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A quadratic can give the same output for two different inputs because its graph is symmetric about a vertical line. For y = (x − 3)² + 2, inputs equally far from x = 3 give equal squared values and therefore equal outputs. Both x = 1 and x = 5 give y = 6. The output alone does not always identify one input.
For G2 Mathematics, find the axis of symmetry, identify the requested output level, and check every possible input against the stated domain. A horizontal level can meet a quadratic graph twice, once or not at all. Symmetry helps explain those possibilities, while the domain and the question’s wording decide which inputs belong in the final answer.
Contents
Sections 1–5
Sections 6–10
Sections 11–15
Sections 16–20
Sections 21–24
For y = (x − 3)² + 2, x = 1 is two units left of the axis x = 3. Input x = 5 is two units right. In both cases, the squared term is 4, so the output is 6. The paired inputs add to 6, twice the axis value.
At the minimum output y = 2, only x = 3 works. Below that minimum, such as y = 1, no real input works because a real square cannot be negative. Above the minimum, the unrestricted real-number model gives two inputs.
Those two inputs are not automatically both valid in a contextual problem. If the domain is 0 ≤ x ≤ 4, x = 1 is permitted but x = 5 is not. If the question asks for the smaller input, report the smaller one even though both solve the unrestricted equation.
This guide focuses on the repeated-output decision. The broader quadratic-graphs guide covers shape, roots, symmetry and turning points together. Here we follow the symmetry relationship through equations, graph readings and domain checks.
The SEAB G2 Mathematics syllabus for 2027 includes quadratic functions and graphs, their symmetry and turning-point properties, and solving quadratic equations. Reading repeated outputs connects those topics rather than treating each as a separate exercise.
The teaching examples here begin with quadratics written in a form that exposes their axis. Some later examples involve completing the square, restricted domains or non-integer inputs. Choose the parts that match the learner’s taught methods and the question’s scope.
The goal is not to insist that every quadratic question requires a sketch. It is to understand why two algebraic candidates can occur and why a graph may show two intersections with the same horizontal line. Algebra and graph describe the same relationship.
For the wider subject labels, use what G1, G2 and G3 Mathematics mean. A learner can build this idea from a numerical table before using a general symbolic form.
Start with one pair of inputs on opposite sides of the axis. Calculate both outputs and compare their distances from the axis. Once that connection is secure, use it to find missing partners and to interpret the number of solutions at different output levels.
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A function assigns one output to each permitted input. It does not require every output to come from only one input. For y = x², inputs 2 and −2 both give output 4, while each input still has just one output.
This distinction often becomes visible when a learner reverses a question. “Find y when x = 2” has one answer for the specified function. “Find x when y = 4” can have more than one answer because several inputs may share that output.
Do not label the function invalid merely because its graph is met twice by a horizontal line. That test concerns whether output alone determines a unique input, not whether the original input-to-output relationship is a function.
For the quadratic in this guide, x = 1 produces y = 6 and x = 5 also produces y = 6. There is no ambiguity about either forward calculation. The ambiguity appears only when the output 6 is supplied and the input is requested without another condition.
The guide to a letter’s consistent meaning supports this reading. Keep x as input and y as output throughout the argument. Quietly reversing their roles can turn a correct relationship into a confusing statement about which quantity is known.
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In y = (x − 3)² + 2, the squared term is smallest when x − 3 = 0, meaning x = 3. Its value is then zero, so the smallest output is 2. The turning point is (3, 2), and the vertical symmetry line is x = 3.
For inputs x = 3 − d and x = 3 + d, the brackets become −d and d. Their squares are both d², so both outputs are d² + 2. This explains the symmetry rather than merely naming it from a sketch.
The distance d is measured horizontally from the axis. It is not the output value and not the distance from the origin unless the axis happens to be x = 0.
That is why the partner of x = 1 is 5, not −1, in this shifted quadratic. The inputs are reflected about 3 rather than about zero. Using opposite signs alone would find the partner only for a suitable axis at the origin.
A short check is to average the paired inputs. The average of 1 and 5 is 3, the axis value. This gives a useful diagnostic when a proposed symmetric partner has been found by inspection.
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For y = (x − 3)² + 2, calculate outputs for inputs from 0 to 6. The values are 11, 6, 3, 2, 3, 6 and 11. The sequence falls towards the minimum and rises in the reverse pattern afterwards.
| Input x | Distance from x = 3 | Output y |
|---|---|---|
| 1 | 2 | 6 |
| 2 | 1 | 3 |
| 3 | 0 | 2 |
| 4 | 1 | 3 |
| 5 | 2 | 6 |
| 6 | 3 | 11 |
The table shows pairs 0 and 6, 1 and 5, and 2 and 4. Each pair has the same horizontal distance from the axis and the same output. The input 3 sits on the axis and does not need a distinct partner for the minimum output.
