A quadratic can be reconstructed from its turning point and one definite-integral condition without knowing three separate graph points. Write the quadratic in vertex form f(x)=a(x−h)²+k, where (h,k) is the given turning point. The integral determines the remaining scale coefficient a, provided the interval has positive width and the condition supplies an independent numerical value.
For example, suppose a quadratic has turning point (1,2), and the definite integral from x=0 to x=2 equals 20/3. Write f(x)=a(x−1)²+2. Integration gives (2/3)a+4=20/3, so a=4. The reconstructed quadratic is f(x)=4(x−1)²+2, or 4x²−8x+6.
Did you know that the same integral condition can also tell us whether the quadratic opens upward or downward? This guide explains why vertex form is efficient, how to check the turning point and area, what happens when the data are insufficient, and why geometric area should not automatically be equated with a signed definite integral.
Why a turning point is powerful information
A general quadratic f(x)=ax²+bx+c has three coefficients. Its turning point supplies two connected facts: a location where the gradient is zero and the actual function value at that location.
Writing the curve as f(x)=a(x−h)²+k automatically satisfies both parts when (h,k) is the turning point and a is nonzero.
Therefore only one unknown coefficient a remains. A single independent condition, such as a particular definite integral over a nonzero interval, may be enough to determine it.
This is a useful change of representation. Instead of solving three simultaneous coefficient equations in ax²+bx+c, use the form that directly incorporates the information the question has already given.
The difference between vertex input and vertex height
If the turning point is (h,k), h is its x-coordinate and k is the function output there. They play different roles.
The expression (x−h)² vanishes at x=h, leaving f(h)=k. The coefficient a determines how steeply the graph rises or falls away from that point.
For the given turning point (1,2), the correct form is f(x)=a(x−1)²+2. Writing (x+1)² would place the turning point at x=−1, not 1.
A quick check is to substitute x=h into the proposed vertex form before any integration. If it does not give k, the formula was misread and later calculations will reconstruct the wrong curve.
Why the coefficient a cannot be zero for a genuine quadratic
The family f(x)=a(x−h)²+k becomes constant when a=0. A constant function does not have the same unique quadratic turning-point geometry.
Therefore a genuine quadratic requires a≠0. Positive a gives an upward-opening parabola with a minimum at (h,k), while negative a gives a downward-opening parabola with a maximum there.
The definite-integral condition can determine the sign as well as the magnitude of a. But if it forces a=0, the stated assumptions may be inconsistent with requiring a genuine quadratic.
Include that structural check after solving. An algebraic coefficient may satisfy one numerical equation while invalidating the model type used to construct the problem.
Set up the opening example in vertex form
Suppose the curve has turning point (1,2) and ∫₀² f(x)dx=20/3.
Write f(x)=a(x−1)²+2, where a is the only unknown.
The definite-integral equation is ∫₀²[a(x−1)²+2]dx=20/3.
Separate the contributions: a∫₀²(x−1)²dx+∫₀²2dx=20/3. The constant term contributes four because the integration interval has width two.
The remaining integral of (x−1)² provides the coefficient that will multiply a. This setup is shorter and more reliable than expanding a general ax²+bx+c before using the turning point.
Integrate the squared-distance term
An antiderivative of (x−1)² is (x−1)³/3. Evaluate it between x=0 and x=2.
At x=2, its value is 1³/3=1/3. At x=0, the value is (−1)³/3=−1/3.
Subtract upper minus lower: 1/3−(−1/3)=2/3. Therefore ∫₀²(x−1)²dx=2/3.
This result is positive, as expected, since the integrand is a square. The negative antiderivative value at the lower bound does not imply a negative definite integral. The orientation and subtraction give the correct non-negative accumulation.
Solve for the leading coefficient
The original integral condition now reads (2/3)a+4=20/3.
Convert four to twelfths? It is simpler to use thirds: 4=12/3. Therefore (2/3)a+12/3=20/3.
