DSE Maths | Quadratic Graphs Lesson-Prep Guide (for tutors) | completing the square, discriminant, four common errors

2026-11-05

Tutor lesson-prep guides · Secondary Maths

Faced with y equals x squared minus 6x plus 5, students start a table of values and fill half a page without finding the vertex, then panic at a negative discriminant. This lesson completes the square to read the vertex, axis and root count, then compares three parabolas.

Objectives — Students can: (1) rewrite y equals x squared minus 6x plus 5 in vertex form and read off the vertex (3, -4) and the axis x equals 3; (2) use the discriminant 16 to conclude there are two distinct real roots and obtain 1 and 5 from the formula; (3) use the sign of a to state the opening direction and the discriminant to state how many times the curve meets the x-axis, and explain why the vertex of y equals minus 2x squared plus 8x minus 3 is a maximum.

01 Prerequisites (3-minute check)

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

Skill	Quick check	Pass mark
Perfect squares	What is x squared plus 6x plus 9	x plus 3, all squared, written at once
Expansion and factorisation	Expand x minus 3, all squared	x squared minus 6x plus 9, signs correct
Quadratic formula	What is b squared minus 4ac called	The discriminant, and the formula starts with minus b

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-6 min	Prior check	Perfect squares, expansion, the name of the discriminant	Answer orally, errors go on the board
6-24 min	Act 1 completing the square	Work three lines from general form to vertex form	Copy once and check the signs
24-38 min	Act 2 discriminant and roots	From discriminant 16 to two real roots	Compute the discriminant and count roots
38-50 min	Act 3 opening and transformations	Compare a positive and a negative leading coefficient	Describe the shift and reflection from vertex form
50-56 min	Question drill	One item from each type, six minutes	Work alone, then swap and mark
56-60 min	Exit check	Collect three items and mark on the spot	Books closed

03 Teaching script (read it out)

Scene 1: Act 1: completing the square is squeezing three terms into one square plus a constant

Start from y equals x squared minus 6x plus 5. Halve the coefficient of x: half of minus 6 is minus 3. Write x minus 3, all squared, then notice the expansion adds 9, so subtract 9 to keep the value. The last line is x minus 3, all squared, minus 4. Three moves only: halve, square, compensate.
Then ask where the vertex is. Answers come back as (3, 4) or (minus 3, minus 4) because the signs blur. Teach the shape of vertex form: y equals a times x minus h, all squared, plus k, so the vertex is (h, k). The bracket holds minus 3, so h is positive 3; the tail is minus 4, so k is minus 4, giving vertex (3, -4) and axis x equals 3.
Give them a self-check: substitute x equals 3 into the original expression, giving 9 minus 18 plus 5, which is minus 4, the same as k. That substitution lets them verify without me. Misconception box: adding the square without subtracting it changes the expression, so whatever you add you must take away.

Students restate: halve, square, compensate; the bracket holds minus h and the tail is plus k.

Scene 2: Act 2: the discriminant decides how many times the curve meets the x-axis

Substitute a equals 1, b equals minus 6 and c equals 5 into b squared minus 4ac: 36 minus 20 gives 16. A positive discriminant means two distinct real roots, and the formula gives 1 and 5, which matches the factorisation into x minus 1 and x minus 5. Two routes, one answer, always cross-check.
Then two contrasts: y equals minus 2x squared plus 8x minus 3 has discriminant 64 minus 24, which is 40, still positive, so two real roots; y equals x squared plus 2x plus 5 has discriminant 4 minus 20, which is -16, negative, so no real roots and the curve never touches the axis. Three curves side by side turn the discriminant from arithmetic into a statement about the graph.
Add the root relations: sum of roots is minus b over a, which is 6, and product is c over a, which is 5, checked instantly with 1 plus 5 and 1 times 5. That pair is the tool when factorisation fails.

Students restate: positive discriminant two points, zero one point, negative no points.

