Additional Mathematics Synthesis Guide 21: Quadratic Functions, Completing the Square, Global Bounds and Modelling

Expression → representation → vertex → roots → bound → parameter condition → model → verification.

1. Three forms, three kinds of information

A quadratic function may be written in expanded form

f(x)=ax²+bx+c, a≠0,

factorised form

f(x)=a(x−r₁)(x−r₂)

when real roots are available, or completed-square form

f(x)=a(x−h)²+k.

The same function is being represented three ways. Expanded form displays coefficients; factorised form displays zeros; completed-square form displays the vertex (h,k), axis of symmetry x=h, and whether k is a global minimum or maximum.

2. Completing the square is a structural transformation

Write

x²−6x+11

in completed-square form.

Use

x²−6x=(x−3)²−9.

Hence

x²−6x+11=(x−3)²+2.

Because (x−3)²≥0 for every real x, the expression has global minimum 2 at x=3.

3. Completing the square proves a global bound

The statement

x²−6x+11≥2

is not an estimate. It follows exactly because

x²−6x+11=(x−3)²+2

and squares are non-negative.

Equality occurs at x=3. A complete bound therefore contains both the bound and its equality condition.

4. Negative leading coefficient gives a global maximum

Consider

g(x)=−2x²+8x−3.

Factor −2 from the quadratic terms:

g(x)=−2(x²−4x)−3.

Complete the square:

g(x)=−2[(x−2)²−4]−3=5−2(x−2)².

Since −2(x−2)²≤0,

g(x)≤5,

with equality at x=2. The vertex is (2,5), and 5 is the global maximum over all real x.

5. Vertex formula from completing the square

For

f(x)=ax²+bx+c, a≠0,

factor out a from the first two terms:

f(x)=a[x²+(b/a)x]+c.

Complete the square:

f(x)=a(x+b/2a)²+c−b²/4a.

Therefore the vertex x-coordinate is

x=−b/(2a).

The formula is not separate from completing the square; it is the same structure compressed.

6. The discriminant classifies x-axis intersections

For ax²+bx+c=0, the discriminant is

Δ=b²−4ac.

• Δ>0: two distinct real roots.
• Δ=0: one repeated real root.
• Δ<0: no real roots.

On the graph y=ax²+bx+c, these are exactly the cases where the parabola cuts, touches or misses the x-axis.

7. The discriminant and vertex tell the same story differently

For a>0, a quadratic has no real roots exactly when its global minimum is positive. It has one repeated root when its minimum is zero. It has two real roots when its minimum is negative.

For a<0, reverse “minimum” to “maximum” and reverse the relevant signs.

Thus Δ and completed-square form are not unrelated techniques. They encode the same geometry through different algebraic summaries.

8. Parameter condition for an always-positive quadratic

Find the real values of k for which

x²−4x+k>0

for every real x.

Complete the square:

x²−4x+k=(x−2)²+k−4.

The minimum is k−4. For strict positivity at every real x,

k−4>0, so k>4.

If the condition were ≥0 instead, k=4 would be allowed.

9. The same parameter condition by discriminant

For x²−4x+k to be positive for all real x, we need a positive leading coefficient and no real roots. Therefore

Δ=16−4k<0.

This gives k>4, agreeing with the completed-square route.

The completed-square method exposes the actual minimum. The discriminant method exposes the root-count condition. Both are valid; choose according to the target.

10. Quadratic inequalities are sign questions

Solve

x²−5x+6≤0.

Factor:

(x−2)(x−3)≤0.

The upward-opening parabola is at or below the x-axis between its roots, so

2≤x≤3.

Factoring revealed the sign-change boundaries. Completing the square would be less direct here because the roots are simple and exact.

11. An inequality with no real roots

Consider x²−4x+7. Completing the square gives

(x−2)²+3>0

for every real x. Therefore

x²−4x+7<0

has no real solution.

A learner who automatically “finds roots and draws a sign chart” should recognise that there are no real roots to chart.

12. Intersections of a line and a parabola

Find where

y=x²−4x+1

meets

y=2x−3.