Subtract 12/3 to get (2/3)a=8/3. Multiply by 3/2 to obtain a=4.
The reconstructed vertex-form equation is f(x)=4(x−1)²+2. The positive coefficient confirms that the turning point is a minimum.
This answers the reconstruction without needing a third ordinary graph point or a quadratic formula.
Write the expanded polynomial if requested
Expand (x−1)²=x²−2x+1. Multiply by four and add two.
Thus f(x)=4x²−8x+4+2=4x²−8x+6.
Both forms represent the same quadratic. Vertex form makes the turning point visible, while expanded form may be more convenient for differentiation or term-by-term integration.
Keep the form that the question requests. Do not regard the two expressions as separate solutions. They are algebraically identical representations of one reconstructed curve.
Verify the turning point with direct substitution
In vertex form f(x)=4(x−1)²+2, the squared term is smallest when x=1, where it equals zero.
Therefore f(1)=2. Since the squared term cannot be negative and its coefficient is positive, no other x gives a smaller output. Thus (1,2) is indeed the minimum turning point.
The expanded form agrees: f(1)=4−8+6=2.
This check establishes both the coordinate and the type of turning point. It is stronger than noting that the function merely passes through (1,2), because an ordinary point is not automatically a stationary minimum.
Verify stationarity by differentiation
Differentiate f(x)=4x²−8x+6 to obtain f′(x)=8x−8=8(x−1).
At x=1, the derivative vanishes, so the tangent is horizontal. The second derivative is f″(x)=8>0, confirming a strict local minimum.
The vertex-form argument already established the global minimum, but the derivative offers an independent calculus check.
A student who reports turning point (1,2) after reconstructing the curve should be able to verify that both f(1)=2 and f′(1)=0. These are distinct mathematical conditions encoded by the vertex form.
Verify the definite integral from the expanded form
Use f(x)=4x²−8x+6. An antiderivative is F(x)=(4/3)x³−4x²+6x.
At x=2, F(2)=32/3−16+12=32/3−4=20/3. At x=0, F(0)=0.
Therefore ∫₀² f(x)dx=20/3, exactly the given condition.
This independent verification catches errors in the coefficient recovery or algebraic expansion. A mathematically convincing final answer should reproduce every original condition rather than merely solve one rearranged equation.
The integral also represents geometric area here
Because f(x)=4(x−1)²+2 is at least two for every real x, it lies strictly above the x-axis.
Therefore the definite integral from 0 to 2 is also the ordinary geometric area between the curve and the x-axis over that interval.
The area is 20/3 square units, provided the coordinate scales represent compatible physical length units.
This equality between signed integral and geometric area is justified by positivity. It does not hold for every quadratic. When a graph crosses the axis, the total geometric area must be calculated using absolute values or appropriately separated regions.
Find the average function value as a check
The average value over the interval [0,2] equals the definite integral divided by the width two.
Thus f_avg=(20/3)/2=10/3.
The function’s minimum value is two and its endpoint values are both six. The average 10/3 lies between those bounds, as a continuous average must.
This does not mean the average equals the midpoint output f(1)=2. The upward curvature makes the mean larger than the vertex height. It also does not equal the arithmetic mean of the two endpoint outputs, which is six.
