Vertex, Axis of Symmetry, Zeros and Extreme Values of a Parabola
The vertex of a parabola is its turning point, the axis of symmetry is the vertical line through it, and the zeros are the points where the parabola meets the x-axis. In brief: for f(x) = ax2 + bx + c, the axis is x = −b/(2a), the vertex is (−b/(2a), f(−b/(2a))) and the extreme value is the y-coordinate of the vertex. For f(x) = x2 − 4x + 3 the vertex is (2, −1), the axis is x = 2 and the zeros are 1 and 3.
These key features describe a parabola completely and let you sketch it without a long table of values. The vertex gives the minimum or maximum, which is the answer to many practical questions about the largest area or the highest point. This guide shows how to find each feature from the standard form, how the features relate and how to use them for extreme value problems.
How do you find the axis of symmetry?
A parabola is symmetric about a vertical line through its vertex. For f(x) = ax2 + bx + c, this line is x = −b/(2a). The reason is that the two zeros (when they exist) are symmetric about it: their average is (x1 + x2)/2 = −b/(2a), a fact that follows from the quadratic formula.
Example: for f(x) = 3x2 − 12x + 5, the axis is x = 12/6 = 2. A point at distance 1 to the left, x = 1, has the same height as the point at x = 3: f(1) = 3 − 12 + 5 = −4 and f(3) = 27 − 36 + 5 = −4 ✓.
How do you find the vertex?
- Compute h = −b/(2a).
- Compute k = f(h) by substitution.
- The vertex is (h, k).
Example 1: f(x) = −2x2 + 8x + 1
h = −8/(2 · (−2)) = 2. k = −2 · 4 + 16 + 1 = 9. The vertex is (2, 9). Since a < 0, it is the maximum point. The vertex form is f(x) = −2(x − 2)2 + 9.
Example 2: from the zeros
If the zeros are 1 and 7, the axis is x = 4. For f(x) = (x − 1)(x − 7), the vertex is (4, 3 · (−3)) = (4, −9).
How do you find the zeros?
The zeros are the solutions of ax2 + bx + c = 0. The discriminant D = b2 − 4ac tells how many there are:
- D > 0: two zeros, so the parabola crosses the x-axis twice.
- D = 0: one double zero, so the vertex lies on the x-axis.
- D < 0: no real zeros, so the parabola stays above or below the axis.
The zeros are x = (−b ± √D)/(2a).
Example 3
f(x) = x2 − 4x + 3: D = 16 − 12 = 4, zeros (4 ± 2)/2 = 3 and 1. The vertex is (2, −1), halfway between the zeros ✓. Also the distance from the axis to each zero is √D/(2|a|) = 1.
How do you find the minimum or maximum value?
The extreme value of a quadratic function is the y-coordinate k of the vertex. For a > 0 it is the minimum, and the range is y ≥ k. For a < 0 it is the maximum, and the range is y ≤ k. A formula in terms of coefficients: k = c − b2/(4a) = −D/(4a).
Example 4: range
f(x) = 2x2 + 4x + 5. Then h = −1 and k = 2 − 4 + 5 = 3. The minimum is 3 and the range is [3, ∞). Because 3 > 0, the graph never meets the x-axis, which agrees with D = 16 − 40 = −24 < 0.
Example 5: restricted domain
The function f(x) = x2 − 4x + 3 on the interval [3, 6] is increasing, because the vertex x = 2 lies to the left. So the minimum is f(3) = 0 and the maximum is f(6) = 15. On a closed interval, the extreme values are at the vertex or at an endpoint.
Features of sample parabolas
| Function | Axis | Vertex | Zeros | Extreme value |
|---|---|---|---|---|
| x2 − 4x + 3 | x = 2 | (2, −1) | 1, 3 | min −1 |
| −x2 + 6x − 5 | x = 3 | (3, 4) | 1, 5 | max 4 |
| x2 + 2x + 1 | x = −1 | (−1, 0) | −1 (double) | min 0 |
| 2x2 + 4x + 5 | x = −1 | (−1, 3) | none | min 3 |
| −2x2 + 8x + 1 | x = 2 | (2, 9) | 2 ± 3√2/2 (about −0.12 and 4.12) | max 9 |
Where do these ideas come from?
Finding the maximum of a quadratic expression is an old problem. Greek geometers solved problems about the largest rectangle for a given perimeter, and Euclid and Apollonius studied the geometry of the parabola. Completing the square gave algebraic access to the vertex long before calculus, which later gave a second method: set the derivative 2ax + b equal to zero to get the same x = −b/(2a).
Where are they used today?
A ball thrown upward with h(t) = −4.9t2 + 14.7t + 2 reaches its highest point at t = 1.5 s, where the height is h(1.5) = 13.025 m. A business with profit P(x) = −2x2 + 80x − 300 earns most at x = 20, with P(20) = 500. Engineers use the vertex to find optimal design values.
Which mistakes are common?
Forgetting the minus sign in −b/(2a)
For b = −4 and a = 1, the axis is x = 2, not x = −2.
Giving only the x-coordinate as the vertex
The vertex is a point. The extreme value is the y-coordinate.
Mixing up maximum and minimum
a > 0 gives a minimum, a < 0 gives a maximum.
Ignoring the domain restriction
On a limited interval the extreme value may be at an endpoint, not at the vertex.