If an Additional Mathematics question gives three points on a quadratic graph, turn the coordinates into three equations for the coefficients of f(x) = ax² + bx + c. Substitute each x- and y-coordinate, solve the simultaneous equations, then check all three points in the completed function.
For example, the points (0, 2), (1, 5) and (2, 12) give c = 2, a + b = 3 and 2a + b = 5. Therefore a = 2, b = 1 and f(x) = 2x² + x + 2. This guide explains why the method works, when the information does not describe a genuine quadratic, and how to use symmetry, differences and graphical checks to work more efficiently.
Start with three unknown coefficients, not guessed roots
The standard form of a genuine quadratic function is f(x) = ax² + bx + c, with a ≠ 0. The three coefficients determine its opening, location and vertical position. Three suitably placed points can provide three independent equations and determine those coefficients.
A point (u, v) on the graph gives the statement f(u) = v. It does not immediately identify a root unless v = 0, nor does it say that the point’s coordinates are coefficients. Correct reconstruction begins by expressing exactly that relationship.
For three points with different x-coordinates, substitute them one at a time into ax² + bx + c. The results are simultaneous linear equations in a, b and c. Solve the equations rather than guessing a parabola that merely looks close.
The method is particularly useful for questions that reverse ordinary function evaluation: instead of giving the quadratic and asking for outputs, the problem provides outputs and asks which quadratic produced them. It strengthens algebra, coordinate interpretation and mathematical modelling together.
Worked example: a graph through three simple points
Find the quadratic through (0, 2), (1, 5) and (2, 12). Begin with f(x) = ax² + bx + c.
Using (0, 2) gives c = 2. Using (1, 5) gives a + b + c = 5, so a + b = 3. Using (2, 12) gives 4a + 2b + c = 12, so 4a + 2b = 10, or 2a + b = 5.
Subtract a + b = 3 from 2a + b = 5. Then a = 2 and b = 1. Combined with c = 2, the function is f(x) = 2x² + x + 2.
Check all points. f(0) = 2; f(1) = 2 + 1 + 2 = 5; and f(2) = 8 + 2 + 2 = 12. These independent checks show that the function meets every stated condition, not merely the easiest point.
Why a point at x = 0 is valuable
When x = 0, the quadratic gives f(0) = c. A known point (0, v) therefore tells us the constant coefficient immediately: c = v.
That is why the opening example is relatively efficient. Instead of solving three full equations, we obtain c directly and are left with two simultaneous equations for a and b. The fact that one point lies on the y-axis is structural information worth recognising.
If a question gives (0, −4), the constant coefficient is −4. This remains true even when the parabola opens downward or has no real x-intercepts.
Students sometimes read the y-value as a root. The point (0, −4) means the graph crosses the y-axis at −4; it does not mean the graph crosses the x-axis at −4. Writing f(0) = −4 first avoids this interpretation mistake.
Translate coordinates using protective brackets
A negative x-coordinate must be substituted into every power with brackets. Suppose the graph passes through (−1, 6). The correct equation is a(−1)² + b(−1) + c = 6, which simplifies to a − b + c = 6.
The squared term is positive a because (−1)² = 1. The linear term is negative b because b(−1) = −b. Confusing these signs can make the entire simultaneous system wrong.
If the input is x = −2, the equation becomes 4a − 2b + c = y. There is no reason for the a-term to be negative, because the square is non-negative.
A reliable method is to write each substituted equation in full before simplifying. When the coefficients are eventually checked, return to the original point coordinates rather than relying solely on the equations derived from them.
A second example with a negative input
Find f(x) = ax² + bx + c through (−1, 6), (0, 1) and (1, 4). The point (0, 1) gives c = 1.
The point (−1, 6) gives a − b + 1 = 6, so a − b = 5. The point (1, 4) gives a + b + 1 = 4, so a + b = 3.
Adding these equations gives 2a = 8, so a = 4. Then b = −1. The reconstructed quadratic is f(x) = 4x² − x + 1.
Check: f(−1) = 4 + 1 + 1 = 6; f(0) = 1; f(1) = 4 − 1 + 1 = 4. This example illustrates how symmetric input coordinates −1 and 1 make elimination easier even though the corresponding outputs are not equal.