Understand a general reconstruction with known vertex
Suppose f(x)=a(x−h)²+k, with a unknown and vertex (h,k) given. Let an independent condition state ∫_L^U f(x)dx=I, where L Separate the terms: I=a∫_L^U(x−h)²dx+k(U−L). Define J=∫_L^U(x−h)²dx. Because a square is non-negative and the interval has positive width, J>0. Therefore a=[I−k(U−L)]/J. This is a unique real value. The resulting function is genuinely quadratic when a≠0. The formula shows exactly what information the integral contributes after the turning point has fixed h and k. The antiderivative of (x−h)² is (x−h)³/3. Hence J=[(U−h)³−(L−h)³]/3. For any real h and L The complete coefficient formula is therefore a=3[I−k(U−L)]/[(U−h)³−(L−h)³]. This formula is useful for checking a general problem, but students should understand its derivation rather than copy it without considering the actual bounds. The turning point need not lie in the interval for the coefficient to be recoverable. Suppose the bounds are h−d and h+d, with d>0. Substitute u=x−h. The squared-distance integral becomes ∫_(−d)^d u²du. By symmetry it equals 2∫₀^d u²du=2d³/3. The interval width is 2d, so the full integral of f becomes I=(2ad³)/3+2kd. This makes coefficient recovery straightforward. It also gives average value f_avg=k+ad²/3. In the opening example h=1,d=1,k=2 and I=20/3. The average relation gives 10/3=2+a/3, so a=4, confirming the earlier result. The expression (x−h)² is even about x=h. Its integral over symmetric limits is twice the positive half-interval integral, not zero. By contrast, an odd function such as x−h integrates to zero over symmetric limits. Students sometimes see positive and negative input offsets around the vertex and expect cancellation. Squaring removes those offset signs, so both halves contribute positively. This distinction is essential for uniqueness: the coefficient multiplying the squared term remains visible in the integral condition. A mistaken cancellation would make the student believe the integral cannot determine a even though it can. Suppose the turning point is (−1,3) and ∫_(−2)^0 f(x)dx=22/3. Write f(x)=a(x+1)²+3. The bounds are symmetric about x=−1, with d=1. Therefore ∫_(−2)^0(x+1)²dx=2/3, while the constant baseline contributes 3×2=6. The condition is (2/3)a+6=22/3. Since six is 18/3, we have (2/3)a=4/3, so a=2. Thus f(x)=2(x+1)²+3. It opens upward and has the specified minimum (−1,3). Direct substitution or integration verifies the supplied data. Keep the turning point (1,2), but suppose ∫_1^3 f(x)dx=28/3. Write f(x)=a(x−1)²+2. Let u=x−1. The squared-distance integral becomes ∫₀²u²du=8/3. The baseline contributes four because the interval width is two. Therefore (8/3)a+4=28/3. Subtract 12/3 to get (8/3)a=16/3, so a=2. The reconstructed curve is f(x)=2(x−1)²+2. This shows that the interval need not be centred at the vertex. Correct evaluation of the squared-distance integral handles any nondegenerate bounds. For f(x)=a(x−h)²+k, the vertex is a minimum if a>0 and a maximum if a<0. The integral formula makes the sign of a depend on whether the observed total I is above or below the baseline contribution k(U−L). Because J>0, we have a>0 exactly when I>k(U−L), and a<0 exactly when I Thus the integral can distinguish an upward-opening from a downward-opening quadratic. If the question has already declared the turning point a maximum, the recovered coefficient must be negative; otherwise the data or arithmetic are inconsistent with that declaration. Suppose a curve has turning point (1,5) and ∫₀² f(x)dx=26/3. Write f(x)=a(x−1)²+5. The squared-distance integral is 2/3, while the baseline contributes 10. Therefore (2/3)a+10=26/3. Write ten as 30/3, giving (2/3)a=−4/3 and hence a=−2. The curve is f(x)=−2(x−1)²+5. Since a is negative, (1,5) is a maximum. The function remains positive on [0,2]: at either endpoint its value is three and at the vertex it reaches five. Thus in this example the signed integral is also positive geometric area. A definite integral measures signed accumulation. A downward-opening quadratic can fall below the x-axis for much or all of the integration interval. If f(x)=−(x−1)²−2, then on [0,2] every output is negative. Its integral is −2/3−4=−14/3. That is mathematically valid for a signed integral. The geometric area between the curve and the x-axis is instead 14/3. A negative integral does not by itself mean the reconstructed coefficient or curve is incorrect. First check whether the question specifies a signed integral, a physical area, displacement or another