Vertex Form of the Parabola

For y = a(x − d)² + e the vertex is S(d|e): d sits inside the bracket, e behind it. The big trap is the sign of d, because the bracket reads x MINUS d. For y = 2(x − 3)² + 1 the vertex is S(3|1), but for y = 2(x + 3)² + 1 it is S(−3|1), since x + 3 = x − (−3). Safe route: ask for which x the bracket becomes 0; the vertex sits there, and e is the corresponding function value. The e, in contrast, is read with its ordinary sign: ... − 4 means e = −4. Quick check: insert the vertex coordinates, y = e must result.

The goal is to turn x² + bx into a perfect square. Recipe: square half of the x-coefficient, add it and subtract it again immediately. Example: y = x² + 6x + 5. Halve 6 to 3, complete with 3² = 9: y = x² + 6x + 9 − 9 + 5 = (x + 3)² − 4, so S(−3|−4). If a factor sits before x², factor it out first: y = 2x² + 8x = 2(x² + 4x) = 2((x + 2)² − 4) = 2(x + 2)² − 8. The two classic mistakes: forgetting to subtract the completion term again and distributing a minus incorrectly when factoring out. Check: expanding must reproduce the original form.

a controls the opening direction and shape of the parabola without moving the vertex. Sign: a > 0 opens upwards (vertex is a minimum), a < 0 downwards (vertex is a maximum). Magnitude: |a| > 1 stretches the parabola in the y-direction, it looks narrower; 0 < |a| < 1 compresses it, it looks wider; a = 1 is the basic parabola. Concretely: from S you go 1 to the right and a upwards (instead of 1 as with the basic parabola), then 2 to the right and 4a upwards. For extreme value tasks this means: the optimal value is e, and a decides whether it is a maximum (a < 0) or a minimum (a > 0).

Set a(x − d)² + e = 0 and solve backwards: (x − d)² = −e/a, so x = d ± √(−e/a). Example: 2(x − 2)² − 3 = 0 leads to (x − 2)² = 1.5 and x = 2 ± √1.5 ≈ 0.78 and 3.22. From −e/a you immediately see the number of roots: positive gives two, zero exactly one (the vertex lies on the x-axis), negative no real root. This is often faster than the detour via the standard form with the pq formula, and geometrically clear: if the vertex lies below the x-axis and the parabola opens upwards (e < 0, a > 0), it must cross the axis twice.

The vertex provides d and e, the additional point determines a. Ansatz y = a(x − d)² + e, insert the point, solve for a: a = (y_P − e)/((x_P − d)²). Example: vertex S(2|−3), point P(4|5): 5 = a·(4 − 2)² − 3 = 4a − 3, so a = 2 and y = 2(x − 2)² − 3. Exactly this scheme underlies modelling tasks: a ball reaches its highest point at S and passes a known point, the flight parabola is required. One vertex plus one further point determine the parabola uniquely; two arbitrary points alone do not suffice, for that you would need three.