Parabola Architecture

Algebra 2 · Axiom Academy

Design a bridge with a parabolic arch — and discover how the vertex form y = a(x-h)^2 + k controls its shape and its position. You're the engineer on a new river crossing. A boat must pass beneath a parabolic arch, so your bridge has to clear 30 m over the water at the centre and span 100 m — 50 m to each bank. Remarkably, one equation, the vertex form y = a(x-h)^2 + k , hands you the entire arch. Watch the arch build itself. Start at the crown and step outward in equal 10-metre strides: the arch drops by 1, then 4, then 9, then 16, then 25 units — the perfect squares — and it falls exactly the same on both banks. That mirrored square-law drop is precisely what a(x-h)^2 draws. Equal steps out, square-law drops down — a parabola is the shape of x^2 . Step 1 — position the arch with h and k The vertex is the crown, the highest point of the arch. In vertex form, h slides it left and right and k sets its height. No algebra needed: drag the crown itself (or the two sliders) until it sits over the river centre with enough clearance for the boat. h and k don't reshape the arch — they carry it. Vertex form places the whole parabola by its crown. Step 2 — set the strength with a Now the width parameter a . It never moves the crown — only how steeply the arch falls to the banks. More negative (near -0.02 ) makes it narrow and steep; less negative (near -0.01 ) makes it wide and gentle. Find an a that lands the arch right on the ground at both edges.

This is the written version of the interactive lesson above. See the full Algebra 2 course.