Math // Algebra

Quadratic Graphs

Recognise that quadratic functions produce curved (parabolic) graphs, distinguish them from linear graphs, and use plotted quadratic graphs to estimate values and find approximate solutions

Ages
13 to 14
Type
Concept
Builds on
1 topic
Leads to
1 topic
Standards
4 codes

Quick check

Can the learner recognise that an equation like y = x² produces a curved graph rather than a straight line, and use a plotted curve to estimate values?

Ask this to see whether the topic is secure.

Topic // Mastery

What mastery looks like

What a learner can do once this topic is secure.
  • Explain why x² produces a U-shaped curve rather than a straight line
  • Plot a simple quadratic such as y = x² − 4 from a table of values
  • Read approximate solutions from a quadratic graph (e.g., where y = 0)

Topic // Map

Where this fits

What to learn first and what this topic opens up, with the reason for each link. Required links come first.
  • RequiredMust come first
  • HelpfulHelps, but not essential

Before this

1 topic
  • Plotting Linear GraphsRequired

    Algebra // Ages 12 to 14

    Understanding quadratic graphs requires prior experience plotting linear graphs

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Quadratic Graphs

Ages 13 to 14 // Concept

Leads to

1 topic

Topic // Research-informed

Teaching notes

From the Marble Skill Taxonomy. These notes have not been validated in classrooms, so treat them as a starting point for your own judgement.

The approach

Help a 13–14 year old see that a quadratic function (one with a nonzero x² term) produces a smooth curve called a parabola, rather than a straight line. It is U-shaped when the x² coefficient is positive and an upside-down U when it is negative. With y = x² (and shifts such as y = x² + c), two linked reasons can be seen: the size of the rate of change grows as x moves away from the vertex at x = 0, and squaring makes +x and −x give the same y, so the curve is symmetric about the y-axis. (Every parabola is symmetric about a vertical line through its vertex. But for quadratics with an x term, such as y = x² + 2x, that line is not the y-axis, and +x and −x do not give equal y.) The core work is building a table of values that includes negative x, plotting the points, joining them with a smooth curve, contrasting directly with a linear graph, explaining the mirror symmetry, and then reading approximate roots (where y = 0) off the curve. Dynamic auto-plotting with coefficient sliders, a static annotated reference, and recognition checks each play distinct, complementary roles.

Common misconceptions

  • Expects a straight line (or draws one), not distinguishing a quadratic from a linear function.
  • Squares a negative wrongly ((−3)² = −9), so the left branch is plotted below the axis and the symmetry is lost.
  • Joins the plotted points with straight line segments rather than a smooth curve.
  • Reads only one root, missing the parabola's second, symmetric crossing.

Useful representations

  • Table of values including negative x → plotted points → smooth parabola
  • The parabola with its axis of symmetry and turning point (vertex) marked
  • Side-by-side straight line (y = 2x) and curve (y = x²) from their equations
  • x-axis crossings shown as the roots / approximate solutions

Teaching practices

  • Build from a table spanning negative and positive x (e.g. x = −3..3 for y = x²) so the symmetry and the turning point emerge from the numbers, not from assertion.
  • Insist on a SMOOTH curve through the points, never straight segments, and discuss why: the gradient changes continuously.
  • Contrast a linear and a quadratic graph side by side from their equations, tying 'straight vs curved' to the presence of the x² term.
  • Make the symmetry explicit: (−3)² = 3² = 9, so the point at x = −3 sits at the same height as x = 3 and the curve folds about its axis of symmetry.
  • Use the graph to read roots (where the curve meets the x-axis) and note that a parabola can cross the axis twice, once, or not at all.

Activities to try

  1. 1

    Plot y = x² from a table that includes negative x

    Why it works: Because (−3)² = 9 = 3², (−2)² = 4 = 2², and so on, the negative-x points land at the same heights as their positive partners, so the graph is a smooth U-shaped parabola symmetric about the y-axis with its lowest point at (0, 0) — not a straight line, because the amount y increases grows as x moves away from zero.

    1. Before plotting, predict the height of y = x² at x = −3 and write down both the value AND your reason.
    2. Complete a table of values for y = x² across x = −3, −2, −1, 0, 1, 2, 3, computing each y yourself.
    3. Plot each point, then join them with a single smooth curve (not straight segments).
    4. Compare the point at x = −3 with the point at x = 3: notice both sit at height 9.
    5. State in your own words why the two sides of the curve mirror each other.
  2. 2

    Line versus curve: contrast y = x with y = x²

    Why it works: A graph is straight only when y changes by a constant amount for each equal step in x; the x² term makes those step sizes vary — shrinking near the vertex and growing as x moves away from zero — so a quadratic must curve. 'Straight vs curved' is tied to the presence of the x² term, not merely to 'being an equation'.

    1. Build one table that holds columns for both y = x and y = x² over the same x-values, including negatives.
    2. Plot both on the same axes and describe how the shapes differ.
    3. Look at how each y-column changes step by step: y = x goes up in constant steps, while y = x² changes by amounts that are not constant — they shrink approaching the vertex at x = 0 and grow as x moves away from zero.
    4. Explain, tying it to the equation, why the x² term forces a curve while the linear term gives a straight line.
    5. Try to invent an equation whose graph would be a straight line, and one whose graph would curve — and say what distinguishes them.
  3. 3

    Read roots and explore hill/valley and shifts with an auto-plotting table and sliders

    Why it works: The axis of symmetry, the turning point (vertex), whether the parabola is a hill or a valley, and the number of roots (two, one, or none, shown where the curve meets the x-axis) are the key features of a quadratic graph; shifting c moves the whole curve while preserving its shape, and roots are the approximate solutions where y = 0.

    1. Fill a table for y = x² with points that plot automatically as you enter negative and positive x; watch the x = −3 and x = 3 points land at the same height.
    2. Toggle the join between straight segments and a smooth curve to see the scattered points resolve into a single U.
    3. Drag the coefficient 'a' below zero and describe what happens to the valley (it inverts into a hill).
    4. With 'a' set back to a positive value (e.g. a = 1), shift the constant 'c' downward to slide the whole parabola down until it cuts the x-axis twice, and read off the approximate roots where y = 0 (shifting upward instead lifts this valley clear of the axis; for a hill with negative 'a' these effects are reversed).
    5. For y = x² − 4, state what changes and what stays the same compared with y = x², then read its roots from the graph.

Topic // Standards

Curriculum standards

Codes for the curriculum statements this topic maps to. Each publisher owns its standard text.
Common Core State Standards for Mathematics
8.F.38.F.5
The national curriculum in England: Key stages 1 and 2 framework document
KS3.Maths.Alg.12KS3.Maths.Alg.9

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