The shape is visible; the equation is missing
A quadratic graph is drawn on an examination page. You can see where it crosses the horizontal axis and where it turns, but the formula has been rubbed out. Can you rebuild the equation? Sometimes yes. Sometimes you need one more piece of information. The first skill is to decide which is true before manipulating symbols. A graph is a picture of relationships; the equation is one algebraic way to record those relationships.
This volume follows Vol 0319 on electric charges and evidence. It switches subject deliberately. Physics asked what observations justify a particle mechanism. Mathematics asks what information determines a model. The shared habit is restraint: a neat answer should not include a coefficient, a charge, or a certainty that the supplied evidence cannot establish.
Check the official G3 Mathematics scope
The 2027 Singapore–Cambridge Secondary Education Certificate G3 Mathematics syllabus is K310, listed in SEAB’s official 2027 directory. The two written papers each last two hours fifteen minutes, carry 90 marks and contribute 50% of the grade. Paper 1 includes about 26 short-answer questions, while Paper 2 has about nine to ten questions, including a final real-world application. Exact questions vary; the exercises here are original learning materials, not reproduced or predicted examination questions.
K310 includes quadratic expressions and equations, the graph of y = ax² + bx + c, and the relationships among the turning point, symmetry and roots. Reconstructing an equation from graph information combines those listed skills. This guide does not imply that every reverse-graph problem appears in a fixed form on the official paper. Nor does it present formal quadratic-inequality solving as a required K310 topic: constraints may be understood through numerical comparisons and valid dimensions without claiming a new syllabus requirement.
Three algebraic forms, three different strengths
The general or expanded form is y = ax² + bx + c, with a not equal to zero. It is useful when the coefficients or y-intercept are given and when we want to compare expressions term by term. The factored form is y = a(x − r)(x − s), where r and s are roots or x-intercepts if they are real. It makes the zeros visible. The turning-point form is y = a(x − h)² + k, with turning point (h, k). It makes the axis of symmetry and the minimum or maximum value clear.
The three forms describe the same kind of curve, but each reveals different information quickly. The competent learner does not force all graph details into y = ax² + bx + c first. If two roots are supplied, begin with factors. If the vertex is supplied, begin with the square form. Expand or factorise later when the question actually needs another view. Choosing the form is often a bigger step than carrying out the calculation.
Know what information determines a quadratic
A quadratic equation of the form y = ax² + bx + c has three unknown coefficients. Three suitable independent constraints will often determine them, but counting pieces of information is not enough: some constraints repeat the same relationship, and others contradict one another. Two roots fix the factor positions but not the vertical scale a. One additional point not already on the x-axis can determine a. A turning point fixes h and k but not the steepness a. Again, one other point away from the vertex can determine the scale.
For instance, roots at x = 2 and x = 8 give a family y = a(x − 2)(x − 8) for every nonzero a. The curves all cross at the same two points. They do not have the same vertex height. Saying that the missing equation is y = (x − 2)(x − 8) assumes a = 1 without evidence. On an examination page, the correct response might be a family of equations until the additional coordinate is supplied.
Reconstruct from two roots and a known point
Suppose a parabola has x-intercepts at 2 and 8 and passes through (0,16). Write y = a(x − 2)(x − 8). Substitute x = 0, y = 16. This gives 16 = a(−2)(−8) = 16a, so a = 1. Therefore y = (x − 2)(x − 8), or y = x² − 10x + 16. Check by substituting x = 2 and x = 8: both give zero. Check the point x = 0: y is 16.
The work is short, but the order is powerful. First use the roots to create the factors. Second use the third condition to determine a. Third verify every supplied datum. If a question gives an integer x-intercept and an awkward y-intercept, do not guess a from the graph’s steepness. The graph may not be drawn to scale; coordinates govern the algebra.
Reconstruct from the turning point and a point
Take a curve whose vertex is (3,−4) and which passes through (1,4). Start with y = a(x − 3)² − 4. Substitute (1,4): 4 = a(−2)² − 4, so 8 = 4a and a = 2. The equation is y = 2(x − 3)² − 4. The turning point is still (3,−4), exactly as specified. To write the expanded form, expand the square to obtain y = 2x² − 12x + 14.
A common mistake writes (x + 3)² for a vertex whose horizontal coordinate is positive 3. The sign inside the bracket is the opposite of h because the square is smallest at x = h. Check by substituting x = 3: only x − 3 becomes zero. The entire formula has a practical interpretation: it reaches its minimum −4 when x = 3 because a is positive.
