Two tangent gradients and one point on a quadratic can determine its entire equation. For f(x) = ax² + bx + c, the derivative is f′(x) = 2ax + b. Two gradient readings at different x-values usually determine a and b; the known point then determines c.
For example, if f′(0) = 3, f′(2) = 11 and the graph passes through (1, 8), then b = 3, 4a + b = 11 and a + b + c = 8. Thus a = 2, b = 3, c = 3, giving f(x) = 2x² + 3x + 3. The lesson is to distinguish evidence about slope from evidence about position and to verify both before accepting a reconstructed curve.
What three unknown coefficients mean
A general quadratic is f(x) = ax² + bx + c, with a non-zero for a genuine parabola. The coefficient a controls its curvature and opening direction; b affects the location of its turning point; c gives the y-intercept.
Three independent numerical conditions can determine these three unknowns. Three points on the curve are one possibility, but they are not the only one. Two slope conditions and a single ordinary point can also be sufficient because differentiation turns the quadratic into a straight-line expression in x.
This distinction matters in Additional Mathematics. A question might not offer three obvious coordinates, yet still provide enough information. The student must translate each statement into either f(u) = v or f′(u) = m before deciding how to solve.
Count unknown coefficients, then count independent constraints. Do not assume that every number in the wording represents a point on the original graph. Some data belong to the gradient function.
Differentiate before forming equations
Differentiate f(x) = ax² + bx + c to obtain f′(x) = 2ax + b. The derivative of c is zero, so the gradient data do not directly reveal the vertical position of the curve.
A known tangent gradient m at x = u means f′(u) = m. Hence 2au + b = m. This is a linear equation in the unknown coefficients a and b.
A known point (v, w) on the curve means f(v) = w, or av² + bv + c = w. That condition introduces the constant c.
Keeping the two kinds of equations separate makes reconstruction efficient: the gradient conditions first determine the shape-related coefficients, and the point condition fixes the vertical translation. A final check with the original curve and derivative confirms that none of the information was misread.
Worked example: gradients 3 and 11
Let f(x) = ax² + bx + c. Suppose its tangent gradient is 3 at x = 0 and 11 at x = 2, and it passes through (1, 8).
From f′(0) = 3, using f′(x) = 2ax + b, we get b = 3. From f′(2) = 11, we get 4a + b = 11. Substituting b = 3 gives 4a = 8, so a = 2.
Now use the ordinary point. Since f(1) = 8, we have a + b + c = 8. Hence 2 + 3 + c = 8 and c = 3.
The complete function is f(x) = 2x² + 3x + 3. Each condition has been used exactly once to obtain an independent equation, avoiding unnecessary simultaneous work in three unknowns.
Verify all three observations
The proposed function f(x) = 2x² + 3x + 3 has derivative f′(x) = 4x + 3. At x = 0 the gradient is f′(0) = 3. At x = 2 the gradient is f′(2) = 8 + 3 = 11.
At the supplied point x = 1, the curve value is f(1) = 2 + 3 + 3 = 8. Therefore (1, 8) lies on the graph.
These three checks correspond exactly to the three original pieces of evidence. Checking only f(1) would not be sufficient: many different quadratics pass through (1, 8). Checking only both gradients would leave a free vertical constant.
Students should complete reconstruction with a short evidence audit. The calculation is not fully established until the same polynomial reproduces every supplied condition.
Why two gradients reveal curvature
For a quadratic, f′(x) = 2ax + b is linear. The change in gradient between two inputs x₁ and x₂ is f′(x₂) − f′(x₁) = 2a(x₂ − x₁).
Thus, when the x-values differ, a = [f′(x₂) − f′(x₁)]/[2(x₂ − x₁)]. This formula says that the slope of the derivative graph is 2a.
In the worked example, the gradient rises from 3 to 11 while x rises from 0 to 2. The gradient change is 8 over a horizontal change of 2. Therefore 2a = 4 and a = 2.
This is more than an efficient formula. It gives geometric meaning to the coefficient a: a positive a makes the tangent gradient increase steadily with x; a negative a makes it decrease. Two gradient observations can therefore reveal the parabola’s opening direction.
