Quadratic Formula Calculator · Practical guide
How to Find the Vertex of a Parabola
The vertex sits at h = −b / 2a on the axis of symmetry; plug h back in to get k = f(h).
At a glance
Quick answer
The vertex of a parabola y = ax² + bx + c sits at h = −b / 2a, with y-coordinate k = f(h). For 2x² − 8x + 3, h = 8 / 4 = 2 and k = 2(2)² − 8(2) + 3 = −5, so the vertex is (2, −5) and the axis of symmetry is x = 2. Enter a, b, and c above to get the vertex, axis, and opening direction instantly.
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Solution
Step-by-Step:
Why the Formula Works
A parabola is symmetric about the vertical line through its vertex, and that line falls exactly halfway between the two roots. Averaging the roots of the quadratic formula, (−b + √D)/2a and (−b − √D)/2a, cancels the square-root terms and leaves −b / 2a. The same line is the axis of symmetry whether or not real roots exist.
Vertex Lookup by Equation
| Equation | h = −b/2a | Vertex (h, k) | Opens |
|---|---|---|---|
| x² − 4x + 1 | 2 | (2, −3) | Up |
| x² + 6x + 5 | −3 | (−3, −4) | Up |
| 2x² − 8x + 3 | 2 | (2, −5) | Up |
| −x² + 2x + 3 | 1 | (1, 4) | Down |
Minimum or Maximum?
The sign of a decides what the vertex means. When a > 0 the parabola opens upward and the vertex is the lowest point — for x² − 4x + 1, the function bottoms out at y = −3. When a < 0 it opens downward and the vertex is the peak: −x² + 2x + 3 maxes out at y = 4. The calculator reports the direction alongside the vertex.
A Real-World Vertex
A ball launched upward at 32 ft/s from a 48-foot platform follows h(t) = −16t² + 32t + 48. The vertex is at t = −32 / (2 × −16) = 1 second, where the height is −16 + 32 + 48 = 64 feet — the exact peak of the throw. Any optimization problem shaped like a parabola (profit vs. price, area vs. dimensions) comes down to this same vertex calculation.
Frequently Asked Questions
What is the axis of symmetry?
The vertical line x = −b/2a that passes through the vertex and splits the parabola into two mirror-image halves. For 2x² − 8x + 3 it is the line x = 2.
Can I find the vertex from the roots?
Yes. Average the two roots to get h, then evaluate f(h) for k. The roots 2 and 3 of x² − 5x + 6 average to h = 2.5, and f(2.5) = −0.25, giving the vertex (2.5, −0.25).
What if the parabola has no real roots?
The vertex formula still works — symmetry does not depend on crossing the x-axis. For x² + 2x + 5, the vertex is (−1, 4) even though the discriminant is negative.
How is vertex form related?
Vertex form, y = a(x − h)² + k, displays the vertex (h, k) directly. Expanding it gives standard form: 2(x − 2)² − 5 expands to 2x² − 8x + 3.