The role of the y-intercept
The y-intercept is the value when x = 0. In expanded form y = ax² + bx + c, the y-intercept is c. In factored form it must be calculated: y(0) = ars, because (−r)(−s) = rs. In vertex form y(0) = ah² + k. It is not enough to copy a number next to the vertical axis if the diagram is schematic. Use the specified coordinate.
Suppose the intercept is 12 and the roots are 2 and 6. We have 12 = a(−2)(−6), so 12 = 12a and a = 1. Now y = (x − 2)(x − 6). If a student instead uses the sum of the roots to claim a = 12/8, the response has no justified relation to the intercept. Remember that the factors are multiplied, not added, in this substitution.
Two roots determine the axis of symmetry
For a quadratic with distinct real roots r and s, the axis of symmetry is halfway between them: x = (r+s)/2. This conclusion follows because the parabola is symmetric about the line through its turning point. Roots 2 and 8 give x = 5. Substituting x = 5 into y = a(x − 2)(x − 8) gives y = −9a. The vertex is (5,−9a). Notice how one still cannot know its vertical coordinate without a.
This relationship is a useful cross-check. If a sketch claims roots at x = 2 and x = 8 but labels the axis x = 6, something is inconsistent. Do not bend your algebra to match mutually incompatible clues. State which property fails. If the question gives an explicit numerical vertex that conflicts with the root midpoint, explain the contradiction and ask for a corrected datum rather than inventing a fourth condition.
Turning point, minimum and maximum
The coefficient a controls whether the parabola opens upward or downward. If a is positive, the turning point is a minimum. If a is negative, it is a maximum. This does not tell you the numerical value of a. Two upward-opening parabolas can have the same roots but very different heights or widths, because a may be 1, 2 or one half.
A learner sometimes writes ‘the turning point is (4,−18) so a = −18’. There is no such general rule. With roots at 1 and 7, y(4) = a(3)(−3) = −9a. If the turning-point y-value is −18, then a = 2, not −18. Solve with the equation; the visible coordinates are constraints, not a recipe for a coefficient.
When the curve touches the x-axis
A parabola that just touches the x-axis at its vertex has a repeated root. Its equation can be written y = a(x − r)². If it touches at x = 2, then y = a(x − 2)². To identify a, use another point. If it passes through (0,12), then 12 = a(−2)² = 4a, so a = 3. The full expression is y = 3(x − 2)².
Touching at a root is not the same as crossing at two different roots. A student who uses (x − 2)(x + 2) because the turning point is at 2 has invented an intercept at −2. Draw a rough graph or substitute key values to check the geometric meaning. The repeated factor reflects a tangency to the x-axis, not a claim about physical contact.
Can three arbitrary points determine the equation?
Given three x-values with distinct coordinates, you can often substitute each point into y = ax² + bx + c and solve simultaneous linear equations for a, b and c. This is an extension of algebraic substitution and simultaneous-equation skills, not a call for a calculator to choose an arbitrary ‘best fit’. If the supplied points lie on one quadratic, the three equations should be consistent with a not equal to zero.
For points (0,3), (1,6) and (2,11), the first gives c = 3. The second gives a+b = 3. The third gives 4a+2b = 8, or 2a+b = 4. Subtracting yields a = 1 and then b = 2, so y = x² + 2x + 3. Substitution into all three points confirms the result. If you get a = 0, the unique matching function may be linear rather than genuinely quadratic, an important classification to mention.
When the information is insufficient or contradictory
Suppose the question says that a curve has roots −1 and 5 and opens upward, but provides no other point. We know y = a(x + 1)(x − 5) with a greater than zero. That is a whole family, not one equation. A student who makes a = 1 by habit has transformed a partial description into an unsupported unique claim. In contrast, if a given point (−1,4) contradicts the stated root −1, no quadratic can satisfy both statements: the point on the x-axis would require y = 0 there.
Another inconsistency arises when a stated vertex lies away from the midpoint of two distinct roots. Explain which supplied facts cannot be simultaneously true. If a diagram is expressly ‘not drawn to scale’, use given numbers rather than measuring a sketch. In performance terms, checking sufficiency and consistency is more advanced than merely recognising a graph shape.
How to recover a missing coordinate
A reconstructed equation makes missing coordinates accessible. If y = (x − 2)(x − 8), the axis is x = 5 and y(5) = (3)(−3) = −9, so the turning point is (5,−9). If asked for the y-intercept, use x = 0 to obtain 16. If the question instead asks for the x-coordinates where y = 7, set (x − 2)(x − 8) = 7, expand to x² − 10x + 9 = 0, and solve to find x = 1 or x = 9.