Use symmetric inputs to recover coefficients efficiently
For the three inputs x = −h, 0 and h, the function outputs are f(−h) = ah² − bh + c, f(0) = c and f(h) = ah² + bh + c.
Subtracting the two outer outputs gives f(h) − f(−h) = 2bh, so b = [f(h) − f(−h)]/(2h) for h ≠ 0. Adding the outer outputs and subtracting twice the middle output gives 2ah², so a = [f(h) + f(−h) − 2f(0)]/(2h²).
For the points (−1, 6), (0, 1) and (1, 4), b = (4 − 6)/2 = −1 and a = (4 + 6 − 2)/2 = 4. The central value gives c = 1.
These short formulas are optional, not replacements for understanding simultaneous equations. They show how the mathematical structure can simplify elimination and how symmetric positions separate the odd linear term from the even square term.
Three consecutive inputs reveal second differences
A quadratic sequence sampled at equally spaced x-values has constant second differences. This can be used as a check or a shortcut for reconstruction when the input spacing is known.
For (0, 2), (1, 5) and (2, 12), the output differences are 5 − 2 = 3 and 12 − 5 = 7. The second difference is 7 − 3 = 4. When the x-spacing is 1, that equals 2a, so a = 2.
With c = 2, the first difference f(1) − f(0) = a + b = 3 yields b = 1. This recovers 2x² + x + 2 without a three-equation elimination.
Be careful: the second-difference shortcut depends on equal spacing. For spacing h, the second difference is 2ah², not simply 2a. If the x-coordinates are irregular, direct substitution into ax² + bx + c is generally safer.
A clear case where a is zero
Suppose three given points are (0, 1), (1, 3) and (2, 5). The output differences are constant at 2, and the second difference is zero. The unique polynomial of degree at most two through the points is f(x) = 2x + 1.
That expression has a = 0. It is linear, not a genuine quadratic. If a question merely asks for a polynomial of degree no more than two, the result is legitimate. If it insists that the graph belongs to a quadratic with a ≠ 0, these three collinear points are inconsistent with that requirement.
This distinction is useful because three points do not guarantee that the fitted expression has a non-zero squared term. A learner who obtains a = 0 should not force a small non-zero coefficient into the answer simply to satisfy an expected pattern.
Check the word “quadratic” against the result. When the data are exactly collinear, no genuine parabola can pass through all three points.
Distinct x-values matter for uniqueness
A genuine function cannot have two different outputs at the same input. Thus the points (1, 3) and (1, 5) cannot both lie on any graph y = f(x), quadratic or otherwise. The data are contradictory.
If the same point is repeated—for example, (1, 3), (1, 3) and (2, 6)—the repetition supplies no new independent information. There are only two distinct points, so generally infinitely many quadratics can pass through them.
For three distinct x-values, however, there is exactly one polynomial of degree at most two through the three points. That is an algebraic uniqueness result, though its coefficient a may turn out to be zero.
An efficient preliminary check is to inspect the inputs and look for repeated coordinates before setting up the simultaneous equations. It can reveal an impossible question or missing independent data at once.
Why three distinct points determine at most one quadratic
Suppose two polynomials P and Q of degree at most two pass through the same three points with distinct x-values r, s and t. Their difference R(x) = P(x) − Q(x) is also a polynomial of degree at most two.
Because the two polynomials agree at the three given inputs, R(r) = R(s) = R(t) = 0. A non-zero polynomial of degree at most two cannot have three distinct roots. Therefore R must be the zero polynomial, and P and Q are the same polynomial.
This explains why a correctly solved simultaneous system produces a unique function, rather than merely one possibility among many. It also clarifies why repeated x-values do not automatically give independent information.
Students do not need to reproduce this argument in every basic reconstruction exercise, but understanding the logic helps with polynomial roots, the factor theorem and more advanced modelling questions.
Use factorised form when two points are x-intercepts
If two of the points lie on the x-axis, the problem may be easier in factored form than in general form. Suppose the points are (1, 0), (3, 0) and (0, 6).