context in which signs carry meaning. Suppose the problem says “the area between the curve and x-axis equals a given value,” rather than explicitly giving ∫ f(x)dx. If the curve is non-negative throughout the interval, area equals the ordinary definite integral. If the curve is non-positive throughout, area equals the negative of that integral. If the curve crosses the x-axis, split the interval at its real zeros and integrate the magnitude of the function on each piece. Therefore a geometric-area condition may produce a piecewise equation in a, rather than the simple linear integral equation used in the worked example. Check the curve’s sign after reconstructing a candidate and before accepting it as an area solution. Geometric area uses absolute values, so it can lose information about the sign or location of portions of the graph. For example, compare two functions with fixed central offset f_a(x)=a(x−1)²−1 over [0,2]. Their absolute areas depend on whether the parabola lies entirely below the axis or crosses it near the endpoints. The total area is not generally a linear function of a once sign changes occur. Different choices of a may produce the same total geometric area even though their signed integrals differ. This explains why a supplied area should not automatically be treated as one linear equation in a. The straightforward uniqueness proof applies to a specified signed integral, not to every possible absolute-area observation. The general formula gives a=0 exactly when the observed integral equals the constant baseline contribution k(U−L). In that case f(x)=k for every x, which is a constant function rather than a genuine quadratic. If the question specifically requires a quadratic with a unique turning point, these data are inconsistent with that requirement. The solution is not to invent a tiny nonzero coefficient or round zero away. If the problem allows a degenerate polynomial of degree at most two, the constant function may be acceptable under that broader wording. Mathematical modelling conditions determine how to interpret the algebraic result. For L That means the coefficient of a in the integral equation cannot vanish. The signed-integral condition therefore determines exactly one value of a. If L=U, the definite integral is zero for every integrable f, regardless of a. Such a zero-width condition supplies no information about the curve’s curvature. This is a useful exception. A problem cannot reconstruct a nonzero quadratic coefficient from an integral over an interval with identical upper and lower bounds unless additional data are supplied. Suppose the question says the vertex occurs at x=1 but does not give its y-coordinate. Then the family is f(x)=a(x−1)²+k, with both a and k unknown. One definite-integral condition gives one linear relationship between a and k. It generally leaves infinitely many solutions. For example, the condition ∫₀² f(x)dx=20/3 becomes (2/3)a+2k=20/3. This can be rearranged to a+3k=10. The pairs (a,k)=(4,2) and (a,k)=(1,3) both satisfy that integral equation, although they describe different curves. A second independent condition is required to fix both parameters. Continue with f(x)=a(x−1)²+k and integral condition a+3k=10. Suppose the graph also passes through (1,2). Since f(1)=k, the point condition gives k=2. Then a+3(2)=10, so a=4. This recovers the original curve f(x)=4(x−1)²+2. The example clarifies what the known full turning point contributed: its horizontal coordinate fixed the centre, while its vertical coordinate fixed k. Losing one coordinate creates an additional unknown requiring independent information. Knowing that f′(1)=0 tells us the quadratic’s axis of symmetry passes through x=1, but it does not give the curve’s height there. For f(x)=a(x−1)²+k, differentiation gives f′(x)=2a(x−1), so f′(1)=0 for every a and k. Therefore the derivative condition supplies no information about k. One area or integral observation may constrain a and k together but usually not determine both. When reconstructing a polynomial, distinguish gradient data, point data and accumulation data. They encode different kinds of information even when all refer to the same location or curve. For the opening function f(x)=4(x−1)²+2, inputs equally spaced around x=1 have the same output. For example, f(0)=4+2=6 and f(2)=4+2=6. The tangent at the midpoint x=1 is horizontal. This symmetry agrees with the stated turning point and with the equal contributions of the squared-distance term on the two sides of the interval. If