Every calculation should answer a stated question. The graph’s zeros are not the same as the x-coordinates where y equals some nonzero target. The turning point is not the y-intercept unless its x-coordinate happens to be zero. A graph task becomes manageable once each requested coordinate is tied to the correct equation.
Multiple representations and the cost of unnecessary expansion
Factor form is excellent for roots. Vertex form is excellent for maxima, minima and symmetry. Expanded form makes comparison of coefficients and substitution at x = 0 immediate. An answer might move between all three: y = 2(x − 1)(x − 7) gives roots 1 and 7; its axis is x = 4; the turning point is (4,−18); its expanded form is y = 2x² − 16x + 14. Each line reveals something new.
Moving forms is not free. Every expansion can introduce a sign error. If the examination question asks only for the equation in factored form and the turning point, expand only if it helps verify or communicate a needed property. If asked for the coefficient b, expand carefully and identify it from ax² + bx + c. Efficient algebra is not about finishing with the most complicated expression.
Accuracy, units and diagrams
For a quadratic used as a model of a school display height, width or cost, define what x and y represent and which units they have. A vertex at (4,12) in a graph of length against time cannot be called a ‘minimum cost’ unless y is cost and the function represents that quantity. Numbers are only meaningful after labels are read. A dimension may also need to be nonnegative or lie within a permitted interval, although the algebraic parabola itself extends over all real x.
Where a numerical answer is not exact, follow the official instruction on rounding and any precision stated in the question. K310 commonly expects numerical answers to three significant figures, or angles to one decimal place, unless otherwise specified. Do not round a coefficient halfway through reconstruction when you can keep fractions exact. Show one correct substitution so the examiner can see where the model came from.
A design example that does not repeat perimeter optimisation
A workshop sign has an arch whose shape, in suitable coordinate units, is represented by a parabola. The curve passes through ground markers at x = 1 and x = 7 and its highest point is four and a half units above the baseline. The axis must be x = 4. Let y = a(x − 1)(x − 7). At x = 4, y = −9a. Setting −9a = 4.5 gives a = −0.5, so y = −0.5(x − 1)(x − 7). The negative coefficient is appropriate for a downward arch.
Now suppose someone proposes y = −(x − 1)(x − 7). This shares both roots and the same symmetry axis, but its maximum height is 9, not 4.5. The design is different despite matching two visible features. Compare with Vol 0316, which owns fixed-perimeter area optimisation. Here the central task is to recover and check the curve itself, not to maximise the area of a rectangle.
A second worked problem: two curves that look alike
Two upward-opening parabolas share the roots −2 and 6. Curve A passes through (0,12); Curve B passes through (0,24). Begin with y = a(x + 2)(x − 6). At x = 0 the product is −12. For A, 12 = −12a, giving a = −1; that conflicts with the stated upward opening. For B, 24 = −12a gives a = −2, also downward opening. Both sets of numerical conditions require a downward-opening curve. The description is inconsistent: an excellent reason to stop rather than produce a flattering sketch.
To repair the example, suppose the graph labels say ‘opens downward’ instead. Then both are valid and differ by vertical scale. Curve A is y = −(x + 2)(x − 6), and B is y = −2(x + 2)(x − 6). The vertex in each is at x = 2; their respective heights are 16 and 32. A small word in the question, ‘upward’, made the original premises impossible. This is exactly why consistency checks belong in a complete solution.
A reliable five-step examination routine
First, list the supplied facts: roots, points, turning point, opening direction or axis. Second, select the form that exposes the most information immediately. Third, substitute a point to determine an unknown scale or coefficient. Fourth, substitute back into every original condition, including the sign of a and the vertex position. Fifth, answer the exact request in suitable form, with coordinates and units where needed. One clear check can save several marks lost to a fluent but wrong calculation.
For an extended response, show the beginning equation before evaluating a. Writing only the final expanded polynomial hides the source of your reasoning and can make a later error hard to trace. The secondary functions-and-graphs owner develops the wider mathematics route. For pupils moving from the PSLE, the PSLE maths problem-solving route reinforces substitution, units and checking.
Thirty-five small reconstructions and judgement checks
Work out each before reading its answer. These are original practice items designed to make method selection, completeness and verification automatic. Several problems have no unique answer on purpose: the examination skill includes detecting insufficiency instead of treating every collection of numbers as an instruction to calculate.
