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Graphing Quadratics Basics

A free Algebra I lesson from the “Quadratic Foundations” unit, with a worked example and practice problems including step-by-step solutions.

A quadratic graph is a parabola, and different algebraic forms reveal different features. In y = a(x - h)^2 + k, the vertex is (h, k), the symmetry line is x = h, and the sign of a tells whether the vertex is a minimum or maximum. This follows because the squared term is never negative. In factored form y = a(x - r)(x - s), real zeros occur at r and s; they give the x-intercepts (r, 0) and (s, 0). In standard form y = ax^2 + bx + c, c gives the y-intercept (0, c). Completing the square connects standard and vertex forms. To sketch, mark the vertex and axis, calculate intercepts when they exist, and find one or two additional symmetric pairs. A zero may touch rather than cross the horizontal axis when it is a repeated root. An upward parabola decreases to its vertex and increases after it; a downward parabola does the reverse. Decreasing describes how outputs change as inputs increase, not whether those outputs are negative. Apply any stated domain restriction before describing the graph's range.

What you'll learn

Why it matters: Quadratic graphs reveal a highest height or lowest cost, while their intercepts answer different questions about when a quantity reaches zero.

Worked example

Problem. Analyze y = x^2 - 4x + 3: find its vertex, axis, intercepts, and where it decreases or increases, then describe a sketch.

  1. Complete the square: x^2 - 4x + 3 = (x - 2)^2 - 1. Therefore the vertex is (2, -1) and the axis is x = 2.
  2. The leading coefficient is positive, so the parabola opens upward and has minimum output -1.
  3. Factor to find zeros: x^2 - 4x + 3 = (x - 1)(x - 3), giving x-intercepts (1, 0) and (3, 0).
  4. At x = 0, y = 3, so the y-intercept is (0, 3). Symmetry also gives (4, 3).
  5. Plot these five points and draw a smooth upward parabola symmetric about x = 2.
  6. Moving left to right, outputs fall until x = 2 and rise after x = 2. The entire real-domain range is y >= -1.

Answer: Vertex (2, -1); axis x = 2; x-intercepts (1, 0) and (3, 0); y-intercept (0, 3); decreases for x < 2 and increases for x > 2.

Lesson preview: this page shows 5 practice items. The interactive activity contains 32 practice items and 17 quiz items (typed response and multiple choice). Counts include repeated prompts and variants.

Practice problems

1. For y = -2(x - 3)^2 + 8, what is the maximum output?

Show solution
  1. The square is nonnegative, so -2(x - 3)^2 is at most 0.
  2. The largest output is therefore 8, attained when x = 3.

Answer: 8

2. Which statement about y = (x + 2)^2 is correct?

Choices: It touches the x-axis at (-2, 0) and has one distinct real zero. · It crosses the x-axis at both (-2, 0) and (2, 0). · It has no x-intercept because a square is nonnegative. · Its vertex is (2, 0).

Show solution
  1. The output is zero exactly when x + 2 = 0, or x = -2.
  2. Every other output is positive, so the graph touches the axis without crossing it.
  3. The repeated factor gives one distinct real zero.

Answer: It touches the x-axis at (-2, 0) and has one distinct real zero.

3. A student says y = (x - 4)^2 - 3 is decreasing wherever y is negative. What is the correct description?

Choices: It decreases for x < 4 and increases for x > 4, regardless of whether y is positive or negative. · It decreases for every real input because -3 is negative. · It increases only where y is positive. · It has no decreasing interval.

Show solution
  1. The upward-opening parabola has vertex input 4.
  2. As x moves toward 4 from the left, (x - 4)^2 decreases; after 4 it increases.
  3. The sign of an output describes its position relative to the horizontal axis, not the direction of change.

Answer: It decreases for x < 4 and increases for x > 4, regardless of whether y is positive or negative.

4. For y = x^2 - 6x + 5, the point (1, 0) lies on the graph. What is the x-coordinate of the other point at the same height, reflected across the symmetry axis?

Show solution
  1. Completing the square gives y = (x - 3)^2 - 4, so the axis is x = 3.
  2. Input 1 lies 2 units left of the axis. The reflected input is 2 units right: 3 + 2 = 5.
  3. Indeed 5^2 - 6(5) + 5 = 0.

Answer: 5

5. The graph of a quadratic is called a...

Choices: Parabola · Line · Circle · Histogram

Show solution
  1. Quadratic graphs are parabolas.

Answer: Parabola

Explain and apply

Complete these tasks on paper. Use the rubric to review your reasoning with a teacher or study partner; these tasks are not automatically scored.

Produce and justify a parabola sketch

Sketch y = -x^2 + 4x + 5. Show the vertex, symmetry axis, both x-intercepts, and the y-intercept. Explain why the graph has a maximum, and state the intervals on which its output increases and decreases.

Review rubric and solution
  • Rewrite as y = -(x - 2)^2 + 9 and label vertex (2, 9) with axis x = 2.
  • Find x-intercepts (-1, 0), (5, 0) and y-intercept (0, 5).
  • Draw a downward parabola consistent with those points and symmetry.
  • Explain maximum 9, increasing for x < 2 and decreasing for x > 2.
  1. Completing the square gives y = -(x - 2)^2 + 9, so the vertex is (2, 9) and the axis is x = 2.
  2. Factoring gives y = -(x - 5)(x + 1), so the x-intercepts are (5, 0) and (-1, 0). Substituting x = 0 gives (0, 5); symmetry gives (4, 5).
  3. Draw the smooth downward parabola through these points. Since a nonnegative square is subtracted from 9, the maximum output is 9.
  4. The graph rises as x approaches 2 from the left and falls after 2; it increases on (-infinity, 2) and decreases on (2, infinity).

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