Why the known point fixes the vertical constant
Two different quadratics can have identical derivatives. For example, 2x² + 3x + 3 and 2x² + 3x − 5 both have derivative 4x + 3, so they produce the same tangent gradient at every x.
They are vertical translations of one another, separated by eight units. Gradient information cannot distinguish them because adding a constant does not change the derivative.
The point (1, 8) selects the first curve: 2(1)² + 3(1) + 3 = 8. The second curve gives zero at x = 1 and fails the point condition.
That is the precise role of the third observation. The slope data determine the derivative, but the point anchors the curve’s vertical position. This connects quadratic reconstruction with the constant of integration that appears later in calculus.
Another example with negative inputs
Suppose f′(−1) = −7, f′(1) = 1 and f(0) = 5. Begin with f′(x) = 2ax + b.
The gradient conditions become −2a + b = −7 and 2a + b = 1. Subtracting gives 4a = 8, so a = 2. Substitute into 2a + b = 1 to obtain b = −3.
The point f(0) = 5 gives c = 5 immediately. Therefore f(x) = 2x² − 3x + 5.
Check its derivative f′(x) = 4x − 3. At −1 the value is −7; at 1 it is 1. At x = 0 the original function equals 5. Negative inputs cause no conceptual change, but protective brackets and correct signs matter.
A stationary point supplies a zero gradient
If a curve has a stationary point at x = p, its tangent gradient is zero there. Thus the information is f′(p) = 0, even before its y-coordinate is known.
Suppose f′(1) = 0, f′(3) = 8 and f(2) = 1. We have 2a + b = 0 and 6a + b = 8. Subtracting gives 4a = 8, so a = 2 and b = −4.
Now f(2) = 1 means 4a + 2b + c = 1, or 8 − 8 + c = 1, so c = 1. The curve is f(x) = 2x² − 4x + 1.
Completing the square gives 2(x − 1)² − 1, confirming that x = 1 is indeed the turning-point input. This example makes the stationary condition and gradient change work together.
A tangent equation gives both a slope and a point
Sometimes the question states that y = 4x + 7 is tangent to a curve at x = 2. The line’s gradient is 4, so f′(2) = 4. It also gives the contact height 4(2) + 7 = 15, so f(2) = 15.
These are two conditions from a single geometric statement. A second independent gradient at a different input would then supply a third condition for a general quadratic.
For example, if f′(0) = 0 as well, then b = 0 and 4a = 4, so a = 1. The contact condition f(2) = 15 becomes 4 + c = 15, giving c = 11. The reconstructed function is x² + 11.
At x = 2 its gradient is 4 and its value 15, matching the stated line. Read tangency as information about both position and local direction, not merely about the line’s intercept.
What happens if two gradient inputs are the same?
Suppose a problem claims f′(1) = 3 and f′(1) = 5 for the same differentiable function at the same input. No polynomial can satisfy both conditions. One derivative at a fixed input cannot have two different values.
If both statements instead say f′(1) = 3, they are compatible but redundant. Repeating an equation does not add independent information about the unknown coefficients.
Two gradients at distinct x-values are what allow the derivative’s slope 2a to be recovered. When the inputs coincide, the expression [f′(x₂) − f′(x₁)]/[2(x₂ − x₁)] would require division by zero.
Check the input coordinates before assuming that two gradient statements determine a unique quadratic. Independent information, not merely the number of statements, is what matters.
Equal gradients at different inputs imply a linear result
For a genuine quadratic, f′(x) = 2ax + b with a ≠ 0, so the derivative is strictly increasing or strictly decreasing. It cannot have the same gradient at two different x-values.
Suppose f′(0) = 2 and f′(2) = 2. The conditions give b = 2 and 4a + b = 2, hence a = 0. This is a linear function rather than a genuine quadratic.
If the graph also passes through (1, 7), then f(1) = b + c = 7 gives c = 5. The unique polynomial of degree at most two is f(x) = 2x + 5.
