General, vertex, and factored — three outfits for one parabola — and how to recover the equation from data.
A parabola does not change, but the way you write it can. The general form shows where it crosses the y-axis; vertex form shows the turning point; factored form shows where it crosses the x-axis. They are three outfits for one and the same curve — and knowing which one to reach for makes the rest easy. Better still, you can run the machine backward: given three points, or a vertex, or two crossings, you can recover the equation by solving for the unknown coefficients. That is the method of undetermined coefficients you met for lines in Stage 21, now grown up for quadratics.
Every quadratic function can be dressed three ways, and each dress shows off one feature for free. You do not have to compute anything — you just read it off.
| Form | Looks like | Reads off at a glance |
|---|---|---|
| General | y = ax2 + bx + c | the y-intercept (0, c) (put x = 0) |
| Vertex | y = a(x − h)2 + k | the vertex (h, k) and axis x = h |
| Factored | y = a(x − x1)(x − x2) | the zeros x1, x2 (x-intercepts) |
All three describe one curve. Take the example from the picture above: the same parabola is
y = x2 − 2x − 3 = (x − 1)2 − 4 = (x − 3)(x + 1).
Multiply any one of them out and you land back on the others. Notice the quiet thread running through all three: the leading coefficient a is the same in every form — here a = 1. The forms re-package b and c, or the vertex, or the roots, but a is fixed by the shape of the bowl and never changes outfit.
One parabola, three equations. General hands you the y-intercept; vertex form hands you the vertex; factored form hands you the zeros. The same a appears in all three.
Suppose all you have is a few points on a parabola and you want its equation. The general form y = ax2 + bx + c has three unknowns — a, b, and c — so you need three independent facts to pin them down. Three points on the curve give exactly three equations: a solvable system.
Find the parabola through (0, −3), (1, −4), and (3, 0).
Plug each point into y = ax2 + bx + c:
• From (0, −3): a·0 + b·0 + c = −3, so c = −3 immediately (the constant is the y-intercept).
• From (1, −4): a + b + c = −4, and with c = −3 that is a + b = −1.
• From (3, 0): 9a + 3b + c = 0, and with c = −3 that is 9a + 3b = 3, i.e. 3a + b = 1.
Subtract the two: (3a + b) − (a + b) = 1 − (−1) gives 2a = 2, so a = 1, then b = −2.
Answer: y = x2 − 2x − 3. Sanity-check a fourth fact — the vertex should be at x = −b/2a = 1, with y = 1 − 2 − 3 = −4 ✓.
Always do the y-intercept point first when one of your points has x = 0 — it gives c for free, leaving a clean 2×2 system in a and b. And after solving, plug in a fourth known fact to catch arithmetic slips.
If you already know the turning point (h, k), do not waste effort on a 3×3 system. Write the parabola in vertex form y = a(x − h)2 + k with h and k already filled in. Only one unknown is left — the leading coefficient a — so one more point pins it down.
A parabola has vertex (1, −4) and passes through (3, 0). Find its equation.
Start from y = a(x − 1)2 − 4. Substitute the known point (3, 0):
0 = a(3 − 1)2 − 4 = 4a − 4, so 4a = 4 and a = 1.
Answer: y = (x − 1)2 − 4. (Multiply out to check: x2 − 2x + 1 − 4 = x2 − 2x − 3 — the same curve as before.)
Vertex + one extra point ⇒ solve for a. Set up y = a(x − h)2 + k, drop in the point, and the only thing left to find is the leading coefficient: a = (y − k) / (x − h)2.
If you know where the curve meets the x-axis — its zeros x1 and x2 — reach for factored form. A parabola that is zero at x1 and x2 must have those as factors, so write y = a(x − x1)(x − x2). Again only the leading coefficient a is unknown, and one more point fixes it.
A parabola crosses the x-axis at −1 and 3, and passes through (0, −3). Find its equation.
Start from y = a(x + 1)(x − 3). Substitute the point (0, −3):
−3 = a(0 + 1)(0 − 3) = a·(1)·(−3) = −3a, so a = 1.