Do not confuse equal outputs with equal inputs. The table contains two different x values for several y values. If the task asks for all inputs giving output 6, both 1 and 5 are required in the unrestricted real-number model.
A table is a representation, not a complete proof of every real input relationship. It illustrates the pattern at the selected inputs. The squared-expression argument explains why the symmetry holds beyond those listed integer values as well.
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To find inputs giving y = 6, write (x − 3)² + 2 = 6. Subtract 2 to obtain (x − 3)² = 4. The bracket can equal either 2 or −2, because both square to 4.
Thus x − 3 = 2 gives x = 5, while x − 3 = −2 gives x = 1. Both values should be checked in the original expression: each gives output 6.
Writing only x − 3 = 2 loses one branch. The square-root symbol √4 denotes the nonnegative square root 2, but solving u² = 4 requires considering both u = 2 and u = −2. These are related but different statements.
You can also rearrange and factorise: x² − 6x + 11 = 6 becomes x² − 6x + 5 = 0, or (x − 1)(x − 5) = 0. The same two inputs appear through another valid method.
The graph interpretation is a horizontal line y = 6 intersecting the parabola at (1, 6) and (5, 6). The algebra identifies the horizontal coordinates of those intersection points. Each method answers the same target when its conditions and accuracy are respected.
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For y = 2, solve (x − 3)² + 2 = 2. This gives (x − 3)² = 0, so x = 3. The two symmetric branches meet at the turning point rather than producing two distinct inputs.
Do not write 3 and −3 merely because the expression contains a square. The bracket must equal zero, and zero has no distinct positive and negative versions. The only input here is 3.
On the graph, the horizontal line y = 2 touches the parabola at its minimum. It has one intersection point. This is the geometric counterpart of a repeated algebraic root after the equation is rearranged.
If a question asks how many distinct real inputs give the turning output, the answer is one. Writing the same input twice does not create two distinct possibilities. The wording distinct matters.
This example helps correct the overgeneralisation that every quadratic equation must have two different real solutions. A quadratic can have two, one or no distinct real solutions, depending on the level being considered. The graph’s symmetry explains the pairing when two exist, but it does not guarantee two separate intersections at every height.
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For output y = 1, solve (x − 3)² + 2 = 1. The result is (x − 3)² = −1. No real square is negative, so there is no real input that gives this output.
The graph says the same thing: its minimum is 2, so the horizontal line y = 1 lies below the whole curve. There is no intersection in the real coordinate plane.
The phrase no real input is important because it specifies the domain being discussed. The G2 examples here concern real-number inputs. Do not invent a real answer by taking a square root of a negative number on a calculator and ignoring the meaning of the result or error display.
This is not the same as a missing answer caused by difficult arithmetic. The expression gives a general reason: the requested output lies outside the function’s range over the real domain.
The fractional-equation guide to no valid solution develops the wider habit of interpreting an empty solution set. The mechanism differs here: no denominator is forbidden; instead, the requested output would require a negative real square.
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For a quadratic with axis x = 3, paired inputs have midpoint 3. If one input is 0.5, the partner is 5.5 because both are 2.5 units from the axis. Their sum is 6.
More generally, if the axis is x = h and one input is p, its reflected partner is 2h − p. This follows from the midpoint equation (p + partner)/2 = h.
For y = (x − 3)² + 2 and p = 0.5, the output is 6.25 + 2 = 8.25. Substituting 5.5 gives the same result. The partner rule saves work while preserving a clear reason.
If p equals h, the reflected input is the same value. The turning point is its own reflection. Do not assume the partner formula always produces a different input.
The formula also needs the domain check. A reflected partner may lie outside the interval allowed by a contextual question. Symmetry of the unrestricted quadratic does not automatically make every reflected point part of a restricted model.
Use the shortcut only after identifying the correct axis. Reflecting about zero instead of h is one of the most common ways a remembered symmetry rule produces the wrong partner in a shifted quadratic.
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Consider y = 12 − (x − 2)². The axis is x = 2 and the turning point is (2, 12). Because a nonnegative square is subtracted, 12 is the maximum rather than the minimum.
