How to Improve Additional Mathematics | Quadratic Graphs and Completing the Square

Quick Read: A Reliable Improvement Method

Improving completing the square and interpreting quadratic graphs in Additional Mathematics starts with understanding the rule rather than copying a solution. Factor the leading coefficient before completing the square, then use the vertex and opening direction to interpret the graph. Students should identify restrictions, show each transformation and verify their answers. This guide gives original worked exercises and a practical plan for independent revision.

Singapore G3 Curriculum and Source

Check the official 2027 SEAB G3 syllabus directory and the K341 Additional Mathematics document for the applicable cohort. The examples below are original teaching illustrations, not past examination questions.

Worked Example 1: x²−6x+11

The exact answer is (x−3)²+2, minimum 2 at x=3. Factor the leading coefficient before completing the square, then use the vertex and opening direction to interpret the graph. Check the result in the original expression and explain any restrictions. Do not rely on a numerical answer alone: show the mathematical reasoning that makes it valid.

For further practice, change one coefficient of x²−6x+11 and solve the new question without hints. Compare the two solutions to identify which parts of the method remain valid. This helps distinguish understanding from memorising a particular example.

Worked Example 2: 2x²+8x+3

The exact answer is 2(x+2)²−5, minimum −5 at x=−2. Factor the leading coefficient before completing the square, then use the vertex and opening direction to interpret the graph. Check the result in the original expression and explain any restrictions. Do not rely on a numerical answer alone: show the mathematical reasoning that makes it valid.

For further practice, change one coefficient of 2x²+8x+3 and solve the new question without hints. Compare the two solutions to identify which parts of the method remain valid. This helps distinguish understanding from memorising a particular example.

Worked Example 3: −x²+4x+1

The exact answer is −(x−2)²+5, maximum 5 at x=2. Factor the leading coefficient before completing the square, then use the vertex and opening direction to interpret the graph. Check the result in the original expression and explain any restrictions. Do not rely on a numerical answer alone: show the mathematical reasoning that makes it valid.

For further practice, change one coefficient of −x²+4x+1 and solve the new question without hints. Compare the two solutions to identify which parts of the method remain valid. This helps distinguish understanding from memorising a particular example.

Worked Example 4: x²+4x+7=0

The exact answer is (x+2)²+3=0, no real roots. Factor the leading coefficient before completing the square, then use the vertex and opening direction to interpret the graph. Check the result in the original expression and explain any restrictions. Do not rely on a numerical answer alone: show the mathematical reasoning that makes it valid.

For further practice, change one coefficient of x²+4x+7=0 and solve the new question without hints. Compare the two solutions to identify which parts of the method remain valid. This helps distinguish understanding from memorising a particular example.

A Fourteen-Day Improvement Programme

Days 1–2: Review Definitions

Complete two short sessions on completing the square and interpreting quadratic graphs. Begin by stating the rule, solve a changed example, check it and record the first wrong line. After a short delay, repeat the skill with unfamiliar coefficients to test independent improvement.

Days 3–4: Repair Algebraic Prerequisites

Days 5–6: Practise the First Two Examples

Days 7–8: Practise the Other Two Examples

Days 9–10: Check Common Errors

Days 11–12: Attempt Unseen Mixed Problems

Days 13–14: Retest Independently

Independent Self-Test and Answers

Question: x²−6x+11. Answer: (x−3)²+2, minimum 2 at x=3. Explain the method and verify the result using the original expression.

Question: 2x²+8x+3. Answer: 2(x+2)²−5, minimum −5 at x=−2. Explain the method and verify the result using the original expression.

Question: −x²+4x+1. Answer: −(x−2)²+5, maximum 5 at x=2. Explain the method and verify the result using the original expression.

Question: x²+4x+7=0. Answer: (x+2)²+3=0, no real roots. Explain the method and verify the result using the original expression.

Parents and Tutors: What to Check

Parents can ask the learner to explain the first decision, the applicable condition and how the final result is checked. In a small three-student tutorial, a tutor can identify different weaknesses and provide targeted practice. A delayed unseen retest is more informative than repeatedly copying one worked example.

Related Learning

Continue through the Additional Mathematics improvement guide, the subject hub and eduKate Singapore.

Further Worked Practice and Mathematical Verification

Complete the square to write f(x)=a(x−h)²+k; the vertex is (h,k) and the sign of a decides maximum or minimum. The correct technique is chosen from the structure of the original question, not from a remembered final answer. Students should state all assumptions before transforming an expression and verify each proposed solution. The following examples provide fresh numerical variations for independent retrieval.

