Stretching a Parabolic Arch to Fit a Bridge Deck
A landscape architect models a decorative concrete arch with the base function y = x^2. The arch must span a walkway 12 m wide and rise exactly 4 m at its centre. She plans to transform the base parabola using a vertical stretch by factor a, a horizontal stretch by factor b, and a translation, writing the result as y = a((x - h)/b)^2 + k. Site constraints require the vertex to sit on the walkway centreline, and the arch must meet the ground exactly at both edges of the walkway. A surveyor notes that if she instead applies the vertical stretch first and then a horizontal compression by factor 1/3, the arch will be too narrow to reach the walkway edges, so the order and the value of b matter.
1. Determine values of a, b, h, and k that satisfy all constraints, and explain how the mapping of the vertex and one edge point fixes each parameter.
Read the worked answer
Place the origin at the walkway centre, so the vertex is at (0, 0) and the arch opens upward. Then h = 0 and k = 0, leaving y = a(x/b)^2.
The arch meets the ground at x = ±6 (half of the 12 m width), and at those points y = 0, which is automatically satisfied by the vertex form only if the edges are the x-intercepts. A parabola with vertex at (0,0) opening upward has its only zero at the vertex, so the arch cannot meet the ground at the edges with k = 0. Instead the walkway edges correspond to the arch's base level, so set the vertex at the top: vertex (0, 4), giving y = a(x/b)^2 + 4.
At an edge, y = 0 when x = 6: 0 = a(6/b)^2 + 4, so a(36/b^2) = -4, requiring a < 0 (arch opens downward). Choose b = 6 and a = -4: y = -4(x/6)^2 + 4 = 4 - x^2/9.
Check: at x = 0, y = 4 (correct height); at x = ±6, y = 4 - 4 = 0 (correct span). The horizontal stretch factor is b = 6 and the vertical stretch factor is a = -4 (a reflection plus stretch). The vertex mapping (0,0) → (0,4) fixes h and k; the edge mapping (6,0) → (6,0) fixes the ratio a/b^2 = -1/9.
2. The surveyor's alternative uses a horizontal compression by factor 1/3 before the vertical stretch. Show why this produces an arch that fails the 12 m span, and state what horizontal scale factor would be needed to repair it.
Read the worked answer
Starting from y = x^2, a horizontal compression by factor 1/3 replaces x with 3x, giving y = (3x)^2 = 9x^2. A vertical stretch by factor a then gives y = a·9x^2, and translating the vertex to (0,4) gives y = 9a x^2 + 4.
Requiring y = 0 at the edge x = 6: 0 = 9a(36) + 4, so a = -4/324 = -1/81. That value of a is fine numerically, but the effective horizontal scale is fixed at 1/3, meaning the parabola's shape is y = -(4/36)x^2 only if the coefficient matches. Here the coefficient 9a = -1/9, which does match 4/36 = 1/9, so the curve actually passes through (±6, 0). The real failure is that the compression factor 1/3 sets the horizontal scale, so the span is determined by where y = 0: x^2 = 4/(9|a|). With a chosen for a 4 m rise at the vertex, the span is 2·sqrt(4/(9|a|)) = (4/3)/sqrt(|a|).
To force a span of 12 m (half-span 6), we need sqrt(4/(9|a|)) = 6, so 4/(9|a|) = 36, giving |a| = 4/324 = 1/81. Thus the repaired vertical stretch is a = -1/81 with horizontal compression 1/3, equivalent to b = 3 in the earlier form: y = -(1/81)(3x)^2 + 4 = -x^2/9 + 4. The lesson is that the horizontal scale factor b and the vertical stretch a are linked by a/b^2 = -1/9; changing b without recomputing a breaks the span.