Practice 1: Roots 3 and 9, and y-intercept 27.
Write y = a(x−3)(x−9). At x=0, 27=27a, so a=1 and y=x²−12x+27. The vertex lies at x=6. Substitute x=3 and x=9 to verify the zeros, and use the intercept to confirm the vertical scale. Neither the roots alone nor the drawing’s apparent width would have fixed a.
Practice 2: Roots 1 and 5, and point (2,−9).
The factor form is y=a(x−1)(x−5). At x=2, −9=a(1)(−3), giving a=3. Hence y=3(x−1)(x−5). Check the point and the zeros. Notice the curve opens upward despite y being negative at x=2; positive a controls its opening, not the sign of one particular output.
Practice 3: Vertex (−2,3) and point (0,11).
Start y=a(x+2)²+3. Substitution gives 11=4a+3, so a=2 and y=2(x+2)²+3. The minimum is 3 at x=−2. A mistaken bracket x−2 would make the vertex occur at positive 2, so substitution at the stated vertex is the quickest geometry check.
Practice 4: Vertex (4,−6) and another point (6,2).
Use y=a(x−4)²−6. Insert (6,2): 2=4a−6, so a=2. The correct expression is y=2(x−4)²−6. The graph opens upward and has minimum −6. This problem supplies an independently useful point, not merely another description of the same vertex.
Practice 5: Roots −4 and 2 with no other point.
Only the family y=a(x+4)(x−2), a nonzero, is determined. The midpoint of the roots, x=−1, gives the axis but cannot determine a. A statement that the graph opens upwards restricts a to positive values but still does not choose a unique value.
Practice 6: Roots −2 and 6 with y-intercept −24.
At x=0, −24=a(2)(−6)=−12a, so a=2. Therefore y=2(x+2)(x−6). The axis is x=2 and the vertex value is 2(4)(−4)=−32. Checking this negative minimum against the positive a confirms the expected opening direction.
Practice 7: Roots 2 and 8 with turning-point height −18.
The axis is x=5, and y(5)=a(3)(−3)=−9a. So −18=−9a gives a=2. The curve is y=2(x−2)(x−8), with vertex (5,−18). The turning-point height supplies the third piece of information that two roots alone lacked.
Practice 8: A curve just touches the x-axis at x=3 and passes through (0,18).
A repeated root is required: y=a(x−3)². Then 18=9a and a=2. The curve is y=2(x−3)². Do not manufacture a second distinct root; tangency means the vertex lies on the horizontal axis, so its vertical coordinate is zero.
Practice 9: A parabola has vertex (1,4) and opens downward.
The family y=a(x−1)²+4 with negative a satisfies the information. The vertex and direction do not determine steepness. A further point away from x=1 or another independent condition is required for a unique equation.
Practice 10: A proposed equation is y=3(x+2)(x−6). Give roots and axis.
The roots are x=−2 and x=6; their midpoint is x=2, so the axis is x=2. The vertex height is y(2)=3(4)(−4)=−48. Express the point as (2,−48), not ‘x=−48’, which would confuse horizontal and vertical coordinates.
Practice 11: A graph has roots 1 and 7 but axis labelled x=5.
The midpoint of 1 and 7 is 4, so the label x=5 is inconsistent with an ordinary quadratic having those two roots. No choice of nonzero a repairs that mismatch. State the specific contradiction instead of replacing one supplied coordinate silently.
Practice 12: A sketch gives roots at x=−1 and x=3 and point (−1,5).
The statement is impossible: a root at x=−1 means the function value there must be zero, but the given point claims it is 5. Before solving simultaneous equations, compare repeated x-values and the mathematical meaning of an x-intercept.
Practice 13: An equation is y=−2(x−1)(x−5).
The roots are 1 and 5, axis x=3, and vertex y(3)=−2(2)(−2)=8. Because a is negative, (3,8) is a maximum. The y-intercept is −2(−1)(−5)=−10. These checks separate four graph properties using one compact factor form.
Practice 14: Points (0,2), (1,5) and (2,10) lie on a quadratic.
Let y=ax²+bx+c. The first point gives c=2. The next two give a+b=3 and 4a+2b=8, or 2a+b=4. Subtraction gives a=1, b=2. Thus y=x²+2x+2; substitute all three points to verify the reconstruction.
Practice 15: Points (0,1), (1,3) and (2,5) are supplied as a ‘quadratic’.
