A simple quadratic can model a symmetric arch’s outline. Learn how a span and maximum rise determine a curve, and why a geometric arch equation is not a structural safety calculation.
Why this mathematical connection matters
A simple quadratic can model a symmetric arch’s outline. Learn how a span and maximum rise determine a curve, and why a geometric arch equation is not a structural safety calculation.
A worked mathematical example
For a 20-metre span with endpoints (0,0) and (20,0) and vertex (10,5), y=5-(x-10)²/20. At x=5, y=3.75.
Which concepts students should connect
Quadratic vertex forms, symmetry, gradients, exact height calculations, domain restrictions and engineering limitations.
What the mathematical model assumes
Start by defining every variable and its units. Identify the domain, the fixed quantities and the information that the question does not supply. A correct equation is only as useful as the assumptions supporting it. A classroom geometric or calculus model does not independently establish real-world engineering performance.
How to check the calculation
Substitute the proposed result into the original conditions, use exact arithmetic before rounding, and compare the sign and size of the answer with the diagram or graph. If a derivative is involved, interpret its units as output change per input change. If a geometric scaling is involved, check whether the required quantity is one-, two- or three-dimensional.
A practical revision routine
First draw and label the mathematical situation. Next write the governing equation and its permitted domain. Solve one example with exact values. Change one input and explain how the output changes. Finally, explain a limitation of the model and solve a related question independently without copying the worked answer.
What parents can ask
Ask which mathematical idea makes the example work, which assumptions are necessary, what the answer’s units mean and how the learner would check a proposed solution. Ask whether the result is a mathematical prediction or a verified real-world measurement. These questions build reasoning rather than rote memorisation.
Singapore SEC G3 Additional Mathematics connection
The application draws on algebra, functions, graphs, trigonometry or calculus depending on the specific example. Advanced engineering or computing interpretation is enrichment, not a claim of a separately examined topic. Check the official 2027 G3 syllabus at SEAB 2027 G3 syllabus directory.
Small-group learning and further reading
A true three-student tutorial can distinguish an algebra error from a misconception about the graph or the model. The immutable tutorial reference is Secondary 1 Mathematics small-group tutorial in Clementi. Related Additional Mathematics reading is available at Additional Mathematics library.
Build the mathematical relationship
A symmetric arch outline can be fitted by a quadratic, but an arch outline alone does not establish structural strength. Begin with y=5-(x-10)²/20. State the variables, coordinate convention and units. Explain how each coefficient changes the output. A model is not an observation; it is an assumed relationship whose consequences can be checked.
Calculate and verify an example
At x=0 and x=20, y=0. At x=10, y=5. At x=5, y=3.75. Substitution into the original equation verifies the numerical result. Where the problem involves a graph, check symmetry, intercepts, turning points and signs. A second method is valuable because it may reveal an arithmetic or interpretation error.
Go further with the mathematics
Differentiate y=5-(x-10)²/20 to get y’=-(x-10)/10. At x=5 the gradient is 0.5, while at x=15 it is -0.5. The gradients are mirror images. The model gives the shape’s height and tangent, but not stresses, loading, materials, construction methods or safety. A catenary arch, for example, follows a different mathematical curve.
Explain what would change if the input doubled
Start from y=5-(x-10)²/20. Replace the independent input with twice its original value, keeping all other parameters fixed. Simplify before making a numerical prediction. Do not assume that doubling an input always doubles an output: the exponent, function family and any fixed offsets determine the result.
Check units and the permitted domain
A length must be measured in compatible units. Area uses squared units and volume cubed units. A derivative has units of output per input. Identify whether the formula permits negative coordinates, only positive physical lengths or a restricted time interval. Exclude answers that violate the stated physical interpretation.
Separate an exact result from an approximation
The central worked example is At x=0 and x=20, y=0. At x=10, y=5. At x=5, y=3.75. Exact substitution proves an identity or checks a constructed function within its stated mathematical assumptions. It does not establish that a real manufactured object, physical measurement or engineered system matches the ideal formula. When a derivative approximation is used, state its approximate nature explicitly.
What students should be able to explain
A strong learner should identify the independent and dependent variables, construct the governing expression, substitute exact values, explain the graph and describe one limitation. They should be able to solve a modified problem with changed inputs, not merely repeat the same numerical steps.