Additional Worked Example 1: x²−8x+19

The correct calculation gives (x−4)²+3; minimum 3. Write the governing identity or condition first, substitute the coefficients carefully, and check the final result in the original expression. This avoids accepting an answer that is algebraically plausible but violates a restriction or has a lost sign.

A second check is to change one coefficient in x²−8x+19 and repeat the same reasoning. Compare the two problems: the rule remains Complete the square to write f(x)=a(x−h)²+k; the vertex is (h,k) and the sign of a decides maximum or minimum. Only the arithmetic and, when relevant, the domain boundaries change. A learner who can explain both solutions without copying is demonstrating transferable understanding.

Additional Worked Example 2: 3x²+12x+8

The correct calculation gives 3(x+2)²−4; minimum −4. Write the governing identity or condition first, substitute the coefficients carefully, and check the final result in the original expression. This avoids accepting an answer that is algebraically plausible but violates a restriction or has a lost sign.

A second check is to change one coefficient in 3x²+12x+8 and repeat the same reasoning. Compare the two problems: the rule remains Complete the square to write f(x)=a(x−h)²+k; the vertex is (h,k) and the sign of a decides maximum or minimum. Only the arithmetic and, when relevant, the domain boundaries change. A learner who can explain both solutions without copying is demonstrating transferable understanding.

Additional Worked Example 3: −2x²+8x−3

The correct calculation gives −2(x−2)²+5; maximum 5. Write the governing identity or condition first, substitute the coefficients carefully, and check the final result in the original expression. This avoids accepting an answer that is algebraically plausible but violates a restriction or has a lost sign.

A second check is to change one coefficient in −2x²+8x−3 and repeat the same reasoning. Compare the two problems: the rule remains Complete the square to write f(x)=a(x−h)²+k; the vertex is (h,k) and the sign of a decides maximum or minimum. Only the arithmetic and, when relevant, the domain boundaries change. A learner who can explain both solutions without copying is demonstrating transferable understanding.

Additional Worked Example 4: x²+2x+5=0

The correct calculation gives (x+1)²+4=0; no real roots. Write the governing identity or condition first, substitute the coefficients carefully, and check the final result in the original expression. This avoids accepting an answer that is algebraically plausible but violates a restriction or has a lost sign.

A second check is to change one coefficient in x²+2x+5=0 and repeat the same reasoning. Compare the two problems: the rule remains Complete the square to write f(x)=a(x−h)²+k; the vertex is (h,k) and the sign of a decides maximum or minimum. Only the arithmetic and, when relevant, the domain boundaries change. A learner who can explain both solutions without copying is demonstrating transferable understanding.

Diagnostic Practice and Retesting

Common difficulties in quadratic graphs include misreading the question, choosing an invalid transformation, dropping a minus sign and failing to check the final result. Rather than assigning more identical questions, identify the earliest incorrect line, explain why it fails, and create a replacement question with different numbers. Retest the skill after several days to distinguish durable improvement from immediate imitation.

A Self-Check Before Submitting

Read the exact command word. State the relevant formula and any domain restrictions. Carry out the transformation line by line. Substitute each answer into the original expression and verify that it satisfies the required interval, sign or geometric condition. If the result is a function or derivative, use an independent algebraic or graphical check where possible. These checks make working easier to audit.

Additional Fully Worked Problems

Extended Example 1: Complete square for 3x²−18x+20

3(x²−6x)+20=3[(x−3)²−9]+20=3(x−3)²−7. Vertex (3,−7), minimum −7.

After obtaining the result, identify the mathematical condition that makes the method valid. Check the original expression, including any sign, interval or domain restrictions. For a second attempt, change one coefficient and solve independently rather than copying the displayed numbers.

Extended Example 2: Complete square for −2x²−8x+1

−2(x²+4x)+1=−2[(x+2)²−4]+1=−2(x+2)²+9. Vertex (−2,9), maximum nine.

Extended Example 3: Find range of y=2x²−12x+15

Complete square: y=2(x−3)²−3, so y≥−3 for real x.

Extended Example 4: Find intersections of y=x²−4x+1 with x-axis

Solve (x−2)²−3=0. The roots are x=2±√3, so the graph crosses the x-axis twice.