For output y = 3, solve 12 − (x − 2)² = 3. This gives (x − 2)² = 9, so x − 2 = 3 or −3. The inputs are 5 and −1, equally far from the axis x = 2.
At output 12, only x = 2 works. Above 12, no real input works. Below 12, the unrestricted real-number model gives two inputs. The two-one-none pattern remains, but its position relative to the turning level is reversed.
Do not apply the upward-opening statement “above the minimum gives two” without checking the sign of the squared term. The graph’s opening direction determines whether the turning output is a lower or upper boundary.
You can check the maximum directly: subtracting any positive squared value from 12 makes the output smaller. This gives a reason before any sketch. A graph can then illustrate the same relationship and help locate the paired inputs visually.
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The expression y = x² − 6x + 11 is the same as y = (x − 3)² + 2, because expanding the latter gives x² − 6x + 9 + 2. The axis x = 3 is easier to read from the completed-square form.
To construct that form, take half the coefficient of x, which is −3, and square it to obtain 9. Then write x² − 6x + 11 = (x − 3)² + 2. The added constant outside compensates for the constant already inside the square.
This is not a new function. It is a different expression for the same relationship. A numerical table or substitution should give identical outputs in both forms.
If the task asks for paired inputs at output 6, the completed-square form leads to (x − 3)² = 4. The expanded form leads to a factorisable equation. Choose a valid route that matches the question and your taught methods.
The algebra-instruction guide helps distinguish rewriting an expression from solving an equation. Completing the square reveals structure; setting the expression equal to a target output creates the equation whose inputs you must find.
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The roots or x-intercepts of a quadratic correspond to y = 0. They are not every possible pair of inputs that share an output. A horizontal level other than zero can also meet the graph twice.
For y = (x − 3)² + 2, there are no real roots because the minimum output is 2. Nevertheless, output 6 occurs at two real inputs. A quadratic with no x-intercepts can still have repeated outputs above its minimum.
This distinction prevents a graph-reading error in which a learner searches only the x-axis when the question asks for inputs at y = 6. Draw or imagine the requested horizontal line, not automatically the zero level.
If the quadratic is y = (x − 3)² − 4, its roots are 1 and 5 because setting y = 0 gives (x − 3)² = 4. The same paired input numbers can therefore represent roots for one quadratic and nonzero-output intersections for another.
Keep the output level attached to the calculation. The phrase solve the quadratic may be used informally, but the actual equation determines which horizontal level matters. A clear first line sets the function expression equal to the supplied target.
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Return to y = (x − 3)² + 2, but allow only 0 ≤ x ≤ 4. The unrestricted equation for output 6 gives x = 1 and x = 5. The domain admits 1 and excludes 5, leaving one valid input.
The algebra was not wrong to find 5. It found a solution of the unrestricted equation. The domain check then determines whether that candidate belongs to the stated model.
If the domain is x ≥ 3, every output above 2 has only its right-branch input permitted. If the domain is x ≤ 3, only the left-branch input remains. The restriction changes how uniquely an output identifies an input.
Do not choose a branch simply because it is on the right side of the page or because it gives a positive input. The domain must supply the reason. Both candidates can be positive, as 1 and 5 demonstrate.
The valid-solution guide develops this broader distinction between a calculated candidate and an accepted answer. In quadratic questions, the final check can be a contextual interval, a whole-number requirement or a statement selecting an earlier or later occurrence.
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An invented model uses y = (t − 4)² + 3 for t ≥ 0 over a stated interval. For output y = 7, the equation becomes (t − 4)² = 4, giving t = 2 and t = 6.
Both times are nonnegative, so positivity alone does not select one. If both lie within the model’s stated time interval, both are valid occurrences of the output. A learner should not discard the later one merely because an earlier time was found first.
If the question asks for the first time, answer 2. If it asks for the time after t = 4, answer 6. If it asks for all times in the domain, report both. The quantifier in the question determines the final selection.
Do not assume that every quadratic model of time has a physical interpretation outside its stated interval. The example is invented to teach repeated outputs. Contextual models often have limited domains, and those limits should be applied before giving a real-world conclusion.
This case shows why the instruction “reject the negative answer” is not a complete method for contextual quadratics. Sometimes one candidate is negative; sometimes both are positive; sometimes neither lies in the permitted interval. Read the actual condition rather than applying a remembered rejection pattern.
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For y = (x − 3)² + 2, output y = 4 requires (x − 3)² = 2. The real inputs are 3 − √2 and 3 + √2. They are valid real-number solutions of the equation.
