vertex and opening direction.

Correct answer: D) y ≥ 0

A square-root output is always nonnegative. The smallest value occurs at x = 5, where f(x) = 0.

SAT trap: For a basic square-root function, the range often starts at 0 even when the domain starts at another number.

Correct answer: B) y ≥ -4

The square-root part is at least 0. Subtracting 4 shifts the range down, so y ≥ -4.

SAT trap: Vertical shifts affect the range endpoint.

Correct answer: D) y ≤ 6

The square-root part is nonnegative. Multiplying by -1 makes it nonpositive, then adding 6 gives outputs less than or equal to 6.

SAT trap: A negative sign before the square root reverses the range direction.

Correct answer: B) y ≥ 0

Absolute value is always nonnegative. The smallest value is 0 when x = 4.

SAT trap: The value inside the absolute value changes the x-location of the vertex, not the minimum output.

Correct answer: D) y ≥ -3

The minimum value of |x + 2| is 0. Subtracting 3 makes the minimum value -3.

SAT trap: The number outside the absolute value controls the range boundary.

Correct answer: B) y ≤ 8

The negative sign makes the absolute value graph open downward. The largest value occurs when |x – 5| = 0, so the maximum output is 8.

SAT trap: A negative absolute value has a maximum, not a minimum.

Correct answer: D) y ≥ 9

The squared term is at least 0. Adding 9 makes the minimum output 9.

SAT trap: Do not use the x-coordinate -4 as a range value.

Correct answer: B) y ≤ 5

This can be written as -(x – 1)² + 5. The graph opens downward and has maximum value 5.

SAT trap: Rewrite the expression in vertex form before deciding the range.

Correct answer: D) y ≥ 1

Complete the square: x² – 6x + 10 = (x – 3)² + 1. The minimum output is 1.

SAT trap: Standard form does not show the range boundary directly. Convert or use the vertex formula.

Correct answer: B) {-2, 0, 3, 6}

The domain is the set of input values, which are the x-coordinates: -2, 0, 3, and 6.

SAT trap: The domain comes from x-values, not y-values.

Correct answer: D) {1, 4, 5}

The range is the set of output values. The y-values are 5, 1, 4, and 1. Listing unique values gives {1, 4, 5}.

SAT trap: Repeated output values are listed only once in a set.

Correct answer: B) {2, 8, 9}

The range is made from the output values. The unique y-values are 2, 8, and 9.

SAT trap: The range is not the full table. It is only the output set.

Correct answer: D) -4 ≤ x ≤ 2

The x-values covered by the segment begin at -4 and end at 2. Both endpoints are included, so -4 ≤ x ≤ 2.

SAT trap: For a line segment, the domain is restricted by the x-coordinates of the endpoints.

Correct answer: B) 1 ≤ y ≤ 7

The y-values covered by the segment go from 1 to 7. Since the endpoints are included, the range is 1 ≤ y ≤ 7.

SAT trap: Do not report the x-interval when the question asks for range.

Correct answer: D) -1 < x ≤ 4

The open circle at -1 means -1 is excluded. The closed circle at 4 means 4 is included. Therefore the domain is -1 < x ≤ 4.

SAT trap: Open and closed endpoints matter in interval questions.

Correct answer: B) y ≥ -6

The graph opens upward from its lowest point. The smallest output is -6, so the range is y ≥ -6.

SAT trap: A vertex gives the range boundary only when you know whether the graph opens upward or downward.

Correct answer: D) y ≤ 10

The graph opens downward from its highest point. The maximum output is 10, so the range is y ≤ 10.

SAT trap: The range boundary is the y-coordinate of the highest point.

Correct answer: B) {4, 6, 8, 10}

The domain is restricted to the four listed inputs, so the outputs are only 4, 6, 8, and 10.

SAT trap: Do not assume values between the listed table entries are included unless the problem says so.

Correct answer: D) x ≥ 1

The ray starts at x = 1 and continues to the right. The closed endpoint includes x = 1, so the domain is x ≥ 1.

SAT trap: A ray has one endpoint and continues forever in one direction.

Correct answer: B) y ≥ -2

Since the ray starts at y = -2 and increases, all output values are at least -2. The endpoint is included.

SAT trap: For range, track vertical movement, not horizontal movement.

Correct answer: D) {3}

Every point on the segment has the same y-value, 3. Therefore the range is the single value {3}.

SAT trap: A horizontal segment can have many x-values but only one output value.

Correct answer: B) {2}

Every point on the segment has x = 2, so the domain is the single value {2}.

SAT trap: A vertical line segment is not a function of x, but its domain can still be described as a set of x-values.

Correct answer: D) {-1, 0, 8, 15}

Evaluate the function for each allowed input: f(-4)=15, f(-1)=0, f(0)=-1, and f(3)=8. The range is {-1, 0, 8, 15}.

SAT trap: When a domain is explicitly listed, use only those input values.

Correct answer: B) The function has at least one output equal to 5.

The range is the set of output values. If 5 is in the range, then at least one input produces an output of 5.

SAT trap: Range information alone does not determine the domain or the graph type.

Correct answer: D) m ≥ 0

Miles cannot be negative in this context. A ride can be 0 miles in the model, so m ≥ 0 is the reasonable domain.

SAT trap: Real-world contexts often restrict the domain even when the equation alone would allow all real numbers.

Correct answer: B) n = 0, 1, 2, 3, …

The number of notebooks must be a whole number and cannot be negative. The reasonable domain is nonnegative integers.

SAT trap: Counting items usually requires whole-number inputs, not every decimal value.