A four-week study pathway
Week one: revise algebra, indices and coordinates relevant to the model. Week two: derive the governing relationship and complete three numerical checks. Week three: study graph shape, derivatives or scaling as applicable. Week four: solve a new example and write a short explanation of the model’s limitations.
Frequently asked questions
Does the mathematical formula describe every real object? No; its assumptions must hold. Can the formula be checked? Yes, substitute the original constraints and verify the dimensions. Is an attractive graph sufficient proof? No, algebraic verification is required. Is the advanced application a separate SEC examination topic? Not necessarily; the underlying functions, geometry and calculus are the assessed foundations.
Why study this in Singapore
The 2027 SEC G3 Additional Mathematics foundation includes relevant algebra, functions, graph interpretation, trigonometry and calculus. Consult the official SEAB syllabus directory for the exact assessed scope: SEAB 2027 G3 syllabus directory. The engineering, optics and approximation connections here are enrichment rather than examination predictions.
Set up an arch coordinate system
Place the horizontal deck along the x-axis and the centre of a symmetric arch at x=0. Let y measure vertical height above the line joining the two ends. If the arch rises to H at its centre and meets the end line at x=−L/2 and x=L/2, its idealised quadratic outline can be written y=H−ax². This is a model of a visible curve, not a calculation of bridge strength. A real bridge can have different structural geometry and loading.
Derive the coefficient from span and rise
At either end of the ideal arch, x=L/2 and y=0. Substitute into 0=H−a(L/2)². Therefore a=4H/L². The complete quadratic is y=H−(4H/L²)x². It reaches height H at x=0 and zero at the two endpoints. The expression is symmetric because replacing x with −x does not change x². This derivation uses only the stated span and rise, making it easy to check against a drawing.
Work an exact span-and-rise example
Suppose a model arch spans 20 metres and rises 5 metres. Then L=20 and H=5, so a=4×5/400=1/20. Its equation is y=5−x²/20 for −10≤x≤10. At x=10, y=5−100/20=0. At x=−10, the same result holds. At x=0, y=5. All three conditions are satisfied. This equation is a geometric idealisation, not a claim that an actual twenty-metre bridge should use this shape.
Calculate the height at a selected position
For y=5−x²/20, find the arch height five metres horizontally from the centre. Substitute x=5: y=5−25/20=3.75 metres. Because the equation is symmetric, the height at x=−5 is also 3.75 metres. This is a direct evaluation of a function. The answer is a vertical height above the end line, not a clearance measurement unless the end line and clearance reference have been explicitly defined.
Find where a specified height occurs
For the same arch, find the horizontal positions where the height is 3.2 metres. Solve 3.2=5−x²/20, so x²/20=1.8 and x²=36. Thus x=−6 or x=6. Both lie within the model domain −10≤x≤10. The two answers arise from the arch’s symmetry. A student who reports x=6 alone has ignored the left-hand side of the arch. If the question asks for distance from the centre, the answer is six metres, not two signed coordinates.
Convert centre coordinates to a left-end origin
Some diagrams measure horizontal distance from the left endpoint rather than from the centre. Let X=x+10 in the twenty-metre example, so x=X−10. Substitute into y=5−x²/20 to obtain y=5−(X−10)²/20 for 0≤X≤20. Expand if needed: y=X−X²/20. The two equations describe the same curve in different coordinates. The vertex is at X=10, y=5, and the ends are X=0 and X=20. Clear coordinate definitions prevent apparently conflicting formulas.
Use three points to determine a quadratic
A general quadratic y=ax²+bx+c has three coefficients. Three suitable independent point conditions can determine them. For a symmetric arch with endpoints (0,0), (20,0) and centre point (10,5), substitute (0,0) to get c=0. Substitute (20,0) to get 400a+20b=0. Substitute (10,5) to get 100a+10b=5. Solving yields a=−1/20 and b=1, so y=X−X²/20. This method does not require the vertex form at the beginning.
Why the quadratic opens downward
In y=H−(4H/L²)x², positive rise H and positive span L make the coefficient of x² negative. The graph therefore opens downward, giving a maximum at the centre. If the coefficient were positive, the centre would be a minimum and the curve would dip below its end heights instead. The sign of the leading coefficient is a quick graphical check. It also distinguishes an arch rising above the supports from a valley-shaped profile.