How to Diagnose the First Incorrect Line

A reliable correction starts before the final answer. Identify whether the student selected the wrong method, copied a sign incorrectly, applied a rule outside its domain, or omitted a verification step. Write one sentence explaining the mathematical reason. Then solve a changed problem without prompts and compare the results. This process distinguishes conceptual repair from simply rewriting the correct answer.

Short Practice for the Next School Assessment

Start with two prerequisite questions, then one straightforward worked-type question and one unfamiliar variation. Ask the learner to state the intended rule before calculating and to check the answer in the original problem. After two or three days, retest the same decision with different coefficients. A short independent explanation provides stronger evidence of improvement than a long sequence of copied examples.

Deeper Problem-Solving and Verification

These additional Quadratic Graphs examples focus on the decisions that make an answer mathematically valid. Before calculating, identify all domain restrictions and the meaning of the required result. After calculating, check each candidate in the original problem. A transformed equation may contain a root that is not permitted by the original conditions.

Advanced Worked Example 1: Complete square for 4x²−8x+7

4(x−1)²+3, vertex (1,3), minimum three.

Check the final conclusion against the original question. If the problem is an inequality, confirm whether boundary values are included; if it is a function, retain its domain; if it is differentiation, compare the derivative structure with the original factors. Then vary one coefficient and repeat the method without hints. This builds transferable reasoning rather than memorisation.

Advanced Worked Example 2: Find vertex of −3x²+12x−8

−3(x−2)²+4, vertex (2,4), maximum four.

Advanced Worked Example 3: Solve x²−6x+5=0 by completing square

(x−3)²−4=0, so x=1 or 5.

Advanced Worked Example 4: Find parameter k for x²−4x+k to be nonnegative for all x

(x−2)²+k−4≥0 for all x exactly when k≥4.

An Independent Examination-Readiness Check

Give the learner an unfamiliar Quadratic Graphs problem and ask for three things: the rule that applies, a complete chain of justified working, and a verification that does not simply repeat the same calculation. When an error occurs, record its first line and choose a shorter corrective problem that targets that specific decision. After a delay, retest with new numbers. This is the clearest evidence that a weak method has been repaired.

Why Completing the Square Works

For x²+6x+5, take half the coefficient of x, which is three, and square it to get nine. Add and subtract nine: x²+6x+9−9+5=(x+3)²−4. The completed-square form shows the vertex immediately: (−3,−4). Since the square is nonnegative, the minimum value is −4. This transformation is an exact algebraic identity, not a numerical approximation. Expanding (x+3)²−4 recovers the original quadratic, providing a direct check.

A Leading Coefficient Greater Than One

Complete the square for 3x²−12x+7. Factor three from the quadratic and linear terms: 3(x²−4x)+7. Complete the inner square: x²−4x=(x−2)²−4. Thus the expression becomes 3(x−2)²−12+7=3(x−2)²−5. The vertex is (2,−5), and the minimum is −5 because the coefficient three is positive. The common error is to forget to multiply the subtracted four by the outside factor three.

A Negative Leading Coefficient

For −2x²+8x+1, factor −2: −2(x²−4x)+1. Then x²−4x=(x−2)²−4, so the quadratic becomes −2(x−2)²+9. The vertex is (2,9). Because the squared term is multiplied by a negative coefficient, the graph opens downwards and has a maximum of nine. The expression can decrease without bound, so nine is not a minimum. The sign of the leading coefficient decides which extremum the vertex represents.

Finding the Axis of Symmetry

In y=4(x+1)²−3, the vertex is (−1,−3) and the vertical axis of symmetry is x=−1. Points equally far from −1 have equal y-values: at x=0 and x=−2, y=1 in both cases. This symmetry is a useful graph check. In general, for y=a(x−h)²+k, the axis is x=h, even if h is negative. Students sometimes report x=+1 when the bracket is x+1, confusing the sign inside with the vertex coordinate.

Recovering the Vertex Formula

For y=ax²+bx+c with a≠0, factor a from the first two terms and complete the square. The result is y=a[x+b/(2a)]²+c−b²/(4a). Therefore the vertex x-coordinate is −b/(2a), and its y-coordinate is f(−b/(2a)). This derivation explains the familiar vertex formula rather than treating it as unrelated memorisation. It also shows why the axis depends on a and b, while c shifts the graph vertically without changing its axis.