If the question restricts x to integers, neither candidate is permitted. There is no integer input giving output 4. That does not mean there is no real input; the domain-specific conclusion must be stated accurately.
For output y = 3, the equation gives (x − 3)² = 1, so x = 2 or 4. Both are integers. Similar-looking output levels can therefore have different outcomes under the same integer restriction.
Do not round 3 ± √2 to nearby integers and call those exact solutions. Substituting the rounded values generally changes the output. A whole-number domain is a condition to satisfy, not an instruction to round any real candidate until it fits.
The guide to two valid examples and insufficient information explains why domains matter when judging possibilities. Here the equation determines real candidates, while the stated domain controls which of those candidates count as answers to the particular question.
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If a question supplies a plotted quadratic and asks for inputs at a horizontal level, read the intersections with that level and project down to the x-axis. There may be two, one or no visible intersections within the displayed and stated domain.
Graph readings are generally estimates at the precision supported by the scale and drawing. Do not invent exact integers merely because an intersection looks close to a grid line. If an exact equation is supplied and exact values are required, use the appropriate algebraic method.
Symmetry can check the readings. If the axis is x = 3 and the estimated inputs are 0.8 and 5.2, their midpoint is 3. If the readings are 0.8 and 5.8, inspect the scale, curve or projection because that pair is not symmetric about 3.
This check does not prove that both readings are accurate. A pair such as 1 and 5 is also symmetric about 3, but it belongs to a particular output level. You still need to read the correct horizontal line.
The exact-values and graph-estimates guide explains the wider distinction. Symmetry is a consistency check; it does not turn a rough graph measurement into an exact solution without additional information.
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A student solves (x − 3)² = 4 by writing x − 3 = 2 only. The repair is to test both 2 and −2 as bracket values. A short numerical square example can make the missing branch visible before returning to algebra.
Another student gives partner −1 for input 1 when the axis is x = 3. The repair is to measure distance from the correct axis rather than from zero. Use midpoint checking: the correct pair must average to 3.
A third finds both inputs correctly but ignores 0 ≤ x ≤ 4. The repair is a domain check after solving. A fourth rejects x = 1 because the turning point lies at 3, even though the domain includes the left branch. The repair is to distinguish the axis from a restriction.
The marked-paper guide helps identify these different next steps. More factorisation practice will not automatically repair a forgotten domain or a mistaken reflection line.
Ask the learner to state the cause of each accepted or rejected candidate. The final answer should be a reasoned selection from the possibilities, not simply whichever root looks most familiar.
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Use y = (x − 3)² + 2 for Questions A to D. Question A: find all real inputs giving output 11. Check the midpoint of your pair against the axis.
Question B: find the input giving output 2. Explain why it is one distinct input rather than two different values. Then decide whether any real input gives output 0.
Question C: find all inputs giving output 6 when the domain is 0 ≤ x ≤ 4. Show the unrestricted candidates before applying the domain.
Question D: one input giving a particular output is 1.5. Find the other input using symmetry and calculate their common output. State the axis used in the reflection.
Question E: for y = 12 − (x − 2)², find inputs giving output 3, then state the maximum output and where it occurs. Explain what happens for output 13.
Question F: an invented model y = (t − 4)² + 3 has domain 0 ≤ t ≤ 5. Find the time or times giving output 7. Explain why a candidate can be a correct solution of the unrestricted equation but not a valid time in this model.
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For A, (x − 3)² = 9 gives x = 0 and 6. Their midpoint is 3, and both give output 11. For B, the minimum output 2 occurs only at x = 3. Output 0 is below the minimum, so no real input gives it.
For C, the unrestricted candidates are 1 and 5. Only 1 lies in 0 ≤ x ≤ 4. State x = 1 as the accepted input rather than reporting both without the condition.
For D, the partner is 6 − 1.5 = 4.5. Each is 1.5 units from the axis, giving output 1.5² + 2 = 4.25. The reflection is about x = 3, not about the y-axis.
For E, the inputs are −1 and 5. The maximum output is 12 at x = 2. Output 13 would require a negative squared value after rearranging, so there is no real input.
For F, the unrestricted candidates are 2 and 6, but only 2 lies in the stated interval. The domain excludes 6 even though substitution into the unrestricted formula gives the requested output. The formula and its allowed interval together define the model used in the question.
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Consider y = 2(x − 1)² − 3. Its axis is x = 1 and its minimum output is −3. Find the inputs at outputs 5 and −1, then compare how far the intersections lie from the axis.
