From general form to the vertex: the axis x = −b/2a, the vertex formula, and reading a, b, c off the curve.
In vertex form, the turning point is handed to you on a plate — y = a(x − h)² + k wears its vertex (h, k) right on its sleeve. But quadratics rarely arrive that tidy. Most show up in the everyday general form y = ax² + bx + c, where the same information is hidden inside three coefficients. This lesson teaches you to extract the three things that pin a parabola down — the opening (from a), the axis of symmetry, and the vertex — straight from a, b, and c, using either completing the square or a ready-made formula. By the end you will glance at a graph and name a, b, c, and glance at a, b, c and picture the graph.
Hold a parabola up to the light and you will see it is perfectly matched, left and right, as if folded against an upright mirror. That mirror is a vertical line called the axis of symmetry, and it runs straight through the turning point of the curve.
The mirror has a useful consequence: any two points of the curve that sit at the same height are the same distance from the axis, one on each side. If (−1, 0) and (3, 0) are both on the curve, the axis sits exactly halfway between them, at x = (−1 + 3)/2 = 1. The whole curve folds onto itself across that line.
Every parabola has a vertical axis of symmetry. The curve is its own reflection across that line, so two points of equal height are always mirror images — equal distances left and right of the axis — and the axis passes through the turning point.
The single point where the curve turns is the vertex. It is the place where the parabola stops going one way and starts going the other — and which way that is depends entirely on the sign of a.
When the bowl opens upward (a > 0), the vertex is the lowest point the curve ever reaches. When it opens downward (a < 0), the vertex is the highest point. Its coordinates (h, k) fix where the whole curve sits on the plane, and — because the curve folds about the axis — the vertex always lies on the axis of symmetry. In fact h, the x-coordinate of the vertex, is the axis: the line x = h.
The vertex (h, k) is the turning point of the parabola — the minimum when a > 0 and the maximum when a < 0. It lies on the axis of symmetry x = h, so finding the vertex and finding the axis are the same job.
How do we recover (h, k) from a, b, c? One sure method is to reshape the general form y = ax² + bx + c back into the vertex form a(x − h)² + k we met in 23.2. The technique is completing the square — the same move you used when solving quadratic equations. Once the square is built, the vertex falls right out.
Take y = x² − 2x − 3. Look at the x² and x terms: x² − 2x. Half of −2 is −1, and (−1)² = 1, so adding and subtracting 1 builds a perfect square without changing the value:
y = (x² − 2x + 1) − 1 − 3 = (x − 1)² − 4.
So the vertex form is (x − 1)² − 4, and the vertex is (1, −4) — read it straight off h = 1, k = −4.
When a ≠ 1 there is one extra step: factor a out of the x-terms first. Take y = 2x² + 8x + 5:
y = 2(x² + 4x) + 5 = 2(x² + 4x + 4) − 8 + 5 = 2(x + 2)² − 3.
(Notice the −8: we added 4 inside the bracket, but the bracket is multiplied by 2, so we really added 2 × 4 = 8 — and must subtract it back outside.) The vertex is (−2, −3), a minimum since a = 2 > 0.
When a ≠ 1, factor a out of the x² and x terms before halving. Whatever you add inside the bracket is secretly multiplied by a, so subtract a × (that amount) back outside to keep the value unchanged.
Completing the square always works, but it is a lot of writing. If you carry it through once with letters instead of numbers, a shortcut drops out. The x-coordinate of the vertex — and therefore the axis of symmetry — is always
x = −b2a .
This is the most useful single formula in the lesson. Once you have h = −b/2a, you do not need to memorize a separate formula for k — just plug h back in: k = f(h). (For the record, the closed form is k = (4ac − b²)/4a, but f(h) is easier to compute and harder to get wrong.)
Check it on our running curve y = x² − 2x − 3 (a = 1, b = −2, c = −3):
h = −b2a = −(−2)2(1) = 1, k = f(1) = 1 − 2 − 3 = −4 ⇒ vertex (1, −4) ✓
Same answer as completing the square — and far quicker.
The axis of symmetry is the vertical line x = −b/2a. The vertex is (h, k) with h = −b/2a and k = f(h). Find h first, then substitute to get k — do not memorize the k-formula.
The axis does more than place the vertex: it splits the curve into a falling side and a rising side. Walk left-to-right along an upward bowl and you go down until you reach the vertex, then up afterward.
For a > 0 (a valley): the function is decreasing for x < h, then increasing for x > h. We can state these stretches as intervals — decreasing on (−∞, h), increasing on (h, ∞) — using the interval language of Stage 19. For a < 0 (a hill), it is the other way round: increasing first, then decreasing. The turn always happens exactly at the vertex.