Vertex form reveals the maximum immediately
The equation y=5−(x²/20) is already in vertex form y=a(x−h)²+k with h=0 and k=5. The vertex is (0,5), and because a<0 it is the maximum. In left-end coordinates, y=−(X−10)²/20+5 has vertex (10,5). Vertex form is often more informative than an expanded quadratic because it displays the centre and rise directly. Both forms remain mathematically equivalent.
The domain matters to the model
The quadratic y=5−x²/20 exists algebraically for every real x, but the modelled arch occupies only −10≤x≤10. Outside that interval, the same parabola continues downward and can produce negative heights. Those points are not part of the stated physical outline. A mathematical formula and its intended modelling domain must be given together. Ignoring the domain can lead to answers describing a curve that extends beyond the actual structure being discussed.
Differentiate to find the arch gradient
For y=5−x²/20, differentiate to obtain dy/dx=−x/10. At the centre x=0, the gradient is zero, so the tangent is horizontal. At x=5, the gradient is −1/2; at x=−5, it is +1/2. The sign changes across the centre, consistent with the rising left half and descending right half. The derivative gives the slope of the geometric outline, not the force carried by a structural member.
Find a tangent line on the right half
At x=5, the height is 3.75 and the derivative is −1/2. The tangent through (5,3.75) is y−3.75=−(1/2)(x−5), or y=6.25−x/2. Substitute x=5 to confirm y=3.75. This tangent approximates the curve near the selected point. It does not coincide with the full arch, and extending it far from x=5 can give misleading heights. A tangent is a local linear description.
Find the normal to the arch outline
At x=5, the tangent gradient is −1/2, so a perpendicular normal has gradient 2. Through (5,3.75), its equation is y−3.75=2(x−5), or y=2x−6.25. Tangent and normal gradients multiply to −1, confirming perpendicularity. These lines are geometric constructions. A structural force direction cannot be inferred solely from the normal without a loading and structural model.
Use the second derivative to describe curvature
For y=5−x²/20, the first derivative is −x/10 and the second derivative is −1/10. The negative second derivative is constant, confirming that the graph is concave downward throughout its domain. However, the geometric curvature of the curve is not constant, because curvature also depends on the first derivative. A parabola is not a circular arc. This distinction matters when comparing quadratic outlines with arches modelled by circles.
Compare a parabola with a circular arc
A circular arch segment is part of a circle, whereas a parabolic arch is part of a quadratic curve. Both can look similar over a short span, but their equations and curvature differ. A circle with centre (0,c) has equation x²+(y−c)²=R². Solving for its upper or lower branch involves a square root, not a quadratic in x. Three points may determine either model, but fitting the same points does not make the curves identical between them. A good modeller identifies the intended shape rather than assuming every arch is parabolic.
Calculate the area beneath the ideal arch
The area under y=5−x²/20 between x=−10 and x=10 is ∫(5−x²/20)dx over that interval. An antiderivative is 5x−x³/60. At x=10 this gives 50−1000/60=100/3; at x=−10 it gives −50+1000/60=−100/3. Subtract to obtain 200/3 square metres, approximately 66.67 m². This is the two-dimensional area between the outline and the end line. It is not a volume of concrete or a structural cross-sectional area without further specifications.
Derive a general area formula
For an ideal symmetric parabolic arch y=H−4Hx²/L² over −L/2≤x≤L/2, integrate the height. The integral of H across the span is HL. The integral of 4Hx²/L² across the symmetric interval is HL/3. Therefore the area under the arch is 2HL/3. For L=20 and H=5, this gives 2×20×5/3=200/3, matching the direct calculation. The formula depends on the parabolic shape and the chosen baseline; it is not universal for all arch profiles.
Compare the area with its bounding rectangle
The rectangle spanning the full width L and maximum rise H has area LH. The ideal parabolic region beneath the arch has area 2LH/3, exactly two-thirds of that rectangle. In the twenty-by-five example, the rectangle is 100 m² and the parabolic region is 66.67 m². This ratio can be checked visually: the arch reaches the rectangle’s top only at its centre and tapers to zero at both ends. The comparison helps students assess whether an integration answer is reasonable.
Find the width at a specified height
For y=H−4Hx²/L², solve for x²: x²=(L²/4)(1−y/H). The two horizontal positions are x=±(L/2)√(1−y/H). Thus the horizontal width of the arch at height y, for 0≤y≤H, is L√(1−y/H). For L=20, H=5 and y=3.2, the width is 20√(1−0.64)=20×0.6=12 metres. This agrees with the earlier positions x=±6. The width decreases as the selected height approaches the crown.