Finding Roots from the Completed Square

Solve x²−6x+5=0. Complete the square: (x−3)²−4=0, so (x−3)²=4. Taking both square roots gives x−3=±2 and x=1 or five. The ± is essential: a positive square equals four at two real values. The vertex is (3,−4), so the graph crosses the x-axis symmetrically two units to the left and right of x=3. This geometric check supports the algebraic roots.

A Quadratic with No Real Roots

Solve x²+4x+7=0. Completing the square gives (x+2)²+3=0, or (x+2)²=−3. A real square cannot be negative, so there are no real roots. The graph y=(x+2)²+3 has minimum value three, entirely above the x-axis. A student who writes x=−2±√(−3) has moved outside the real-number setting; for the school graph interpretation, the correct conclusion is no real x-intercepts.

A Repeated Root and Tangency

Consider y=x²−4x+4=(x−2)². Its minimum is zero at x=2, so the parabola touches the x-axis at (2,0) without crossing it. The equation y=0 has a repeated root x=2. Its discriminant is b²−4ac=16−16=0, consistent with exactly one real intercept counted as a double root. This links completing the square with the discriminant and the geometry of tangency.

The Discriminant and the Vertex Height

For y=ax²+bx+c, the completed-square vertex height is k=c−b²/(4a)=−D/(4a), where D=b²−4ac. If a>0 and D<0, then k>0, so the upward parabola lies above the x-axis and has no real roots. If D>0, the vertex is below the x-axis and the parabola crosses twice. For a<0 the corresponding signs reverse. This relationship explains why the discriminant predicts the number of x-intercepts.

A Parameter for a Nonnegative Quadratic

Find all real k such that x²−4x+k≥0 for every real x. Complete the square: (x−2)²+k−4. Its smallest value occurs at x=2 and equals k−4. For the expression to remain nonnegative everywhere, require k−4≥0, giving k≥4. At k=4 the parabola touches the x-axis; for k>4 it stays strictly above; for k<4 it dips below zero. This is a precise interpretation of a parameter inequality.

A Parameter for Two Distinct Roots

For x²−6x+k=0, complete the square to obtain (x−3)²=9−k. There are two distinct real roots exactly when 9−k>0, so k<9. There is one repeated root when k=9, and no real roots when k>9. This is equivalent to the discriminant condition D=36−4k. The completed-square form makes the classification visible without first writing the quadratic formula.

A Quadratic’s Range from Its Vertex

For y=−3(x−2)²+7, the square is nonnegative and the coefficient is negative, so y≤7. Every real value at or below seven is attained, because the square can take any nonnegative value. Thus the range is (−∞,7]. The vertex (2,7) is the maximum point. A student who writes y≥7 has ignored the negative multiplier. Range statements should use the graph’s opening direction, not the sign of the constant alone.

Domain Restrictions Can Change the Extremum

For f(x)=(x−3)²+2 over all real x, the minimum is two at x=3. But if the domain is restricted to x≥5, the vertex lies outside the permitted interval. On x≥5 the square increases as x moves away from three, so the minimum occurs at the boundary x=5 and equals six. This shows why the unrestricted vertex is not always the answer to a constrained optimisation question. The allowed x-values must be read before selecting an extremum.

A Quadratic from a Vertex and a Point

Suppose a parabola has vertex (2,−3) and passes through (0,5). Its equation has the form y=a(x−2)²−3. Substitute the point: 5=4a−3, giving a=2. Thus y=2(x−2)²−3=2x²−8x+5. Both forms are useful: the first displays the vertex and opening direction, while the second gives standard coefficients. Checking x=0 reproduces y=5, confirming the parameter.

Comparing Two Quadratics Graphically

The graphs y=(x−1)² and y=(x−1)²+4 have the same shape and axis x=1, but the second is shifted four units upwards. Their minima are zero and four. The graphs y=2(x−1)² and y=(x−1)² have the same vertex, but the first rises more steeply away from it. A student should distinguish vertical translation, horizontal translation, reflection and vertical scaling by reading the coefficients and brackets rather than guessing from a rough sketch.

Completing the Square with Fractions

For y=2x²+3x+1, factor two from the first two terms: 2[x²+(3/2)x]+1. Half of 3/2 is 3/4, so x²+(3/2)x=(x+3/4)²−9/16. Therefore y=2(x+3/4)²−18/16+1=2(x+3/4)²−1/8. The vertex is (−3/4,−1/8). Fractional values do not change the method; they simply require careful arithmetic and a final expansion check.