The changes in y are both 2 for equal x-steps, consistent here with the line y=2x+1. Solving the three-coefficient equations yields a=0, so the resulting function is not a genuine quadratic. A student’s answer must not disguise this classification problem by retaining a zero x² term as though a were nonzero.
Practice 16: A vertex is (0,−7) and the graph passes through (2,1).
Write y=ax²−7. Since 1=4a−7, a=2, giving y=2x²−7. Here the vertex lies on the y-axis, so the y-intercept is indeed −7; that coincidence should not be generalised to vertices at arbitrary horizontal coordinates.
Practice 17: A curve y=a(x−4)²+3 passes through (4,3).
The supplied point is the vertex already encoded in the expression, so it provides no new information about a. Substituting yields 3=3 for every a. Another point away from x=4 would be required. The number of stated conditions matters less than whether each one is independent.
Practice 18: The roots are 2 and 6. Does y=x²−8x+12 fit?
Factor the polynomial as (x−2)(x−6); yes, both zeros match. The axis is x=4, and the vertex height is (2)(−2)=−4. This inverse-to-forward check shows a good reconstruction can be verified by returning to graph properties.
Practice 19: The roots are 2 and 6. Does y=2x²−8x+12 fit?
No. Substituting x=2 gives 8−16+12=4, not zero. The coefficient of x² was altered without multiplying the whole factored polynomial by the same scale. Factor-form reconstruction avoids this inconsistent scaling error.
Practice 20: A vertex is (5,9) and the curve passes through (4,7).
Use y=a(x−5)²+9. Insert the point: 7=a+9 so a=−2. The equation is y=−2(x−5)²+9, with maximum 9. The negative sign is not a mistake; it matches the described downward opening.
Practice 21: A graph has two distinct x-intercepts but its turning point is on the x-axis.
Those claims cannot both describe a non-degenerate quadratic. Touching the x-axis at the vertex produces one repeated root, not two distinct roots. Ask whether the diagram perhaps meant ‘turning point below the x-axis’ rather than forcing a false equation.
Practice 22: At what x-value is the vertex when roots are −3 and 11?
The midpoint is (−3+11)/2=4, so the axis is x=4. The vertical coordinate depends on a. Giving (4,0) as the vertex would be an unsupported additional claim; only its x-coordinate is determined.
Practice 23: For y=(x−2)(x−8), find x when y=7.
Expand x²−10x+16=7, giving x²−10x+9=0. Factor as (x−1)(x−9)=0, so x=1 or x=9. These are not the original roots because the target is y=7, not y=0. Include both solutions unless a domain restriction removes one.
Practice 24: A model gives height y=−0.5(x−1)(x−7) metres for x in [1,7].
The arch meets the baseline at x=1 and x=7, and its midpoint x=4 is the maximum. Substitute x=4 to get y=−0.5(3)(−3)=4.5 metres. The stated domain matters physically: it describes only the intended span, not the entire infinite mathematical parabola.
Practice 25: A proposed arch y=−(x−1)(x−7) has the correct roots but the wrong height.
At x=4 the height would be 9, not 4.5. To match a 4.5-high arch with the same roots, the coefficient must be −0.5. The exercise demonstrates why two intercepts determine the horizontal positions but not the vertical scale of a design.
Practice 26: The vertex is (2,−3). Which form should you write first?
Write y=a(x−2)²−3, because the vertex information appears directly. This is a choice of method, not an answer for a. If the problem later gives another point, substitute it to find a. Starting with three unknown coefficients adds unnecessary algebra.
Practice 27: The roots are −1 and 4. Which form should you write first?
Write y=a(x+1)(x−4). These factors make both zeros immediately visible. If a y-intercept is later supplied, set x=0 to determine a. Avoid writing a=1 merely because the brackets contain small integers.
Practice 28: The curve opens upward but has a point with negative y-coordinate.
That is entirely possible: an upward parabola may lie below the horizontal axis over part of its domain. The sign of a tells the opening direction, not whether every y-value is positive. Identify a through reconstruction or supplied shape information, not by inspecting one output sign.
Practice 29: A graph is expressly ‘not drawn to scale’ and appears to cross near 1.7.
If labelled intercepts are 2 and 6, use those numbers instead of measuring the appearance. If no coordinates are given, you may only make qualitative observations; precise algebraic recovery is not justified from a schematic. The exam assesses mathematical relationships, not the reliability of a ruler on a printed sketch.
