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How do you determine the domain and range of a quadratic function, and connect its representations to the situation it models?

Determine the domain and range of quadratic functions and represent them using inequalities, and describe representations of quadratic functions in relation to their solutions and the real-world situations they model (TEKS A.6A, A.6B).

A STAAR Algebra I answer on the domain and range of quadratic functions (TEKS A.6A, A.6B), why the range is bounded by the vertex, representing with inequalities, and connecting representations to real-world models.

Generated by Claude Opus 4.810 min answer

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  1. What this topic is asking
  2. Domain: all real numbers (unless restricted)
  3. Range: bounded by the vertex
  4. Representations and real-world models
  5. How STAAR examines this topic
  6. Finding the range when only standard form is given
  7. Why the domain and range describe different axes
  8. Try this

What this topic is asking

TEKS A.6A and A.6B sit in the Quadratic Functions category and ask two connected things: find the domain and range of a quadratic (with inequalities), and describe its representations in relation to its solutions and the real-world situation it models. The defining feature is that a parabola's range is bounded by its vertex, unlike a line.

Domain: all real numbers (unless restricted)

For a bare quadratic f(x)=ax2+bx+cf(x) = ax^2 + bx + c, you can input any real number, so the domain is all real numbers. The parabola never stops extending sideways. This is the same as a line, and it is why the domain is rarely the interesting part of a quadratic question, except in a context.

Range: bounded by the vertex

The range is where quadratics differ from lines. Because a parabola turns around at its vertex, the outputs reach a single extreme and then reverse.

  • Opens up (a>0a > 0). The vertex is the lowest point. The range is yky \ge k, all outputs at or above the minimum.
  • Opens down (a<0a < 0). The vertex is the highest point. The range is yky \le k, all outputs at or below the maximum.

So f(x)=x24f(x) = x^2 - 4 (vertex (0,4)(0, -4), opening up) has range y4y \ge -4, and g(x)=(x1)2+5g(x) = -(x - 1)^2 + 5 (vertex (1,5)(1, 5), opening down) has range y5y \le 5.

Representations and real-world models

A quadratic can be shown as an equation, a graph, or a table, and each highlights different features: the graph shows the vertex and intercepts; the factored equation shows the zeros; vertex form shows the maximum or minimum. A.6B asks you to connect these to a situation. For a projectile, the vertex is the peak height, the zeros are launch and landing times, and the range is the set of achievable heights, restricted to non-negative values. Matching the algebra to the physical meaning is the assessed skill.

How STAAR examines this topic

  • Multiple choice. Choose the range (the vertex-bounded inequality), with the domain offered as a distractor.
  • Inline choice. Identify where a maximum occurs (the vertex), and the restricted range in context.
  • Drag and drop. Match a representation to a feature or a situation.

A clarifying idea is that the vertex is the gatekeeper of the range: find the vertex and the opening direction, and the range follows immediately as "at or above" or "at or below" the vertex's height.

Finding the range when only standard form is given

If a quadratic arrives in standard form rather than vertex form, you still need the vertex's yy-coordinate to state the range. Compute the axis of symmetry x=b2ax = \frac{-b}{2a}, substitute it back to get the minimum or maximum value kk, then write yky \ge k or yky \le k depending on the opening direction. For f(x)=x26x+5f(x) = x^2 - 6x + 5, the axis is x=3x = 3, and f(3)=918+5=4f(3) = 9 - 18 + 5 = -4, so the range is y4y \ge -4. Skipping the substitution and guessing the range from the constant term cc is a common error, because cc is the yy-intercept, not the vertex value, unless the vertex happens to lie on the yy-axis.

Why the domain and range describe different axes

Keeping domain and range straight is easier when you remember they answer different questions. The domain asks which horizontal positions the curve occupies, and a parabola spans the whole xx-axis, so the answer is all real numbers. The range asks which heights the curve reaches, and a parabola only reaches heights on one side of its vertex, so the answer is a half-line in yy. Because both can be written with inequalities that look similar, the discriminator on a multiple-choice item is usually whether the variable is xx (domain) or yy (range) and whether the bound is the vertex value.

Try this

Q1. State the range of f(x)=(x+2)2+3f(x) = -(x + 2)^2 + 3. [1 point]

  • Cue. Vertex (2,3)(-2, 3), opens down: y3y \le 3.

Q2. What is the domain of f(x)=x2+5x1f(x) = x^2 + 5x - 1 with no restriction? [1 point]

  • Cue. All real numbers.

Exam-style practice questions

Practice questions written in the style of TEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

STAAR (style)1 marksMultiple choice. What is the range of f(x)=x24f(x) = x^2 - 4? (A) y4y \ge -4 (B) y4y \le -4 (C) all real numbers (D) x4x \ge -4
Show worked answer →

The correct answer is (A).

The parabola opens up (a=1>0a = 1 > 0), so it has a minimum at its vertex. The vertex of f(x)=x24f(x) = x^2 - 4 is (0,4)(0, -4), so the lowest output is 4-4 and outputs go upward from there: the range is y4y \ge -4. Choice (C) is the domain (all real numbers), a frequent mix-up; a quadratic's range is bounded by the vertex while its domain is unrestricted.

STAAR (style)2 marksInline choice. A ball's height is h(t)=16t2+32th(t) = -16t^2 + 32t for t0t \ge 0 until it lands. The maximum height occurs at the [vertex / y-intercept], and the range of heights is [0 to 16 / all real numbers], in feet.
Show worked answer →

The maximum height occurs at the vertex, and the range of heights is 0 to 16 feet.

The parabola opens down, so the vertex is the maximum. The axis is t=322(16)=1t = \frac{-32}{2(-16)} = 1, and h(1)=16(1)+32=16h(1) = -16(1) + 32 = 16 feet, the peak. Heights run from 0 (ground, at launch and landing) up to 16, so the range is 0h160 \le h \le 16. Real-world quadratics restrict the range to physically possible values, here non-negative heights up to the peak.

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