Set the y-values equal:

x²−4x+1=2x−3.

So x²−6x+4=0, giving

x=3±√5.

Then y=2x−3, so the two intersection points are

(3±√5, 3±2√5).

The quadratic formed by equating the functions is the algebraic object that controls the intersection count.

13. Tangency as a repeated intersection

For the line y=mx+c to be tangent to a parabola, the simultaneous equation formed by substitution has one repeated root. Therefore its discriminant is zero.

This is the same repeated-root logic used for line-circle tangency in Guide 20. Tangency is not owned by one chapter; it is a geometric contact condition that several algebraic representations can detect.

14. Worked tangency parameter

Find k so that the line

y=2x+k

is tangent to

y=x²−4x+7.

Equate:

x²−4x+7=2x+k

so

x²−6x+7−k=0.

Tangency requires

36−4(7−k)=0.

Hence 36−28+4k=0, so

k=−2.

The repeated root is x=3, and the tangent point is (3,4).

15. Verify tangency by calculus

For y=x²−4x+7,

dy/dx=2x−4.

At x=3, the curve gradient is 2, exactly the gradient of the tangent line y=2x−2.

The point (3,4) also satisfies both equations. Discriminant and derivative routes therefore verify the same tangency independently.

16. Quadratic models need a domain

Suppose a height model is

h(t)=−5t²+20t+3.

As a pure quadratic function, h is defined for every real t. As a physical height model, negative time may be meaningless, and the model may stop being relevant when the object reaches the ground.

Completing the square gives

h(t)=23−5(t−2)².

The model’s maximum height is 23 at t=2, provided t=2 lies inside the physically relevant time interval.

17. Restricted-domain maximum can differ from global maximum

Let f(x)=10−(x−4)². Globally, the maximum is 10 at x=4.

But if x is restricted to 7≤x≤9, the vertex is not feasible. On that interval, f decreases as x moves farther from 4, so the maximum occurs at the nearer endpoint x=7:

f(7)=1.

Never transfer a global bound into a restricted-domain optimisation problem without checking whether the equality point is allowed.

18. A rectangle optimisation model

A rectangle has perimeter 40. Let one side be x. Then the other side is 20−x, with 0<x<20.

Area

A=x(20−x)=−x²+20x.

Complete the square:

A=100−(x−10)².

Hence A≤100, with equality at x=10. The maximum-area rectangle is therefore a 10 by 10 square.

The physical domain confirms x=10 is feasible.

19. Parameter family and changing number of roots

Consider

x²−2kx+k+2=0.

The discriminant is

Δ=4k²−4(k+2)=4(k²−k−2)=4(k−2)(k+1).

Therefore:

• two distinct real roots when k<−1 or k>2;
• one repeated real root when k=−1 or k=2;
• no real roots when −1<k<2.

The parameter changes the geometry of the graph family by moving it across root-count boundaries.

20. Recover a quadratic from its vertex and one point

A parabola has vertex (2,−3) and passes through (5,15). Its equation has form

y=a(x−2)²−3.

Use (5,15):

15=9a−3, so a=2.

Thus

y=2(x−2)²−3.

The vertex representation made the reconstruction almost immediate.

21. Recover a quadratic from roots and one point

A parabola has roots −1 and 4 and passes through (0,8). Write

y=a(x+1)(x−4).

At x=0:

8=a(1)(−4), so a=−2.

Therefore

y=−2(x+1)(x−4).

Again, choose the representation that contains the supplied information directly.

22. Symmetry gives a quick root check

For a quadratic with roots r₁ and r₂, the axis of symmetry lies halfway between them:

x=(r₁+r₂)/2.

This equals −b/(2a), consistent with the sum-of-roots formula r₁+r₂=−b/a.

If a computed vertex x-coordinate is not the midpoint of two computed roots, at least one calculation is wrong.

