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Study Guide11th GradeLevel 2 · Standard

Free 11th Grade Functions, Domain and Range

A senior study guide to function notation, the vertical line test, and finding the domain and range of common functions. Worked examples cover evaluating f(a + 1), square-root and reciprocal restrictions, completing the square to find a range, and restricted domains, with a six-question self-check and answers.

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Grade
11th Grade
Subject
Math
Topic
Functions
Difficulty
Level 2 · Standard
Estimated time
25 minutes
Curriculum
Texas grade levels (TEKS)
Answers
Not applicable
Format
PDF (A4) + print

Students will practice

  • using function notation to evaluate and substitute expressions
  • deciding whether a relation is a function using the vertical line test
  • finding the natural domain of square-root and rational functions
  • finding the range of a function by completing the square or using a restricted domain

Curriculum: Texas grade levels (TEKS). We show specific outcome codes only where they have been verified against the official curriculum document.

What's next?

Completed: Functions, Domain and Range

  1. 1Polynomials: Division, Factor and Remainder Theorems
  2. 211th Grade Function Notation, Composites and Inverses Worksheet — Level 1
  3. 311th Grade Function Notation, Composites and Inverses Worksheet — Level 2
  4. 411th Grade Function Notation, Composites and Inverses Worksheet — Level 3
Study guide

How to use this study guide

  1. Read it together first, pausing at each worked example to try the step before reading the answer.
  2. Attempt the "Check yourself" questions at the end without looking back.
  3. Then practice with a worksheet from the pathway above and finish with the topic test.

Common questions

Who is this study guide for?

11th Grade students (typically ages 16–17) working on functions. It is pitched at level 2 · standard.

Are the answers included?

This is a study guide, so there is no separate answer sheet; the 'Check yourself' questions include answers.

How long does it take?

About 25 minutes. Short, regular sessions work best: two or three a week beats one long one.

Do I need to sign up to download?

No. Click Download Free PDF and it opens immediately. It is free for personal, classroom and homeschool use.

What should we do next?

Try Polynomials: Division, Factor and Remainder Theorems.

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Success TutoringWorksheets11th Grade · Mathematics · Functions

Functions, Domain and Range

What you need to know

A relation is any set of ordered pairs (x, y). A function is a relation in which every x-value is paired with exactly one y-value. On a graph, use the vertical line test: if any vertical line cuts the graph more than once, it is not a function. For example, y2 = x is not a function because x = 4 gives both y = 2 and y = −2. We write functions as f(x), read "f of x"; f(3) means "substitute x = 3 into the rule".

The domain is the set of all allowed x-values (inputs). The range is the set of all y-values the function actually produces (outputs). Unless a question restricts the domain, use the natural domain: every real x for which the rule makes sense. Two things break a rule: dividing by zero and taking the square root of a negative number.

FunctionNatural domainRange
y = x2all real xy ≥ 0
y = √xx ≥ 0y ≥ 0
y = x ≠ 0y ≠ 0
y = |x|all real xy ≥ 0
y = 2xall real xy > 0
y = ln xx > 0all real y

Evaluate f(x) = x2 − 3x + 1 at x = 2, x = −1 and x = a + 1

  1. f(2) = 22 − 3(2) + 1 = 4 − 6 + 1 = −1
  2. f(−1) = (−1)2 − 3(−1) + 1 = 1 + 3 + 1 = 5
  3. f(a + 1) = (a + 1)2 − 3(a + 1) + 1
  4. = a2 + 2a + 1 − 3a − 3 + 1
  5. = a[sup 2] − a − 1

Find the domain and range of f(x) = √(x − 3)

  1. The expression under the square root must not be negative: x − 3 ≥ 0.
  2. So the domain is x ≥ 3.
  3. A square root never gives a negative output, and it equals 0 when x = 3, so the range is y ≥ 0.

Find the domain of g(x) =

  1. The denominator must not be zero: x2 − 4 ≠ 0.
  2. x2 ≠ 4, so x ≠ 2 and x ≠ −2.
  3. Domain: all real x except x = ±2. (On the graph, these are vertical asymptotes.)

Find the range of h(x) = x2 − 4x + 7

  1. The parabola opens upwards, so the range starts at the minimum value. Complete the square to find it.
  2. x2 − 4x + 7 = (x − 2)2 − 4 + 7 = (x − 2)2 + 3
  3. (x − 2)2 is never negative and equals 0 when x = 2, so the smallest value of h is 3.
  4. Range: y ≥ 3.

Restricted domain: f(x) = x2 for −1 ≤ x ≤ 3

  1. Do not just substitute the endpoints. f(−1) = 1 and f(3) = 9, but the turning point x = 0 lies inside the domain and gives f(0) = 0.
  2. Smallest value: 0 (at x = 0). Largest value: 9 (at x = 3).
  3. Range: 0 ≤ y ≤ 9.
Common mistake: mixing up the two kinds of restriction. A square root needs the inside to be greater than or equal to zero (≥), but a denominator needs to be not equal to zero (≠). For √(x − 3) the domain is x ≥ 3; for the domain is x ≠ 3.
Common mistake: working out f(a + 1) by adding 1 to f(a). You must replace every x in the rule with (a + 1), parentheses included, and then expand carefully.

Check yourself

  1. If f(x) = 2x2 − 5, find f(−3).
  2. State the domain of y = √(5 − x).
  3. State the domain of y = .
  4. State the range of y = x2 + 1.
  5. Find the range of f(x) = x2 − 6x + 10.
  6. f(x) = 3x − 2 for 0 ≤ x ≤ 4. Find the range.

Answers: 1. f(−3) = 2(9) − 5 = 13 2. x ≤ 5 3. x ≠ −2 4. y ≥ 1 5. (x − 3)2 + 1, so y ≥ 1 6. f(0) = −2 and f(4) = 10, so −2 ≤ y ≤ 10

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Last reviewed
1 October 2026
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