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DSE Maths: Quadratic Functions and Graphs — Tutor Lesson Plan

DSE Maths: Quadratic Functions and Graphs — Tutor Lesson Plan

  • 2026-10-27

Students solve equations but freeze at graphs. Three switches fix it: a opens the curve, the vertex gives the extreme value, the discriminant counts the intercepts.

Objectives — the student can complete the square, state the vertex and axis of symmetry, and use the discriminant to count x-intercepts.

01 Prerequisites (3-minute check)

If any of these fails, fix it before the lesson starts.

Skill Quick check Pass mark
Quadratic equations Solve x² − 5x + 6 = 0 x = 2 or 3
Completing the square Write x² − 4x as (x − 2)² − 4 Correct
Function notation What does f(2) mean? Substitute x = 2

02 Lesson flow (60 minutes)

Time Stage Tutor does Student does
0-8 min Three switches a, vertex, discriminant Repeat them
8-22 min Completing the square y = x² − 4x + 3 Two alone
22-36 min Reading graphs Vertex and axis from a graph Write both
36-50 min Discriminant Three cases of intercepts Match five
50-56 min Applications Maximum area problem Set up and solve
56-60 min Exit check Three questions marked Finish alone

03 Teaching script (read it out)

Scene 1: a decides the opening

In y = ax² + bx + c, a > 0 opens upwards with a minimum; a < 0 opens downwards with a maximum.
The larger |a|, the narrower the curve.

Student states the opening for three equations.

Scene 2: The vertex comes from completing the square

y = x² − 4x + 3 = (x − 2)² − 1, so the vertex is (2, −1) and the axis is x = 2.
The h is the value that makes the bracket zero; k is the extreme value.

Student does y = x² + 6x + 5: vertex (−3, −4).

Scene 3: The discriminant counts intercepts

Δ = b² − 4ac. Positive: two intercepts (two real roots). Zero: one touching point. Negative: none.
Equations and graphs are two views of the same thing.

Note: no real roots does not mean no solution at all.

04 Question types and worked steps

Type Example Steps Answer
Completing the square y = x² − 4x + 3 (x − 2)² − 1 Vertex (2, −1), axis x = 2
Opening and extreme value y = −2x² + 4x a < 0, maximum Maximum 2 at (1, 2)
Discriminant y = x² + 2x + 5 Δ = 4 − 20 = −16 No x-intercepts
Application 20 m of fence, largest rectangle Area x(10 − x) 25 m² at x = 5
Sanity check (teach this): The vertex y-value is the extreme value: a minimum when a > 0, a maximum when a < 0.

05 Independent practice (3 questions)

Question Answer What to watch
Vertex of y = x² + 6x + 5 (−3, −4) Completing the square
Axis of y = 2x² − 8x + 1 x = 2 Use −b/2a
Intercepts of y = x² − 2x + 3 None, Δ = −8 Sign of Δ

06 Common mistakes and what to say

Mistake Student writes Say this
Constant dropped x² − 4x + 3 written as (x − 2)² + 3 Subtract the 4 back: −1.
Vertex sign flipped y = (x − 2)² − 1 gives (−2, −1) Minus two inside means h = +2.
Minimum claimed when a < 0 y = −x² said to have minimum 0 Downward opening gives a maximum.
Two roots claimed when Δ < 0 negative discriminant, two roots No real roots.

07 Homework (paper, section and question numbers)

08 Exit check (last 5 minutes)

Question Answer Decision
Vertex of y = x² − 4x + 3 (2, −1) All right: move to logarithms
Extreme value when a < 0 Maximum One wrong: re-teach the three switches
What Δ = 0 means One repeated root Two wrong: repeat scene 3

09 Tutor FAQ

Memorise the vertex formula?

Learn x = −b/2a to check, but completing the square is needed for maximum and minimum problems.

How accurate must the sketch be?

Mark the vertex, axis and intercepts; no need to plot points.

How is this different from quadratic equations?

That chapter finds roots; this one reads shape and extreme values.

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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.