23. Common failure patterns

• Completing the square but forgetting to compensate for the number added inside the square.
• Calling the vertex value a maximum when a>0.
• Using Δ<0 alone to claim “always positive” without checking the leading coefficient.
• Forgetting the difference between >0 and ≥0 at a repeated root.
• Using global maximum/minimum without checking a restricted domain.
• Reporting an intersection x-value without the corresponding y-coordinate when a point is required.
• Using a physical quadratic model outside the interval where its assumptions make sense.
• Expanding a useful factorised or vertex form before extracting the information already visible.

24. A reliable quadratic routine

1. Identify the target: roots, vertex, bound, intersection, inequality or model parameter.
2. Select expanded, factorised or completed-square form accordingly.
3. Use the discriminant for root-count conditions and repeated-root tangency.
4. Use completed square for global bounds and vertex structure.
5. For inequalities, locate sign-change boundaries and include endpoints only when allowed.
6. For modelling, state the feasible domain before interpreting extrema.
7. Check roots against the axis of symmetry and coefficients where possible.
8. Verify parameter results at boundary cases.

25. Practice set

1. Write x²−8x+21 in completed-square form.
2. State its global minimum and where it occurs.
3. Write −3x²+12x+1 in completed-square form.
4. State its global maximum.
5. Find the vertex of 2x²+8x−5.
6. Classify the real roots of x²−4x+8=0.
7. Find k so x²−6x+k>0 for every real x.
8. Find k so x²−6x+k≥0 for every real x.
9. Solve x²−7x+10<0.
10. Solve x²−4x+7≤0.
11. Find intersections of y=x² and y=4x−3.
12. Find k so y=x+k is tangent to y=x²−2x+4.
13. Find the tangent point in Question 12.
14. For f(x)=−x²+6x−5, find the maximum on all real x.
15. Find the maximum of the same function on 5≤x≤8.
16. A rectangle has perimeter 24. Find its maximum possible area.
17. A parabola has vertex (−1,2) and passes through (1,10). Find its equation.
18. A parabola has roots 2 and 5 and passes through (0,20). Find its equation.
19. For x²−2kx+k+2=0, state the k-values giving no real roots.
20. Explain why Δ=0 corresponds to tangency in a line-parabola intersection problem.

Answers

1. (x−4)²+5.
2. 5 at x=4.
3. 13−3(x−2)².
4. 13.
5. 2(x+2)²−13, vertex (−2,−13).
6. Δ=16−32<0, so no real roots.
7. (x−3)²+k−9>0 for all x, so k>9.
8. k≥9.
9. (x−2)(x−5)<0, so 2<x<5.
10. No real solution.
11. x²=4x−3 gives (x−1)(x−3)=0, so (1,1) and (3,9).
12. x²−2x+4=x+k gives x²−3x+4−k=0. Tangency: 9−4(4−k)=0, so k=7/4.
13. Repeated root x=3/2; y=13/4.
14. 4 at x=3.
15. The vertex x=3 is outside the interval and the function decreases for x>3, so maximum at x=5 is 0.
16. Let sides x and 12−x: A=36−(x−6)², maximum 36.
17. y=a(x+1)²+2; 10=4a+2, so a=2.
18. y=a(x−2)(x−5); 20=10a, so a=2.
19. −1<k<2.
20. The simultaneous equation has one repeated x-coordinate, meaning the line and curve meet at one contact point rather than two crossing points.

26. What mastery looks like

Mastery means the learner can choose the best quadratic representation for the target, prove global bounds through completed squares, use discriminants as geometric root-count information, distinguish strict and non-strict inequalities, interpret tangency as a repeated intersection, and restrict modelling conclusions to the feasible domain.

The transfer test is to give the same quadratic family through different information: a vertex and point, two roots and a point, an “always positive” condition, or a tangent-line condition. If the learner changes representation rather than applying one fixed procedure to every version, the quadratic system is connected.

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Continue through Batch 06

• Guide 22: Simultaneous Equations, Line–Curve Intersections, Discriminants and Tangency Conditions
• Guide 23: Binomial Theorem, General Terms, Coefficient Extraction and Parameter Problems
• Guide 24: Cubic Polynomials, Factor and Remainder Theorems, Roots, Multiplicity and Graph Structure
• Additional Mathematics Directory
• BTT Mathematics Hub