For Integrated Programme students: Your current school materials, teacher instructions, and assessment scope take precedence because IP topic sequence and depth vary by school. This is an Eclat IP guide, not the O-Level / SEC G3 exam-track guide.
How this chapter applies
Eclat core: factor form, completed-square form, roots, turning points, symmetry, sketching, quadratic equations, and simple modelling form the main route.
School-sensitive extension: discriminant conditions, parameter problems, and optimisation models can be taught at A-Math depth in some schools.
2027 national comparison: K310 Topics N6 and N7 include quadratic graphs and their properties, sketches from completed-square or factor form, and equations solved by factorisation, formula, completing the square, or graph.
Check your school: confirm how much discriminant reasoning and modelling is assessed in E-Math.
Q: What does IP EMaths Notes (Upper Sec, Year 3-4): 06) Quadratic Functions and Graphs cover? A: Complete the square, locate turning points, and sketch parabolas with intercepts and symmetry clearly marked.
The core idea is simple: A quadratic graph is a parabola; its vertex and intercepts tell most of the story.
Use it as a working check: Completed-square form gives the turning point quickly, the discriminant tells how many real roots exist, and the sign of the squared term shows whether the curve opens up or down.
Then go one layer deeper: Work through the examples to connect algebra to the sketch: complete the square, mark the axis of symmetry, find intercepts, and use the turning point for maximum or minimum questions.
Quadratic graphs model projectile paths, profit curves, and optimisation problems. Completing the square reveals turning points instantly and helps evaluate maxima or minima.
Completed square: y=a(x−h)2+k where (h,k) is the vertex.
Axis of symmetry: x=h. Discriminant Δ=b2−4ac indicates intercept behaviour.
Quadratic route-choice checkpoint
Before expanding or factorising, decide what the question is really asking for. The same quadratic can need a different first move depending on whether the target is a sketch, an intercept, a maximum or minimum value, or the number of roots.
Question cue
First move
What to check before answering
"Sketch the graph"
Find the opening direction, turning point, axis of symmetry, and intercepts.
The vertex, axis, and intercepts must agree visually. If the roots are symmetric about the axis, the sketch is more reliable.
"Find the maximum or minimum value"
Use completed-square form or x=−2ab.
Decide maximum or minimum from the sign of a, then substitute the correct x-value to get the value of y.
"How many real roots?"
Compute Δ=b2−4ac.
Δ>0 means two real roots, Δ=0
"Find the x-intercepts"
Set y=0, then factorise or use the quadratic formula.
Intercepts are points on the graph, so write them as (x,0) when the question asks for coordinates.
"Find the y-intercept"
Set x=0.
This is usually (0,c) in standard form, but still check the requested output form.
Misconception check: the turning point is not automatically an x-intercept. A parabola can have a clear maximum or minimum even when it never crosses the x-axis.
Completed-square interpretation checkpoint
Once a quadratic is written as y=a(x−h)2+k, read the graph features directly before expanding again.
Feature to read
How to read it
Example from y=2(x−3)2−5
Turning point
Use (h,k).
Turning point is (3,−5).
Axis of symmetry
Use x=h.
Axis is x=3.
Opening direction
Check the sign of a.
Since a=2>0, the graph opens upward.
Minimum or maximum value
If a>0, minimum is k; if a<0, maximum is k.
Minimum value is −5
Worked check: y=−3(x+2)2+7 has turning point (−2,7), not (2,7). The graph opens downward because a=−3, so the maximum value is 7.
Common trap: the sign inside the bracket is opposite to the x-coordinate of the turning point. Write x−h mentally before reading h.
Worked example - Sketch from completed square
Sketch y=x2−6x+5, stating the axis of symmetry and intercepts.
Complete the square: x2−6x=(x−3)2−9. So y=(x−3)2−4.
After finding the vertex and intercepts, check that all features can belong to the same parabola before drawing the final curve.
Feature found
Consistency check
Common trap
Two x-intercepts
Their midpoint should lie on the axis of symmetry.
Plotting roots that are not balanced around the vertex.
Completed-square form
The vertex should sit on the axis and match the maximum or minimum value.
Reading x+2 as vertex x=2 instead of x=−2.
y-intercept
It must lie on the left or right branch at x=0.
Marking (c,0) instead of (0,c)
Discriminant
It must agree with the number of x-axis contacts.
Drawing two crossings when Δ<0, or no crossing when Δ>0.
Worked check: for y=x2−6x+5, the roots are 1 and 5, so their midpoint is 3. This agrees with the axis x=3 from (x−3)2−4. The y-intercept is (0,5), so the left branch must pass through that point while the curve opens upward.
Misconception check: a sketch is not just a list of calculated points. The axis, roots, vertex, opening direction, and intercepts must fit one coherent curve.
Worked example - Use the discriminant and quadratic formula
Determine the nature of the roots of y=2x2−5x+2, then find the x-intercepts exactly.
Identify coefficients: a=2, b=−5, c=2.
Compute the discriminant: Δ=b2−4ac=(−5)2−4(2)(2)=25−16=9.
Since Δ>0, there are two distinct real roots.
Apply the quadratic formula: x=2a−b±Δ=45±3
Solutions: x=45+3=2 and x=45−3=21
Therefore the parabola crosses the x-axis at (2,0) and (21,0).
Worked example - Optimise a quadratic model
A projectile is launched so that its height above the ground after t seconds is given by h(t)=−5t2+20t+3. Find the time at which it reaches maximum height and the corresponding height.
The quadratic is in the form h(t)=at2+bt+c with a=−5, b=20.
The turning point occurs at t=−2ab=−2(−5)20=2
Substitute t=2 into h(t): h(2)=−5(2)2+20(2)+3=−20+40+3=23
Because a<0, this turning point is a maximum.
The projectile reaches its maximum height of 23 metres at t=2 seconds.
Practice Quiz
Test your grip on sketching, discriminants, and quadratic optimisation with guided questions.
Try this
Rewrite y=2x2+8x+3 in completed-square form and state whether the curve has a minimum or maximum value.