Practice 30: A student uses y=a(x−3)²+4 to represent vertex (−3,4).
Correct the bracket to x+3: it vanishes at x=−3. The corrected family is y=a(x+3)²+4. This sign rule is easy to verify independently, which is safer than memorising a slogan about ‘changing signs’ without testing x=h.
Practice 31: A quadratic y=−2(x−1)(x−7) is expanded.
Multiply (x−1)(x−7)=x²−8x+7, then multiply by −2 to obtain y=−2x²+16x−14. Both the x² and linear terms change signs under the negative scale. Re-factor or substitute x=1 and x=7 to detect a missed sign.
Practice 32: Two students give equivalent answers y=2(x−3)²−4 and y=2x²−12x+14.
Expand the squared bracket: 2(x²−6x+9)−4=2x²−12x+14. They represent exactly the same graph. The first shows the minimum (3,−4); the second shows the intercept 14. An examination response should use the form requested, not reject a valid alternative solely for looking different.
Practice 33: A reader treats the lowest value on the printed page as the parabola’s minimum.
Only the actual turning point determines the mathematical minimum of an upward quadratic. A drawing cropped by a page boundary may hide the vertex; the bottom edge of paper is not a mathematical feature. Work from formula or specified coordinates instead of page layout.
Practice 34: The equation fits two roots and one point; is checking the second point unnecessary?
The second point or other supplied condition is exactly what tests the model. After finding a, substitute every condition back. This is especially important when the question contains a diagram that could be misleading. Verification is part of the explanation, not an optional flourish.
Practice 35: What should an answer say if no unique quadratic exists?
Give the family with a free nonzero parameter, state the constraints already known and identify the extra independent condition needed. For example, roots 2 and 8 imply y=a(x−2)(x−8) for any a≠0; another point off the x-axis determines a. That is more informative than writing ‘not enough information’ alone.
A full worked audit: restoring a science-club sign
A fictional science club commissions a curved sign. The specified span begins at horizontal marker x=1 and ends at x=7, where height is zero; the centre height must be 4.5 units. The team submits a draft equation y=−0.5(x−1)(x−7). First, check the end points: substitute x=1 and x=7 to obtain zero in both cases. Second, find the axis: (1+7)/2=4. Third, calculate y(4)=−0.5(3)(−3)=4.5. Fourth, the negative coefficient indicates a maximum at that midpoint. All design constraints are consistent with the equation.
Now audit a competing version y=−0.6(x−1)(x−7). The endpoints still fit, but the height at x=4 is 5.4, so it fails the 4.5-unit requirement. The team cannot accept the second arch simply because its formula is quadratic and its sketch looks smooth. A third proposal, y=−0.5(x−2)(x−7), fails the start marker, even though its height might seem plausible in a diagram. Every constraint contributes a separate test.
If a reader asks whether the shape is safe for construction, the mathematics here does not answer that engineering question. The curve model sets a geometric target; material strength, mounting, public access and other factors require competent design checks beyond this school exercise. An advanced learner can present a mathematically valid equation without converting that correctness into an unsupported real-world guarantee.
How to revise it without memorising one pattern
Day one: convert between expanded, factored and turning-point forms with verification. Day two: reconstruct from roots and a point. Day three: reconstruct from a vertex and a point. Day four: identify missing information. Day five: find inconsistent conditions. Day six: use a real-world-style curve with clearly defined variables and domains. Day seven: mixed review. Over the next week repeat the same categories in a different order and finish with a timed structured solution requiring an explanation of why a proposed graph must be rejected.
Keep a small error record with the first broken step. Did you write x+3 for a vertex at positive 3? Did you assume a=1 from the roots? Did you mistake the axis for an x-intercept? Fix the precise rule. The most useful model answer is not necessarily the longest; it gives an equation, shows where each coefficient came from and checks whether the final curve fulfils the supplied facts.
How this mathematics connects without taking over another topic
This guide is the owner of reverse reconstruction from roots, vertices and points. For choosing quadratic solving methods, follow Vol 0135. For how a fixed perimeter creates a quadratic area optimisation, use Vol 0316. Those chapters ask different questions. The next applied EMS guide will use the ability to audit a proposed exhibit shape without declaring that its design is a predicted SEC examination task.
If you remember only a final principle, remember this: information is not sufficient because there are many numbers; it is sufficient when the conditions determine the object uniquely. Your working should make that distinction visible. An equation chosen from evidence is useful. An equation chosen from habit is only a guess dressed as algebra.