Scene 3: Act 3: a controls the opening, the discriminant controls the crossings, the vertex controls the extreme value

Line the three up: y equals x minus 3, all squared, minus 4 has a equal to 1, opens upward and the vertex (3, -4) is the minimum; y equals minus 2 times x minus 2, all squared, plus 5 has a equal to minus 2, opens downward and the vertex (2, 5) is the maximum; y equals x plus 1, all squared, plus 4 has vertex (-1, 4) and sits entirely above the axis, hence no real roots. Three features, three jobs, and students must keep them apart or every question blurs.
Then transformations: from y equals x squared, three to the right is minus 3 inside and four down is minus 4 outside, so the graph shifts right 3 and down 4; changing a from 1 to minus 2 reflects and stretches vertically by two. Make them narrate the order, shift then stretch, because reversing it gives a different curve.
Close with one question: what does the graph look like when the discriminant is zero? The answer, one touching point with the vertex on the axis, tests both ideas at once and makes a good exit item.

Students restate: a sets the opening, the discriminant sets the crossings, vertex form gives the extreme value directly.

04 Question types and worked steps

Type	Example	Steps	Answer
Complete the square	Write y equals x squared minus 6x plus 5 in vertex form	1 halve the x coefficient 2 form the square 3 compensate the constant	y equals x minus 3, all squared, minus 4, vertex (3, -4)
Discriminant and root count	What does discriminant 16 tell you	1 compute b squared minus 4ac 2 compare with zero 3 state the number of crossings	Positive, so two distinct real roots and two crossings
Solve the equation	Solve x squared minus 6x plus 5 equals zero	1 factorise or apply the formula 2 check the sum of roots 3 state both	x equals 1 or 5
Transformation	Turn y equals x squared into y equals minus 2 times x minus 2, all squared, plus 5	1 read h and k for the shift 2 read a for reflection and stretch 3 state the vertex	Right 2, up 5, reflected and stretched by two, vertex (2, 5)

Sanity check (teach this): Check plausibility: the y value of the vertex must equal the original expression evaluated at x equals 3, and both routes give -4. The sum of the roots must equal minus b over a (6) and the product c over a (5); anything else is a sign slip. The discriminant and crossing count must agree: 16 positive means two crossings, -16 negative means none, so a claim of two roots from a negative discriminant is a reversed formula. Opening direction depends on a alone, never on b or c, which is a fast way to eliminate wrong options.

05 Independent practice (3 questions)

Question	Answer	What to watch
Write y equals x squared minus 6x plus 5 in vertex form and state the vertex and axis.	y equals x minus 3, all squared, minus 4, vertex (3, -4), axis x equals 3	Constant compensated, vertex signs correct
Find the discriminant of y equals x squared plus 2x plus 5 and state how many times the curve meets the x-axis.	Discriminant -16, negative, so no crossings	Negative discriminant not mistaken for two roots
State the vertex of y equals minus 2x squared plus 8x minus 3 and say whether it is a maximum or minimum.	Vertex (2, 5), a negative so it opens down and the vertex is a maximum	Sign of a used rather than a remembered picture

06 Common mistakes and what to say

Mistake	Student writes	Say this
Adding the square without compensating	x squared minus 6x written as x minus 3, all squared, with the 5 left alone	The expansion adds 9, so take 9 back or the expression no longer balances.
Vertex signs reversed	Reading vertex (minus 3, 4) from x minus 3, all squared, minus 4	Whatever is subtracted inside is h; the tail keeps its own sign.
Discriminant and root count mismatched	Claiming two real roots from discriminant -16	A negative discriminant means the curve never reaches the axis, so there are no real roots.
Judging opening by the constant	A positive c read as opening upward	Only a decides the opening; c only moves the y-intercept.

07 Homework (paper, section and question numbers)

Redo this chapter — Redo practice items 1 to 3 (questions 1-4), drawing the axis and marking the vertex each time
Maths chapter index — Pick the quadratic equations chapter and do question 2, using the discriminant to count roots
Build a comparison chart — Sketch the six combinations of positive or negative a with three discriminant cases

08 Exit check (last 5 minutes)

Question	Answer	Decision
Vertex of y equals x minus 3, all squared, minus 4	(3, -4)	A wrong answer means replay act one
How many real roots when the discriminant is -16	None, the curve does not meet the axis	Answering two means replay act two
What is the vertex when a is negative	The maximum point	Answering minimum means replay act three

09 Tutor FAQ

Why learn completing the square when a graph works?

Plotting only estimates; completing the square gives the exact vertex and axis, which parameter questions demand.

How many roots when the discriminant is zero?

One repeated root, with the vertex sitting exactly on the x-axis.

Vertex form or general form?

Vertex form for the extreme value and axis, general form for the intercept and discriminant, and students should move between them.

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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.