The roots are x = 1 and x = 3, so write f(x) = a(x − 1)(x − 3). Substitute the third point: 6 = a(−1)(−3) = 3a, so a = 2.
Therefore f(x) = 2(x − 1)(x − 3) = 2x² − 8x + 6. This is the same curve we would obtain by solving three simultaneous coefficient equations.
Method selection matters. General form works for any three distinct input coordinates; factorised form is especially efficient when roots are explicitly supplied. Read the type of information before deciding which representation to use.
Use vertex form if one point is the turning point
Suppose a parabola has known turning point (2, −3) and passes through (0, 5). Vertex form gives f(x) = a(x − 2)² − 3. Substituting (0, 5) yields 5 = 4a − 3, hence a = 2.
The function is f(x) = 2(x − 2)² − 3. This requires only one additional point because the vertex provides two structural pieces of information: its horizontal and vertical coordinates.
It would be inefficient to ignore the turning-point condition and attempt to guess three arbitrary points. The quadratic’s best algebraic form depends on whether the data describe roots, a vertex or general coordinates.
This example also shows that not all information should be counted merely as individual points. A vertex supplies both a location and a special geometric condition on the curve.
The discriminant checks root claims, not the original point fit
For f(x) = 2x² + x + 2, the discriminant is b² − 4ac = 1 − 16 = −15. The parabola has no real x-intercepts, consistent with its positive leading coefficient and positive minimum.
But computing the discriminant does not prove that the graph passes through (0, 2), (1, 5) and (2, 12). That claim requires direct substitution or equivalent coefficient equations.
Different checks test different properties. The discriminant tests the real-root structure, completed-square form tests the vertex, and point substitution tests the specified data.
A student should avoid lengthy extra calculation when the examination asks only for coefficients. However, a quick independent check can prevent an incorrect final equation from surviving into later parts of a multi-step question.
Find the turning point after reconstruction
The reconstructed example f(x) = 2x² + x + 2 can be written as 2(x + 1/4)² + 15/8 by completing the square. Its vertex is (−1/4, 15/8).
Because a = 2 > 0, the vertex is a minimum. This result is also consistent with the derivative f′(x) = 4x + 1, which vanishes at x = −1/4.
The vertex is not necessarily inside the range of input coordinates originally supplied. Here the observed points use x = 0, 1 and 2, while the vertex lies a little to the left of the observations.
That matters for interpretation. Reconstructing a formula gives a full mathematical curve, but the original data may describe only a restricted domain. If a practical problem limits the permitted x-values, use that restriction before claiming a minimum is physically attained.
A point can have a negative y-value even for an upward parabola
An upward-opening quadratic does not have to be positive everywhere. It only has a positive leading coefficient. Its position depends on the remaining coefficients.
For example, f(x) = x² − 4x + 3 passes through (0, 3), (1, 0) and (2, −1). It opens upward but its vertex is (2, −1), below the x-axis.
The three-point system confirms this: c = 3; a + b = −3; and 2a + b = −2. Therefore a = 1 and b = −4.
A student who rejects a negative output because a is positive is confusing upward opening with positive values. The graph may dip below zero and later rise again. Such conceptual checks are important after reconstructing the coefficients.
What a quick sketch can and cannot tell you
Plot the three supplied points and estimate whether the parabola appears to open upward or downward. The relative output differences can provide a first clue about the leading coefficient when the inputs are ordered.
For (0, 2), (1, 5) and (2, 12), the increasing first differences 3 and 7 are consistent with an upward-opening quadratic. For equally spaced points, the positive second difference confirms a positive a.
A sketch alone cannot determine exact coefficients from imprecisely read graph positions unless the coordinates are explicitly given. Algebra should supply exact values while the diagram checks the shape and relative positions.
Use a sketch to challenge an implausible answer, not to override correctly stated coordinates. For example, a negative leading coefficient in the opening example would contradict its positive second difference.
What happens when the inputs are fractions
A three-point quadratic problem does not require integer inputs. Consider the points (0, 1), (1/2, 2) and (1, 5). Start with f(x) = ax² + bx + c and set c = 1.