At output 5, 2(x − 1)² − 3 = 5 gives (x − 1)² = 4. The inputs are −1 and 3, each two units from the axis. At output −1, the equation gives (x − 1)² = 1, so the inputs are 0 and 2, each one unit from the axis.
The higher output level lies farther from the axis in this upward-opening quadratic. The paired inputs at each level still average to 1. The coefficient 2 changes the relation between vertical height and horizontal distance but does not move the axis.
If the domain is x ≥ 1, only inputs 3 and 2 remain for the two respective levels. If the domain is −2 ≤ x ≤ 4, both pairs remain. State the domain before interpreting how many crossings belong to the actual question.
This example separates three controls: the axis locates the centre of the pairing, the vertical coefficient changes the height-distance relationship, and the domain selects permitted points. Keeping them separate makes a changed quadratic easier to understand without rebuilding the whole idea from scratch.
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Suppose a quadratic gives the same output at x = 2 and x = 8, and the two inputs are distinct. The axis lies halfway between them, at x = 5. The paired points occupy equal horizontal distances on opposite sides of the symmetry line.
This conclusion concerns a quadratic relationship, not every arbitrary function with two matching outputs. Many other functions can share two output values without having that midpoint as a global symmetry axis. The function type is a condition of the argument.
Once the axis is known, an input of 3 has symmetric partner 7, and an input of 4.5 has partner 5.5. Their common output values still depend on the particular quadratic. The axis identifies the pairing, not the entire formula.
For example, y = (x − 5)² and y = 2(x − 5)² + 1 share the same axis and the same input-pair pattern. But at x = 3 and 7, their outputs are 4 and 9 respectively. Knowing an axis does not determine the vertical scale or shift.
This is a useful sufficiency check. The supplied equal-output pair can locate the axis, while additional information is needed to identify a unique quadratic formula. State the conclusion that follows without importing the rest of a familiar parabola.
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Solve x² − 4x + 1 = 6. Completing the square gives (x − 2)² − 3 = 6, so (x − 2)² = 9. The inputs are x = −1 and x = 5. Their midpoint is 2, matching the axis of the quadratic expression on the left.
Substitution checks the requested level: at −1, the expression is 1 + 4 + 1 = 6; at 5, it is 25 − 20 + 1 = 6. The two points share output 6 despite lying on opposite sides of the axis.
If the question instead asks for the roots of y = x² − 4x + 1, set the output to zero. The equation becomes (x − 2)² = 3, giving x = 2 − √3 and x = 2 + √3. These are a different pair at a different horizontal level.
The midpoint check works for both pairs. It confirms their symmetry but does not identify which level was intended. A pair can satisfy the axis relationship while solving the wrong equation. Always combine the symmetry check with substitution into the actual requested equality.
This example also shows how to inspect a copied question. Changing the right-hand side from 6 to 0 changes the target even though the left-hand expression stays identical. A learner who reproduces the same roots for both may be solving a remembered template rather than reading the stated equation.
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Begin by giving the axis and one input, then ask for the partner. Next give the completed-square expression and ask the learner to identify the axis before finding the partner. Finally, give an expanded expression and a target output so that the learner must choose a suitable method.
At each stage, ask for the same two checks: do the candidates have the correct midpoint, and do they give the requested output? Add the domain check whenever the question supplies an interval or contextual condition.
The worked-example revision guide describes this gradual removal of prompts. The goal is to make the decisions independent, not to require a permanent checklist longer than the calculation itself.
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Must every output of a quadratic have two inputs?
No. For real inputs, an output can correspond to two, one or no distinct inputs depending on the turning level and opening direction. A restricted domain can further reduce the permitted possibilities.
Are the paired inputs always opposites?
Only when the relevant symmetry axis is x = 0. In general, they are equally spaced about x = h and add to 2h. For axis x = 3, inputs 1 and 5 are paired even though both are positive.
Does a quadratic with no real roots have no repeated outputs?
No. Roots concern output zero specifically. A quadratic whose minimum is above zero can have no real roots while many higher output levels still occur at two inputs.
Can symmetry replace substitution?
It can find or check a partner when the axis and one point are known. Substitution remains useful for checking that the pair gives the requested output and that no copying error has changed the function.
Where should I go next?
Use the Mathematics explanation directory for another focused decision. The Clementi Secondary 1 Mathematics small-group guide gives the local teaching context. The lasting habit is to distinguish a forward input-to-output calculation from the reverse search for inputs, then let symmetry, the target level and the domain determine the full answer.
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