A Complete Graph-Sketch Checklist

For a quadratic y=ax²+bx+c, identify the opening direction from a, the vertex from completing the square, the axis of symmetry, the y-intercept c and any real x-intercepts. If roots are required, use factorisation, the completed square or the quadratic formula as appropriate. Check whether the vertex lies above, on or below the x-axis. Label coordinates exactly before drawing a smooth parabola. A good sketch communicates relationships; it is not just a decorative U-shaped curve.

A Short Independent Test

Complete the square for x²+2x−8, 2x²−4x+7 and −x²−6x−5. The answers are (x+1)²−9, 2(x−1)²+5 and −(x+3)²+4 respectively. Their vertices are (−1,−9), (1,5) and (−3,4). The first two open upwards and have minima; the third opens downwards and has a maximum. Expand each completed square to verify the coefficients. This self-test checks signs, outside factors and vertex interpretation together.

Why the Sign of a Matters

For y=a(x−h)²+k, the squared expression is always nonnegative. If a>0, the term a(x−h)² is nonnegative and the vertex gives the minimum k. If a<0, that term is nonpositive and the vertex gives the maximum k. The size |a| affects how quickly y changes away from h, while h and k locate the vertex. This interpretation should come before drawing the graph or describing its range.

A Quadratic with a Restricted Interval

Find the minimum and maximum of f(x)=x²−4x+1 on 0≤x≤5. Complete the square: f(x)=(x−2)²−3. The vertex x=2 lies in the interval and gives minimum −3. At the endpoints, f(0)=1 and f(5)=6, so the maximum is six at x=5. A student who reports no maximum because an upward parabola is unbounded above has ignored the restricted domain; on a closed finite interval, both extrema exist.

A Quadratic Inequality from a Graph

Solve x²−4x−5≤0. Factor as (x−5)(x+1)≤0. The upward parabola crosses the x-axis at −1 and five, so it lies on or below the axis between those roots. Thus −1≤x≤5. Completing the square gives (x−2)²−9≤0, or (x−2)²≤9, producing the same interval. The graph and algebra agree. The endpoints are included because the inequality allows equality.

An Outer Quadratic Inequality

Solve −x²+2x+8<0. Multiply by negative one and reverse the sign: x²−2x−8>0. Factor as (x−4)(x+2)>0. The product is positive outside the roots, so x<−2 or x>4. Alternatively complete the square in the original: −(x−1)²+9<0, so (x−1)²>9, giving the same outer rays. Both methods show why the solution is not the middle interval.

A Vertex Check Using Differentiation

For y=2x²−8x+5, completing the square gives y=2(x−2)²−3, so the vertex is (2,−3). Differentiation independently gives dy/dx=4x−8, which vanishes at x=2; the second derivative is four, positive, confirming a local minimum. The two methods agree because a quadratic’s stationary point is its vertex. Completing the square is often simpler for graph interpretation, while differentiation connects the result to later calculus.

Recovering a Quadratic from Its Roots

Suppose a quadratic has roots x=−1 and x=3 and passes through (0,−6). Its equation is y=a(x+1)(x−3). Substituting x=0 gives −6=a(1)(−3), so a=2. Thus y=2x²−4x−6=2(x−1)²−8. The vertex is (1,−8), exactly midway between the two roots along the x-axis. This midpoint symmetry gives an independent check on the completed-square result.

Common Sign Mistakes

For (x+3)²−4, the vertex x-coordinate is −3, not +3. For −2(x−1)²+5, the graph has maximum five, not minimum five. For 3(x+2)²−7, the square term cannot be negative, so y≥−7. These are three different interpretation errors: bracket sign, opening direction and range inequality. A tutor should correct the first mistaken interpretation before giving the student more complicated algebra.

A Final Mixed Self-Test

Determine the vertex and range of y=(x−4)²+2, y=−3(x+1)²+5 and y=2x²+8x+9. The first has vertex (4,2) and y≥2. The second has vertex (−1,5) and y≤5. The third becomes 2(x+2)²+1, with vertex (−2,1) and y≥1. For each, explain the sign of the leading coefficient and expand the completed-square expression to verify that the original coefficients are recovered.

What Parents Can Ask

Parents can ask, “Where is the vertex?”, “Does the graph open upwards or downwards?”, and “Does the question restrict x to an interval?” These questions identify whether the child understands the geometry behind the algebra. A learner who can explain the vertex and range before plotting has a stronger foundation than one who only reproduces a completed-square formula. Use changed coefficients and a delayed unseen test to check that the skill transfers.