At x = 1/2, a/4 + b/2 + 1 = 2, so a + 2b = 4 after multiplying by 4. At x = 1, a + b + 1 = 5, so a + b = 4.
Subtract the equations: b = 0, giving a = 4. Thus f(x) = 4x² + 1. Checking at x = 1/2 gives 4(1/4) + 1 = 2.
The method is identical to the integer-coordinate case. The main extra skill is accurate fraction substitution and elimination. Do not round a half to a decimal unless it genuinely simplifies the work.
Avoid an unnecessary quadratic-formula detour
The ordinary quadratic formula solves ax² + bx + c = 0 for its roots. It does not directly determine unknown a, b and c from three arbitrary points.
A child may attempt to set each given x-coordinate equal to a root, even when its y-coordinate is not zero. That changes the meaning of the data and produces the wrong polynomial.
Use function evaluation instead: for each point (u, v), write au² + bu + c = v. Solve the resulting linear system for the coefficients. Only after the function is known should you use factorisation or the quadratic formula if the question asks for x-intercepts.
This distinction separates reconstruction from root solving. Both involve the same quadratic expression, but their unknown quantities are different.
A compact school-examination solution format
Write f(x) = ax² + bx + c. Substitute the three points, label the resulting equations, and simplify. Solve for a, b and c using elimination or substitution. Write the completed function and verify all three point conditions.
When one point has x = 0, obtain c first. When two points are roots, consider factored form. When the vertex is supplied, consider completed-square form. These method choices reduce unnecessary algebra without weakening the justification.
A final sentence may say “The required quadratic is f(x) = 2x² + x + 2,” followed by one or two explicit substitution checks if time permits.
If a = 0, say whether the requested object is a genuine quadratic or a polynomial of degree at most two. Being precise about that difference is a stronger response than presenting a straight line as though it were a parabola.
Practice questions with checked answers
Question one: find the quadratic through (0, 2), (1, 5) and (2, 12). Answer: 2x² + x + 2.
Question two: find the quadratic through (−1, 6), (0, 1) and (1, 4). Answer: 4x² − x + 1.
Question three: find the quadratic through (0, 3), (1, 0) and (2, −1). Answer: x² − 4x + 3.
Question four: find the quadratic through (0, 1), (1/2, 2) and (1, 5). Answer: 4x² + 1.
Question five: can a genuine quadratic pass through (0, 1), (1, 3) and (2, 5)? No. The unique degree-at-most-two polynomial through them is the straight line 2x + 1, with zero quadratic coefficient.
Try each without the answers, then check every point. The fifth question is important because it tests whether students understand the word “quadratic” rather than just execute a simultaneous-equation procedure.
Parents and tutors: what progress should look like
A student who understands reconstruction can translate coordinates into equations, distinguish roots from ordinary points, choose an efficient representation, solve coefficients accurately and verify every given condition.
If a child copies a model solution but cannot handle a negative x-coordinate, the prerequisite is likely substitution with signs. If the substitution is correct but elimination fails, practise simultaneous equations. If the algebra is correct but x- and y-intercepts are confused, work on coordinate interpretation.
Three-pax tutorials near Sixth Avenue MRT, generally 1.5 hours weekly at eduKateSG subject to class placement, allow a tutor to inspect the point where the first wrong decision occurs instead of responding to every error with another undifferentiated worksheet.
The ultimate test is an independent problem with a new set of coordinates, not simply a neat reproduction of the same three-point example.
Syllabus scope and connected learning
Quadratics, simultaneous equations, graph features and algebraic modelling form a useful bridge across secondary mathematics. Whether a specific reconstruction extension is examined depends on the learner’s programme and year.
Check the appropriate official documents for 2026 O-Level Additional Mathematics 4049, 2027 SEC G2 K232 or 2027 SEC G3 K341. Use the school’s topic sequence to decide which exercises are core and which are enrichment.
For related reading, see How to Reconstruct a Quadratic from Its Roots and a Point, How to Find the Range of a Quadratic on a Restricted Interval and How to Build a Weekly A-Math Question Queue.
Browse the Additional Mathematics archive, or enquire with eduKate Singapore about suitable small-group support. Properly taught kids shine a bright light into the future.