Lesson library
A.6(A)·Algebra II·Quadratic FunctionsReadiness

Quadratic Domain and Range

Determine reasonable domains and ranges of quadratic functions and interpret their solutions.

Students distinguish the unrestricted domain of a quadratic rule from a reasonable contextual domain, then use the vertex and direction of opening to determine range.

Lesson plan and worked examples by TEKSForge · not a TEA-issued lesson. Videos by independent educators.

Archived course: TEKS code and readiness label from TEA's 2013 Algebra II key. Algebra II is not one of Texas's current required STAAR EOC assessments, and this lesson does not change the Biology and Algebra I Study Plans.

Learn the concept

Start with the context

A polynomial quadratic is defined for every real input. A model can have a smaller reasonable domain. For a rectangle with perimeter 60 units, l+w=30l+w=30, so l=30−wl=30-w. Both w>0w>0 and 30−w>030-w>0 must hold.

Use the vertex for range

For y=a(x−h)2+ky=a(x-h)^2+k, the squared term is nonnegative. With unrestricted inputs, a>0a>0 gives y≥ky\ge k, while a<0a<0 gives y≤ky\le k. The vertex output k is included.

Check the endpoints

A rectangle with a zero side length is collapsed, so its width endpoints are excluded. The maximum area can still be included if the vertex input lies inside the reasonable domain. Restricting a domain may also change the range.

Worked example

A rectangle has perimeter 60 meters. Write its area in terms of width w and determine the reasonable domain and range.

  1. From 2l+2w=602l+2w=60, obtain l=30−wl=30-w.
  2. Area is A(w)=w(30−w)=30w−w2A(w)=w(30-w)=30w-w^2.
  3. Positive dimensions require w>0w>0 and 30−w>030-w>0, so 0<w<300<w<30.
  4. Rewrite A(w)=225−(w−15)2A(w)=225-(w-15)^2. The maximum is 225 at w=15w=15; every positive area up to 225 occurs.

A(w)=30w−w2A(w)=30w-w^2; domain 0<w<300<w<30 meters; range 0<A≤2250<A\le 225 square meters.

Watch the lesson

Texas public-school, Texas educator-created, and TEKS-aligned lessons, hosted by their creators. Each card names its source.

Texas public school video

Algebra 2 Analyzing Attributes of Graphs of Quadratic Functions

UT IPSI · Kid2Kid / Texas Education Agency

Use the vertex and direction of opening to determine a quadratic function's range.

Watch on YouTube

Try it yourself

TEKSForge practice · unscored

  1. 1. Find the range of g(x)=−(x−4)2+9g(x)=-(x-4)^2+9 for all real x.

    Show answer and reasoning

    y≤9y\le 9

    The parabola opens downward and has maximum output 9 at its vertex.

  2. 2. A rectangle has perimeter 84 inches. What widths are reasonable?

    Show answer and reasoning

    0<w<420<w<42 inches

    The length is 42−w42-w. Requiring both dimensions to be positive excludes 0 and 42.

Released STAAR practice

Texas Education Agency · archived 2013 released EOC items · scored with TEA's answer key; step-by-step explanations written by TEKSForge because this release has no rationales. Original graphs, tables, and diagrams are shown where the released item includes one.

Archived released STAAR Algebra II EOC · 2013 · Item 20 · 1 point

The function A=36w−w2A = 36w - w^2 can be used to model the area of a rectangle in square inches, AA, if the rectangle has a perimeter of 72 inches and a width of ww inches. In this situation, which of the following best describes the domain of the function?

  1. 0<w<60 < w < 6
  2. 0<w<720 < w < 72
  3. 0<w<180 < w < 18
  4. 0<w<360 < w < 36
Show STAAR answer and explanation

0<w<360 < w < 36

Half the perimeter is 36, so the length is 36−w36-w. Both dimensions must be positive: w>0w>0 and 36−w>036-w>0, so 0<w<360<w<36.

0<w<60 < w < 6: 6 is the square root of 36. Widths between 6 and 36 still give positive lengths.

0<w<720 < w < 72: 72 is the full perimeter. A width over 36 inches would make the length negative.

0<w<180 < w < 18: A width of 18 gives the maximum area, but wider rectangles up to 36 inches are still possible.

Archived released STAAR Algebra II EOC · 2013 · Item 43 · 1 point

What is the range of f(x)=x2+1f(x) = x^2 + 1?

  1. All real numbers
  2. All real numbers greater than or equal to 0
  3. All real numbers greater than or equal to 1
  4. All real numbers less than or equal to 1
Show STAAR answer and explanation

All real numbers greater than or equal to 1

Since x2≥0x^2\ge0 for every x, f(x)=x2+1≥1f(x)=x^2+1\ge1. The vertex (0,1)(0,1) is the minimum, and the parabola opens upward.

All real numbers: A parabola that opens upward has a minimum, so it cannot reach every real number.

All real numbers greater than or equal to 0: 0 is the range of x2x^2; adding 1 shifts every output up 1.

All real numbers less than or equal to 1: The parabola opens upward, so its outputs are at least 1, not at most 1.

Start scored STAAR practice

Learning objectives

By the end of the lesson, students can:

  • Determine a reasonable domain from positive physical dimensions.
  • Identify a quadratic function's minimum or maximum output.
  • Write domain and range with correct endpoint inclusion.

Instructional sequence

  1. 1

    Engage

    5 min

    Ask whether a rectangle can have a negative side length even when an area formula accepts that input.

  2. 2

    Direct instruction

    12 min

    Separate algebraic domain from contextual restrictions. Connect range to the vertex and opening.

  3. 3

    Guided practice

    13 min

    Write both dimensions of fixed-perimeter rectangles and require each dimension to be positive.

  4. 4

    Graph connection

    13 min

    Compare ranges of unrestricted parabolas with ranges after contextual restrictions.

  5. 5

    Closure

    7 min

    Students justify open and closed endpoints in the exit ticket.

Misconceptions to address

““Every quadratic has a contextual domain of all real numbers.””

Teacher move: The polynomial accepts every real input, but lengths, times, and counts may impose restrictions.

““The vertex x-coordinate is the greatest possible input.””

Teacher move: The vertex locates an extreme output. Inputs can often continue on both sides of it.