On which interval is y = x² − 2x − 3 increasing? The axis is x = −b/2a = 2/2 = 1, and a = 1 > 0, so the curve rises to the right of the axis: increasing on (1, ∞) (and decreasing on (−∞, 1)).
Now turn the picture around. Given only a, b, and c, what can you say about the graph — without plotting a single point?
a — the opening and the width. Its sign sets which way the bowl opens (up if a > 0, down if a < 0); its size |a| sets how narrow (large |a|) or wide (small |a|) it is.
c — the y-intercept. Set x = 0 and every other term vanishes, leaving y = c. So the curve always crosses the y-axis at exactly (0, c) — c is read straight off, no work needed.
b, together with a — which side the axis is on. The axis is at x = −b/2a, so a and b together place it. When a and b have the same sign (ab > 0), then −b/2a is negative and the axis sits to the left of the y-axis. When they have opposite signs (ab < 0), the axis is to the right. When b = 0, the axis is the y-axis itself.
| You see | You know |
|---|---|
| sign of a | opens up (a > 0) or down (a < 0) |
| size of |a| | large ⇒ narrow · small ⇒ wide |
| c | y-intercept is (0, c) |
| sign of ab | ab > 0 ⇒ axis on the left · ab < 0 ⇒ right · b = 0 ⇒ the y-axis |
The three essentials of y = ax² + bx + c (a ≠ 0):
1. Opening. a > 0 opens up (vertex = lowest); a < 0 opens down (vertex = highest); |a| sets the width.
2. Axis of symmetry. The vertical mirror x = −b/2a. The curve folds onto itself across it.
3. Vertex. (h, k) with h = −b/2a and k = f(h). Get it by completing the square (factor a out first) or the vertex formula.
Reading off a, b, c: the y-intercept is (0, c); the sign of ab tells you which side of the y-axis the axis sits on; for a > 0 the curve decreases on (−∞, h) and increases on (h, ∞).
Find the axis of symmetry and the vertex of y = x² − 6x + 5.
Axis: x = −b/2a = 6/2 = 3. Then k = f(3) = 9 − 18 + 5 = −4. Vertex (3, −4).
Complete the square on y = x² + 4x + 1 and give the vertex.
Half of 4 is 2, and 2² = 4: y = (x² + 4x + 4) − 4 + 1 = (x + 2)² − 3. Vertex (−2, −3).
For y = −2x² + 4x + 1, find the vertex and say whether it is a maximum or a minimum.
h = −b/2a = −4/(2·(−2)) = 1. k = f(1) = −2 + 4 + 1 = 3, so the vertex is (1, 3). Since a = −2 < 0 the bowl opens down, so this is a maximum.
Where does y = 3x² − 12x + 7 cross the y-axis?
Set x = 0: y = c = 7. The y-intercept is (0, 7).
On what interval is y = x² − 2x − 3 increasing?
Axis x = −b/2a = 2/2 = 1, and a = 1 > 0, so the curve rises to the right of the axis: increasing on (1, ∞).
A parabola opens up with axis x = 2. Is it increasing or decreasing at x = 0? Explain.
x = 0 is to the left of the axis x = 2. For an upward bowl the curve falls to the left of the vertex and rises to the right, so at x = 0 it is decreasing.
Six questions to lock it in. Tap the answer you think is right.
The big idea of 23.3 is that the general form y = ax² + bx + c hides exactly the same information that vertex form displays — and that there are two reliable bridges between them: completing the square (always works, builds understanding) and the vertex formula x = −b/2a (fast, once the algebra is trusted). Encourage students to find h = −b/2a and then compute k = f(h) by substitution, rather than memorizing the four-term k-formula; substitution is less error-prone and reinforces what the vertex means.
Watch for four specific misconceptions. (1) Writing the axis as +b/2a — the minus sign is essential. (2) Forgetting to factor a out of the x-terms before completing the square, which throws off the constant correction. (3) Reading the y-intercept off the wrong axis, or confusing it with c when the form is not yet standard. (4) Assuming the vertex is always a minimum — it is a maximum whenever a < 0. The interactive figures are built so the drawn axis, vertex, and y-intercept are computed from a, b, c, never read back off the picture, so the geometry and the numbers can never disagree.
This lesson supports CCSS F-IF.C.8a (complete the square to reveal the vertex and identify a maximum or minimum), F-IF.B.4 and F-IF.B.6 (intercepts, increasing/decreasing intervals, and extrema from a graph), and A-SSE.B.3b (rewriting an expression to reveal the vertex). It builds toward 23.4 (three forms of a quadratic) and 23.5 (function, equation, and inequality as one picture).