Answer: y = (x + 1)(x − 3) = x2 − 2x − 3 — once more the very same curve. (This is the factoring you practiced in Stage 11, read backward.)
The leading coefficient is not always 1. Many curves pass through the same two crossings — wider, narrower, even flipped upside down. The extra point is what fixes a. Skip it and you have only guessed the shape.
There is a fourth way to get a new quadratic: take one you already know and slide it. As you saw in Stage 23.2, sliding the parent bowl y = x2 is nothing more than rewriting h and k in vertex form — and from there you can expand to general form whenever you need it.
Slide y = x2 right 2 and up 3. The vertex lands at (2, 3), so in vertex form
y = (x − 2)2 + 3 = x2 − 4x + 4 + 3 = x2 − 4x + 7.
Now reflect it to open downward through the same vertex — just flip the sign of a:
y = −(x − 2)2 + 3 = −x2 + 4x − 4 + 3 = −x2 + 4x − 1.
Vertex form makes the move obvious (it shows the vertex); general form makes the arithmetic obvious (it shows c). Travel freely between them with b = −2ah and c = ah2 + k.
A transformation is just a different equation for a moved curve. Slide right h and up k ⇒ vertex form y = a(x − h)2 + k; flip the sign of a to open the other way. Expand for general form using b = −2ah and c = ah2 + k.
One parabola wears three outfits, and a is the same thread in each:
To recover an equation, count what you know and match the form: three points ⇒ general (a 3×3 system); a vertex + one point ⇒ vertex form (solve for a); two crossings + one point ⇒ factored (solve for a). And to make a new curve from an old one, slide the vertex — that is all a transformation is.
Which of the three forms shows the vertex at a glance, with no work needed?
Vertex form, y = a(x − h)2 + k — the turning point (h, k) is written right into the equation.
Write y = (x − 1)2 − 4 in general form.
Expand: (x − 1)2 − 4 = x2 − 2x + 1 − 4 = x2 − 2x − 3. (Same curve, general dress.)
A parabola has vertex (2, 1) and passes through (4, 5). Find a and the equation.
Vertex form y = a(x − 2)2 + 1. Plug in (4, 5): 5 = a(2)2 + 1 = 4a + 1, so 4a = 4 and a = 1. Equation: y = (x − 2)2 + 1.
A parabola crosses the x-axis at 1 and 5 with leading coefficient a = 2. Write it in factored form.
Factored form straight away: y = 2(x − 1)(x − 5). (Expanded: 2x2 − 12x + 10.)
Find the quadratic through (0, 0), (1, 3), and (2, 10).
From (0, 0): c = 0. From (1, 3): a + b = 3. From (2, 10): 4a + 2b = 10, i.e. 2a + b = 5. Subtract: a = 2, then b = 1. So y = 2x2 + x.
Write y = x2 shifted left 1 and down 4 in both vertex form and general form.
Vertex lands at (−1, −4): vertex form y = (x + 1)2 − 4. Expand: x2 + 2x + 1 − 4 = x2 + 2x − 3.
Six questions to lock it in. Tap the answer you think is right.
The big idea is representational fluency: general, vertex, and factored form are three equations for the same curve, and a confident student moves between them on demand — choosing the form that already displays whatever feature the problem asks for. The y-intercept lives in general form, the vertex in vertex form, the zeros in factored form, and the leading coefficient a is invariant across all three.
The recovery half of the lesson is the method of undetermined coefficients: set up the form with the right number of unknowns, substitute the given data, and solve. The most common misconceptions to watch for are (1) assuming a = 1 in factored or vertex form — the extra point is precisely what determines a, and skipping it silently fixes the wrong width; (2) reaching for the wrong form for the data on hand (three loose points beg for general form, not vertex); and (3) arithmetic slips while solving the 3×3 system — encourage using a zero-x point first to get c free, then a clean 2×2, and always verifying a fourth fact.
This lesson supports CCSS A-SSE.B.3a–b (use factored form to reveal zeros, vertex form to reveal the vertex/extreme), A-CED.A.2 and F-BF.A.1 (write an equation that fits given conditions), and F-IF.C.8a (rewrite to reveal and interpret key features).