A student should report that the observations determine a straight line and are incompatible with an insistence that a ≠ 0. Do not invent a tiny nonzero a merely to produce a parabola.
Insufficient information leaves a family of curves
If only f′(0) = 3 and f′(2) = 11 are given, the derivative is fixed as 4x + 3 and the corresponding quadratic family is f(x) = 2x² + 3x + c.
Every real c gives the same gradient observations. Without a point, a y-intercept or another independent value condition, the vertical location remains unknown.
For c = 3 the curve passes through (1, 8); for c = 10 it passes through (1, 15). Both satisfy the original gradient statements.
This is a simple example of why the number of unknown coefficients matters. A quadratic has three coefficients, and knowing two gradients generally constrains only the two appearing in its derivative. An additional condition is required to identify the constant.
A known value may come from an intercept
If the additional condition is a y-intercept rather than a general point, the calculation becomes especially quick. The y-intercept is f(0) = c.
Take f′(−1) = −7, f′(1) = 1 and a y-intercept of 5. The gradient conditions give a = 2 and b = −3, while the intercept gives c = 5.
The resulting quadratic f(x) = 2x² − 3x + 5 is identical to the earlier negative-input example. The intercept is simply a specific kind of point condition at x = 0.
Avoid confusing the intercept of a stated tangent line with the y-intercept of the curve. Unless that tangent touches the curve at x = 0, its constant term need not equal c.
A second point can provide a consistency check
After reconstructing f(x) = 2x² + 3x + 3 from two gradients and (1, 8), suppose the question additionally states that the graph passes through (2, 17).
Evaluate f(2) = 2(4) + 3(2) + 3 = 8 + 6 + 3 = 17. The extra condition is consistent with the reconstructed polynomial.
If the extra point were (2, 20), the conditions would be contradictory: the gradients and first point already uniquely determine the quadratic, and the result cannot produce 20 at x = 2.
Do not force a different coefficient set by discarding a previously valid condition. Recheck arithmetic and the question statement, then identify whether the additional observation is a confirmation or a contradiction.
Interpret the sign of a from gradient data
From f′(x) = 2ax + b, an increase in gradient with x indicates a > 0; a decrease indicates a < 0. This is a useful qualitative prediction before calculating the coefficients.
For f′(0) = 3 and f′(2) = 11, the gradient increases, so the curve opens upward. The exact calculation a = 2 agrees.
For f′(0) = 5 and f′(2) = 1, the gradient decreases by four over a two-unit x change. Thus 2a = −2 and a = −1, so a compatible genuine quadratic opens downward.
This check catches a common error in subtracting simultaneous equations. If algebra gives a negative leading coefficient while the supplied gradients clearly increase with x, investigate the sign of the derivative equation.
Find the vertex after reconstructing the formula
For the opening quadratic f(x) = 2x² + 3x + 3, the derivative is 4x + 3. Set it to zero to find the stationary input x = −3/4.
Evaluate the function there: f(−3/4) = 2(9/16) + 3(−3/4) + 3 = 9/8 − 9/4 + 3 = 15/8.
The vertex is therefore (−3/4, 15/8). Because a = 2 > 0, this is a minimum.
This extra analysis is useful as a graph check, but it is not required merely to determine a, b and c. Students should learn to distinguish the quantities requested in a question from optional extensions that deepen understanding.
How to reconstruct using integration instead
Since f′(x) = 4x + 3, an antiderivative is f(x) = 2x² + 3x + C. The supplied point f(1) = 8 gives 2 + 3 + C = 8, so C = 3.
This is equivalent to the coefficient method. Two slope observations first determine the linear gradient function, and one point then fixes the integration constant.
For students who have studied integration, the reverse-differentiation route may make the structure especially clear. For students who have not, simple coefficient comparison is sufficient and avoids using a method beyond their current sequence.
Showing two valid routes illustrates mathematical connection, not an expectation that every exercise should be solved twice under examination time pressure.
Why a slope at one point does not determine the slope everywhere
Knowing f′(0) = 3 gives b = 3 but leaves a unrestricted. The functions x² + 3x + c, 2x² + 3x + c and −x² + 3x + c all have gradient 3 at x = 0.