Correct answer: D) t ≥ 0 until the ball reaches the ground

Time after the throw starts at t = 0 and continues until the ball returns to the ground. Negative time values are not part of the physical situation.

SAT trap: The algebraic quadratic may accept all real t-values, but the model context does not.

Correct answer: B) 0 ≤ p ≤ 600, where p is an integer

The number of people cannot be negative and cannot exceed 600. It must also be a whole number.

SAT trap: Capacity problems usually have both lower and upper domain limits.

Correct answer: D) 0 < x < 20

The length x must be positive, and the width 20 – x must also be positive. Therefore 0 < x < 20.

SAT trap: If both dimensions must be positive, endpoints that make one dimension zero are not included.

Correct answer: B) 0 ≤ h ≤ 12

The student cannot practice a negative number of hours and is limited to 12 hours, so 0 ≤ h ≤ 12.

SAT trap: A variable can be continuous in hours but still limited by the context.

Correct answer: D) {1, 2, 3, …, 12}

The number of months is limited to whole numbers from 1 through 12. The domain is the set {1, 2, 3, …, 12}.

SAT trap: A range of values is not always continuous. The word ‘whole months’ makes the domain discrete.

Correct answer: B) 1800

The function is in vertex form and opens downward. The vertex is (30, 1800), so the maximum profit is 1800.

SAT trap: The maximum output is the y-value of the vertex, not the x-value.

Correct answer: D) 0 ≤ h ≤ 80

The vertex gives the maximum height of 80. Since height cannot be negative in the physical situation, the possible heights run from 0 to 80.

SAT trap: Context can add a lower range limit even when the algebraic equation continues below zero.

Correct answer: B) 0 ≤ s ≤ 120, where s is an integer

The number of open spaces cannot be negative and cannot exceed 120. It also must be a whole number.

SAT trap: Counting variables usually require integer values.

Correct answer: D) 0 ≤ B ≤ 100

Battery percentage cannot be below 0 or above 100 in the context. The range is 0 ≤ B ≤ 100.

SAT trap: Use context to restrict outputs, even when a model might overshoot outside the realistic interval.

Correct answer: B) Multiples of 15 from 0 to 750

Since x is a whole number from 0 to 50, revenue values are 15(0), 15(1), …, 15(50), which are multiples of 15 from 0 to 750.

SAT trap: A discrete domain usually creates a discrete range.

Correct answer: D) 4 ≤ A ≤ 81

Area is s². At s = 2, A = 4. At s = 9, A = 81. Since s is positive on this interval, the area ranges from 4 to 81.

SAT trap: Do not copy the side-length interval as the area range.

Correct answer: B) 10 ≤ p ≤ 40, where p is an integer

The number of parts must be a whole number and is between 10 and 40, inclusive.

SAT trap: Look for whether endpoints are included and whether the quantity must be counted.

Correct answer: D) 13 ≤ C ≤ 53

Evaluate the endpoints: C(1)=8(1)+5=13 and C(6)=8(6)+5=53. Since the function increases, the range is 13 ≤ C ≤ 53.

SAT trap: Find output values by substituting the domain endpoints into the function.

Correct answer: B) y ≤ 5

Use the vertex formula: x = -b/(2a) = -12/(2(-2)) = 3. Then f(3) = -18 + 36 – 13 = 5. Since a is negative, the range is y ≤ 5.

SAT trap: For standard-form quadratics, find both the vertex x-value and the corresponding y-value.

Correct answer: D) y ≥ 7

Complete the square: 3x² + 6x + 10 = 3(x² + 2x) + 10 = 3(x + 1)² + 7. The range is y ≥ 7.

SAT trap: When factoring out 3, remember that the completed-square adjustment is multiplied by 3.

Correct answer: B) x ≤ -4 or x ≥ 4

Require x² – 16 ≥ 0, so x² ≥ 16. That means x ≤ -4 or x ≥ 4.

SAT trap: For x² ≥ a positive number, the solution is outside the interval, not inside.

Correct answer: D) x ≥ -5 and x ≠ -1 and x ≠ 1

The square root requires x ≥ -5. The denominator x² – 1 = (x – 1)(x + 1) excludes x = 1 and x = -1. Combine the conditions.

SAT trap: Combined expressions require combined restrictions.

Correct answer: B) 7

The denominator x – a equals zero when x = a. If the excluded input is 7, then a = 7.

SAT trap: The excluded value is the value that makes the denominator zero.

Correct answer: D) k = -4

For a positive squared function in vertex form, the minimum output is k. Since the range begins at -4, k = -4.

SAT trap: The range boundary in vertex form comes from k, not h.

Correct answer: B) a < 0

The range y ≤ 5 means the vertex y-value 5 is a maximum. A quadratic has a maximum only when it opens downward, so a < 0.

SAT trap: Range direction reveals whether the parabola opens upward or downward.

Correct answer: D) x < 5

The square root is in the denominator, so the radicand must be positive: 5 – x > 0. Therefore x < 5.

SAT trap: A denominator with a square root requires a strict inequality.

Correct answer: B) y ≤ 4

The value |x + 3| is always nonnegative. Subtracting it from 4 gives a maximum value of 4. The range is y ≤ 4.

SAT trap: A negative absolute value expression has no lower bound but has a maximum.

Correct answer: D) Domain: x ≥ 1; Range: y ≥ 7

The square root requires x – 1 ≥ 0, so x ≥ 1. The square-root output is at least 0, so after adding 7, the range is y ≥ 7.

SAT trap: Domain comes from the inside of the square root; range comes from the output after vertical shifts.