What happens when the span doubles
Keep the rise H fixed while doubling the span from L to 2L. The quadratic coefficient 4H/L² becomes 4H/(4L²), one-quarter as large. The curve is broader and flatter. At a fixed fraction of the span, however, the relative height profile is unchanged. The area beneath the curve doubles because the width doubles while the height remains fixed. These are geometric scaling relationships; they do not establish how a bridge’s structural demands change.
What happens when the rise doubles
Keep the span L fixed while doubling the rise H. The coefficient 4H/L² doubles in magnitude, and the height at every fixed horizontal fraction of the span doubles. The area 2HL/3 also doubles. The slopes of the curve at corresponding x-values double because dy/dx=−8Hx/L². This is a mathematical sensitivity calculation. It should not be interpreted as a direct prediction of material quantity or safety, because actual structural geometry involves thickness, members and other details.
Express the arch in dimensionless coordinates
Let u=2x/L, so u runs from −1 to 1. Let v=y/H, so v measures height as a fraction of maximum rise. Substitute into y=H−4Hx²/L² to get v=1−u². Every symmetric ideal parabola defined by a span and rise has this same normalised profile. This is a powerful modelling idea: changing physical size changes the units and coefficients, but the dimensionless shape is unchanged. Students can compare two arches without mixing metres and centimetres.
A model arch with asymmetric end heights
Not every arch outline is symmetric about a horizontal baseline. Suppose a quadratic passes through (0,0), (10,6) and (20,2). Let y=ax²+bx+c. The first point gives c=0. The others give 100a+10b=6 and 400a+20b=2. Doubling the first equation and subtracting gives 200a=−10, so a=−0.05. Then 10b=11, so b=1.1. The model is y=−0.05x²+1.1x. Its vertex is at x=11, not at the midpoint x=10. This illustrates why symmetry must be stated rather than assumed.
Check the asymmetric model
For y=−0.05x²+1.1x, at x=10 the height is −5+11=6. At x=20 it is −20+22=2. At x=0 it is zero. The derivative is y′=−0.1x+1.1, which vanishes at x=11. The maximum height is y(11)=−0.05×121+12.1=6.05. A model fitted to three specified points can have its maximum at a different location from the highest listed point. This is a reason to calculate the vertex instead of assuming it is among the data points.
What an arch equation cannot prove
A quadratic outline does not specify material strength, cross-sectional thickness, reinforcement, connections, foundations, wind conditions, vehicle loading, buckling resistance or failure modes. It cannot establish whether a bridge is safe or how much load it carries. The model can answer geometric questions such as height, width, gradient and enclosed area. Structural design requires additional physical assumptions, standards, loads and professional analysis. A helpful article keeps this boundary explicit rather than turning a school graph into unsupported engineering advice.
The difference between a visual arch and a structural arch
An arch-shaped decorative element and a load-bearing structural arch may have similar outlines but very different mechanical roles. Even among load-bearing arches, the line of thrust under loading need not coincide with a simple quadratic profile. A parabolic arch can be associated with certain idealised load conditions, but that fact does not mean a parabolic outline is automatically appropriate for any bridge. The mathematics lesson here concerns geometry and elementary calculus, not a universal construction recommendation.
An exact word problem with a clearance question
A model arch is 30 metres wide and 6 metres high at its centre. Set x=0 at the centre and use y=6−(4×6/30²)x²=6−(2/75)x². Find the height at x=9: y=6−(2/75)×81=6−162/75=3.84 metres. If a hypothetical object requires 4 metres of vertical clearance above the chosen baseline, this location would not meet that geometric requirement. This conclusion is conditional on the model and baseline; actual clearance requires surveying and appropriate safety margins.
Find the interval meeting a chosen clearance
For the thirty-metre model y=6−(2/75)x², ask where y≥4. Rearrange (2/75)x²≤2, giving x²≤75 and −5√3≤x≤5√3. The horizontal width of this model region is 10√3 metres, approximately 17.32 metres. The interval lies within the full model domain −15≤x≤15. State that this is the mathematical region at or above four metres, not an independently verified real-world clearance envelope.