Their tangent directions differ at other inputs because their derivative functions have different slopes. That is why the second gradient reading at x = 2 matters.
In a physical interpretation, an instantaneous rate measured at one moment does not reveal how quickly that rate itself changes. The second measurement detects the change in gradient across the interval.
This concept prepares learners to understand why derivatives are functions rather than isolated answers. A curve is reconstructed from a pattern of local changes together with an anchoring value.
A neat examination layout
Write f(x) = ax² + bx + c and f′(x) = 2ax + b at the top of the working. Translate the two gradient statements into equations in a and b.
Solve these first, since c does not appear. Then substitute the known point into f(x) and determine c. Write the full quadratic with its coefficients.
Complete a three-line verification: compute the derivative at both stated inputs and the function value at the known point. If the question asks for a graph feature, continue to that calculation separately.
This layout makes each logical step visible to a marker and helps identify whether an error came from differentiation, interpreting the question or solving simultaneous equations.
Five independent tasks with answers
Task one: f′(0) = 3, f′(2) = 11 and f(1) = 8. Answer: f(x) = 2x² + 3x + 3.
Task two: f′(−1) = −7, f′(1) = 1 and f(0) = 5. Answer: f(x) = 2x² − 3x + 5.
Task three: f′(1) = 0, f′(3) = 8 and f(2) = 1. Answer: f(x) = 2x² − 4x + 1.
Task four: f′(0) = −1, f′(4) = 7 and f(2) = 3. Answer: f(x) = x² − x + 1.
Task five: f′(0) = 2, f′(2) = 2 and f(1) = 7. Answer: f(x) = 2x + 5; it is linear, so no genuine quadratic satisfies these observations.
For each, write the two derivative equations before finding c and explain whether a ≠ 0.
Common questions about slope evidence
Can one gradient and one point determine a quadratic? Not generally, because a, b and c contain three unknowns. Another independent condition is usually needed.
Are the gradients the y-values of the original graph? No. They are values of its derivative, measuring tangent slope.
Why does c disappear when differentiating? Vertical translation does not change a curve’s tangent gradients.
Can both gradients be negative? Yes. The important information is how their values change between the two distinct x-inputs.
What if the gradient is zero? That identifies a stationary input and still provides a valid derivative equation.
Do we need integration to solve the problem? No. Differentiation and simultaneous equations are enough. Integration offers an alternative interpretation when it has been taught.
How parents can recognise secure understanding
A student should be able to explain the three constraints in their own words, derive f′(x), solve a and b from gradients, then obtain c from a curve point. A child who knows the derivative rule but treats a gradient as a coordinate needs conceptual work, not merely more algebra.
Another may translate correctly but miscalculate with a negative x-input. That calls for a short substitution repair. A third may solve the coefficients correctly but not recognise that equal gradients at distinct inputs imply a = 0.
In eduKateSG’s maximum three-student tutorial setting near Sixth Avenue MRT, generally with 1.5-hour weekly lessons where suitable placement is available, the teacher can diagnose the earliest incorrect decision and check transfer on a changed problem.
Independent reconstruction from new data is stronger evidence of progress than reproducing an identical worked solution neatly.
Syllabus scope and connected study routes
Differentiation, tangents and quadratic functions form an important chain of mathematics skills. The exact level of reconstruction and any integration extensions should follow the student’s taught syllabus and school sequence.
For Singapore, consult SEAB’s 2026 GCE O-Level Additional Mathematics list, 2027 SEC G3 K341 list or 2027 SEC G2 K232 list, as appropriate. Avoid assuming every extension is assessed identically.
Read the related guides How to Reconstruct a Quadratic from Three Points, How to Recover Unknown Coefficients from a Tangent Line and How to Reconstruct a Cubic from a Stationary Point and a Root.
Browse the Additional Mathematics archive or arrange a parent–student consultation with eduKate Singapore. The goal is to reconstruct the only quadratic that fits all independent evidence and explain why. Properly taught kids shine a bright light into the future.