a candidate expanded quadratic gives unequal outputs at x=0 and x=2 despite claiming axis x=1, a coefficient sign or bracket error has occurred. Symmetry is a powerful independent check before performing longer calculations. On [h−d,h+d], the mean of f(x)=a(x−h)²+k is k+ad²/3. This follows by dividing the full integral I=(2ad³)/3+2kd by the width 2d. For the opening example h=1,d=1,k=2 and average I/2=10/3. Thus 10/3=2+a/3, immediately giving a=4. The formula also shows why the average is above the vertex value for a>0 and below it for a<0. It connects curvature to continuous averaging and helps explain the recovered coefficient in a visually meaningful way. Work with fractions through the calculation rather than converting 20/3 into a decimal. The equation (2/3)a+4=20/3 has exact rational terms and leads directly to a=4. After reconstruction, substitute the coefficient into both vertex form and expanded form. Check f(1)=2 and ∫₀² f=20/3. A numerical calculator can confirm 20/3≈6.667, but the exact fraction should remain the primary answer when the question supplies exact values. This two-condition verification is more valuable than pressing a calculator repeatedly on the rearranged equation alone. First write the quadratic in vertex form f(x)=a(x−h)²+k using the complete known turning point. Second substitute the specified integration bounds and target total. Evaluate ∫(x−h)² exactly and add the baseline contribution k(U−L). Third solve the resulting linear equation for a. Check whether a is nonzero and whether its sign agrees with a stated maximum or minimum. Fourth expand the polynomial if requested. Finally verify the turning point with f(h)=k and f′(h)=0, and substitute into the original integral. If the condition is geometric area rather than a signed integral, perform the appropriate sign analysis first. Question one: vertex (1,2), ∫₀² f(x)dx=20/3. Answer: f(x)=4(x−1)²+2. Question two: for Question one, find the average value on [0,2]. Answer: 10/3. Question three: vertex (−1,3), ∫_(−2)^0 f(x)dx=22/3. Answer: f(x)=2(x+1)²+3. Question four: vertex (1,2), ∫_1^3 f(x)dx=28/3. Answer: f(x)=2(x−1)²+2. Question five: vertex (1,5), ∫₀² f(x)dx=26/3. Answer: f(x)=−2(x−1)²+5. Question six: vertex input x=1, unknown height k, and ∫₀² f=20/3. Is the quadratic unique? No; the remaining coefficients satisfy a+3k=10. Question seven: if an integral uses identical upper and lower limits, can it recover a by itself? No, because its value is zero for every coefficient. Reconstructing a quadratic from its turning point and a definite integral joins vertex form, quadratic graphs, parameter algebra, integration, average values and mathematical sufficiency. More advanced sign-dependent geometric-area versions should follow the student’s school sequence. For Singapore, consult the relevant SEAB references for 2026 O-Level Additional Mathematics or 2027 SEC G3, as appropriate. Connected eduKateSG guides include How to Reconstruct a Quadratic from Three Points, How to Find an Unknown Parameter from a Definite Integral, How to Find a Function’s Average Value from a Definite Integral and the Additional Mathematics archive. Families may enquire about suitable eduKate Singapore three-student Mathematics tutorials near Sixth Avenue MRT. The enduring skill is to select the polynomial form that captures known geometry and use an independent integral condition to determine the remaining scale without inventing information. Properly taught kids shine a bright light into the future.Evaluate the general squared-distance integral
A symmetric interval around the vertex is especially simple
Why a symmetric interval does not make the square disappear
A second example with a shifted turning point
A nonsymmetric integral interval also works
The coefficient sign identifies a maximum or minimum
A downward-opening reconstruction example
Why the integral may be negative
Why a geometric-area condition is different
Two different coefficients can sometimes give the same total area
What happens when a recovered coefficient is zero
Why a positive-width interval gives independent information
What if only the turning-point x-coordinate is known?
An additional graph point can supply the missing equation
Why the derivative condition alone is not a whole vertex
A quick model check using symmetry
The average-value shortcut for a symmetric interval
How to check exact arithmetic efficiently
A compact examination solution routine
Practice questions with checked answers
Syllabus fit and connected guidance