A tangent-based height approximation
For y=5−x²/20, the derivative at x=4 is −4/10=−0.4. The height at x=4 is 5−16/20=4.2. For a small move to x=4.2, the tangent estimate is y≈4.2+(−0.4)(0.2)=4.12. The exact quadratic gives y=5−(4.2²)/20=5−17.64/20=4.118. The difference is 0.002 metres in this model. This illustrates why tangent approximations are accurate locally but not exact for a curved function.
Calculate the exact tangent approximation error
For a quadratic y=H−ax², the tangent approximation at x=x₀ for a change h is y(x₀)+y′(x₀)h=H−ax₀²−2ax₀h. The exact value is y(x₀+h)=H−a(x₀+h)²=H−ax₀²−2ax₀h−ah². Therefore exact minus tangent estimate equals −ah². For a>0, the tangent overestimates the actual arch height by ah². This formula explains the preceding numerical difference and shows that the error grows with the square of the horizontal step.
An exam-style complete solution
A symmetric model arch spans 24 metres and rises 6 metres. Put the centre at x=0. The equation is y=6−(4×6/24²)x²=6−x²/24, for −12≤x≤12. At x=6, y=6−36/24=4.5 metres. The gradient is y′=−x/12, so at x=6 it is −1/2. The tangent through (6,4.5) is y−4.5=−(x−6)/2, or y=7.5−x/2. The area under the arch is 2LH/3=2×24×6/3=96 m². Each answer follows from a different aspect of the same model.
A second complete example with shifted coordinates
An arch has endpoints at X=0 and X=16 on the baseline, with a maximum rise of four metres at X=8. Write y=a(X−8)²+4. At X=0, y=0, so 0=64a+4 and a=−1/16. Thus y=4−(X−8)²/16. At X=4, y=4−16/16=3. The derivative is y′=−(X−8)/8, giving gradient +1/2 at X=4. The tangent is y−3=(1/2)(X−4), or y=X/2+1. The sign is positive because this point lies on the rising left half.
Practice questions with answers
Question A: A symmetric arch spans 12 m and rises 3 m. Equation y=3−x²/12 for −6≤x≤6. Question B: In that model, height at x=3 is 3−9/12=2.25 m. Question C: The gradient at x=3 is −x/6=−1/2. Question D: Area beneath the arch is 2×12×3/3=24 m². Question E: Height y=2 occurs when 2=3−x²/12, so x=±√12=±2√3 m. Explain each answer in words and identify the coordinate reference.
A harder practice set
Question F: An arch spans 18 m and rises 4.5 m. Its centre-based equation is y=4.5−x²/18. Question G: At x=6, y=4.5−36/18=2.5 m. Question H: The gradient at x=6 is −x/9=−2/3. Question I: The area beneath this ideal arch is 2×18×4.5/3=54 m². Question J: The width at half maximum height is L√(1−1/2)=18/√2=9√2 m. The answers can be checked by substituting them into the general formulas.
Common mistakes and how to repair them
If a student uses the full span in place of the half-span when solving for a, the coefficient is four times too small. If the student gives a positive quadratic coefficient for a downward-opening arch, the graph orientation is wrong. If the student calculates a negative height outside the endpoints, they have ignored the model domain. If they label the area under a curve as a volume, they have confused dimensions. If they describe a calculated outline as structurally safe, they have crossed from a mathematical model into an unsupported engineering claim. Each error has a different correction.
A learning sequence for the Fencing Method
First work with a symmetric arch centred at zero, because the linear term vanishes. Next switch the origin to the left endpoint and show the equivalent expanded quadratic. Then ask for height at a point, followed by positions at a given height. After these are stable, introduce tangent gradients and area under the curve. Finally, use an asymmetric three-point example and ask which assumptions no longer hold. Each stage changes one feature while preserving the earlier mathematical structure.
What parents can ask after the lesson
Ask the learner to explain what the span and rise mean, why the half-span appears in the coefficient, and how to verify the equation at the two endpoints and centre. Ask whether the height at a position is a function value or a structural safety conclusion. Can the learner calculate the tangent gradient and explain its sign on each side? Can they distinguish an area beneath a curve from the material volume of a bridge? These questions reveal understanding without requiring parents to solve every equation themselves.
Singapore syllabus and enrichment boundary
Quadratic equations, graph interpretation and differentiation connect naturally with Additional Mathematics. The arch-bridge application, area calculation and structural discussion are teaching contexts rather than a promise that bridge engineering is an examinable topic. Consult the official 2027 SEAB G3 Additional Mathematics syllabus for prescribed requirements. The mathematical examples are idealised and do not use measured bridge data. The reference teaching approach is the Clementi Mathematics small-group tutorial article, which emphasises clear steps, diagnostic correction and independent verification.
Frequently asked questions about parabolic arches
Is every bridge arch a parabola? No; real arch profiles vary. Why use a quadratic? It gives a simple symmetric curve determined by span and rise. Is the highest point always at the midpoint? Only under the symmetric model. Can a derivative tell us the arch’s slope? Yes, it gives the local geometric gradient. Can an integral give the area under the outline? Yes, when the baseline and interval are defined. Does that area equal the amount of building material? No. Does the equation prove a bridge is safe? No; structural safety requires separate engineering analysis.
Final checklist for an arch model
Define the origin, axes and measurement units. State the span L and maximum rise H. For a symmetric centre-based model, derive y=H−4Hx²/L² and restrict −L/2≤x≤L/2. Verify both endpoints and the vertex. For a height question, substitute the requested x; for a width question, solve for both positive and negative positions. For gradients, differentiate and interpret the sign. For areas, integrate over the stated interval and use square units. Explain clearly that a geometric model is not a structural certification.
Check the full-width formula against the centre-based formula
In centre coordinates, the arch is y=H−4Hx²/L². In left-end coordinates X=x+L/2, so x=X−L/2. Substitute to obtain y=H−(4H/L²)(X−L/2)². Expanding gives y=H−(4H/L²)(X²−LX+L²/4)=4HX/L−4HX²/L². Factor as y=(4H/L²)X(L−X). This factored form makes the two endpoints X=0 and X=L obvious. At the midpoint X=L/2, the value is H. All three forms describe the same ideal curve.
A factorised equation that reveals the supports
Suppose a model arch begins at X=0 and ends at X=30, and its equation is y=kX(30−X). Its highest point occurs at X=15. If the maximum rise is 6, substitute 6=k×15×15 to obtain k=6/225=2/75. Thus y=(2/75)X(30−X). This is equivalent to y=6−(2/75)(X−15)². The factored form is convenient when the two horizontal intercepts are known; the vertex form is convenient when the centre and maximum rise are known.
Why a quadratic can fit three points yet still be the wrong model
A quadratic polynomial can often be fitted exactly through three measured points, but exact interpolation does not prove that the underlying object is truly parabolic. A circular arc or another smooth curve may pass through the same three points and differ between them. Measurements also contain uncertainty, and a real bridge may not be symmetric. The choice of model should therefore follow the question’s stated assumptions or evidence, not simply the convenience of a quadratic formula. For a school exercise, explicitly saying ‘model the outline as a parabola’ supplies that assumption.
How to interpret a slope numerically
For the model y=5−x²/20, the derivative at x=5 is −0.5. This means that, for a small horizontal move of one metre to the right near that point, the height decreases by approximately half a metre. It does not mean that every one-metre step anywhere on the arch changes height by exactly half a metre. The slope varies with x. At the centre the slope is zero; at x=10 it is −1. The derivative is local, which is why tangent approximations are useful near a point but become less reliable farther away.
Why units and rounding belong at the end
Suppose a model uses a span of 18 metres and a rise of 4.5 metres. The coefficient in y=4.5−x²/18 has units of inverse metres, so multiplying it by x² produces a height in metres. If the span is converted to centimetres, the numerical coefficient must change accordingly. Keep exact fractions during algebra and round only the final quantity when a decimal is needed. A small rounding error in the coefficient can affect predicted heights across the full span, particularly near the endpoints where the height is small.
A final independent application
An ideal arch has span 10 metres and rise 2 metres. Its centre-based equation is y=2−(8/100)x²=2−2x²/25 for −5≤x≤5. At x=2.5, the height is 2−(2/25)(6.25)=1.5 metres. The derivative is y′=−4x/25, giving gradient −0.4 at x=2.5. The tangent there is y−1.5=−0.4(x−2.5), or y=2.5−0.4x. The area beneath the arch is 2×10×2/3=40/3 square metres. Checking the endpoints gives y(±5)=0, as required.
One last reason to study this model
The value of the arch example is not that every student will design a bridge. It is that a single mathematical description connects a graph, an equation, a derivative, an integral and a set of clearly stated assumptions. When a student can move between those representations and explain the limits of each answer, Additional Mathematics becomes a way to reason about unfamiliar problems rather than a collection of isolated procedures.