acalculator

What is the vertex of my parabola?

Type a, b and c of y = ax² + bx + c, or a, h and k of the vertex form. The vertex calculator shows the vertex as exact fractions, the axis of symmetry, whether the parabola has a minimum or a maximum, its roots, and the equation in both forms.

Your numbers

My equation is
Vertex (h, k)
(3/2, 5/2)

The vertex is (3/2, 5/2), on the axis x = 3/2.

h (x of the vertex)
1.5
k (y of the vertex)
2.5
Axis of symmetry
x = 3/2
Opens
Up: the minimum is y = 5/2
Standard form
y = 2x² − 6x + 7
Vertex form
y = 2(x − 3/2)² + 5/2
b
-6
c (y-intercept)
7
Real roots (x-intercepts)
None: the parabola does not cross the x-axis.

Vertex (h, k): (3/2, 5/2). The vertex is (3/2, 5/2), on the axis x = 3/2.

The parabola, with its vertex

How to calculate

Computes the vertex (h, k) of a parabola from y = ax² + bx + c or y = a(x − h)² + k, with the axis of symmetry, the minimum or maximum, the y-intercept, the real roots and both forms.

Example with the default inputs (My equation is ax² + bx + c, a 2, b -6, c 7): The vertex is (3/2, 5/2), on the axis x = 3/2.

Method: h = −b ÷ (2a), k = c − b² ÷ (4a); from vertex form, b = −2ah and c = ah² + k; roots x = h ± √(−k ÷ a).

  • a may not be 0 (then the graph is a line with no vertex).
  • Coefficients are read as the exact decimals typed, so the vertex is an exact fraction.

Machine-readable copies: Markdown, JSON.

Worked examples

Each example is checked against the calculator on every build.

  1. My equation is ax² + bx + c, a 2, b -6, c 7 gives Vertex (h, k) (3/2, 5/2), h 1.5, k 2.5, Vertex form y = 2(x − 3/2)² + 5/2, Opens Up: the minimum is y = 5/2.Source: OpenStax, Algebra and Trigonometry 2e, §5.1 Quadratic Functions, Example 3 (vertex (3/2, 5/2); f(x) = 2(x − 3/2)² + 5/2), https://openstax.org/books/algebra-and-trigonometry-2e/pages/5-1-quadratic-functions
  2. My equation is ax² + bx + c, a 1, b -4, c 3 gives Vertex (h, k) (2, −1), Axis of symmetry x = 2, Real roots (x-intercepts) x = 1 and x = 3, c 3.Source: OpenStax, Algebra and Trigonometry 2e, §5.1 Quadratic Functions (h = −b ÷ (2a), k = f(h)), https://openstax.org/books/algebra-and-trigonometry-2e/pages/5-1-quadratic-functions
  3. My equation is a(x − h)² + k, a -0.5, h -2, k 8 gives Standard form y = −(1/2)x² − 2x + 6, b -2, c 6, Opens Down: the maximum is y = 8, Real roots (x-intercepts) x = −6 and x = 2.Source: OpenStax, Algebra and Trigonometry 2e, §5.1 Quadratic Functions (f(x) = a(x − h)² + k; a < 0 opens downward), https://openstax.org/books/algebra-and-trigonometry-2e/pages/5-1-quadratic-functions
  4. My equation is ax² + bx + c, a 1, b 2, c -1 gives Vertex (h, k) (−1, −2), Real roots (x-intercepts) x = −2.414213562 and x = 0.4142135624.Source: OpenStax, Algebra and Trigonometry 2e, §5.1 Quadratic Functions, https://openstax.org/books/algebra-and-trigonometry-2e/pages/5-1-quadratic-functions

How it works

From standard form y = ax² + bx + c (OpenStax calls a(x − h)² + k the standard form of a quadratic function; this page calls it the vertex form):

  • h = −b ÷ (2a)
  • k = c − b² ÷ (4a), which equals the value of ax² + bx + c at x = h

From vertex form y = a(x − h)² + k:

  • b = −2ah and c = ah² + k

For both:

  • Axis of symmetry: x = h.
  • Opens up with a minimum y = k when a > 0, down with a maximum y = k when a < 0.
  • y-intercept = c.
  • Real roots: let q = −k ÷ a. If q < 0 there is no real root. If q = 0 the only root is x = h. If q > 0 the roots are x = h − √q and x = h + √q, smaller first.

Rules. a may not be 0: with a = 0 the graph is a straight line, so there is no answer and the page says so. When h or k is too large to show as a number (beyond about 1.8 × 10³⁰⁸, from an a very close to 0), there is no answer either. Each coefficient is from −1,000,000 to 1,000,000.

Exact arithmetic. Each coefficient is read as the exact decimal you typed (−0.5 is exactly −1/2). h, k, b and c are exact fractions. A root is exact when q is the square of a fraction (q = 1 gives 2 ± 1); otherwise the root uses the square root of q as a decimal number.

Output format. The vertex, the axis, the minimum or maximum and exact roots show as fractions in lowest terms (3/2, −1, 5/2), with a true minus sign. In the two equations, a fraction before x is put in brackets ((1/2)x²), terms with a zero coefficient are left out, and a coefficient of 1 is not written. h, k, b and c also show as decimals with up to 10 significant figures. Roots that are not exact show with 10 significant figures.

Worked examples by hand

y = 2x² − 6x + 7 (OpenStax Example 3). h = −(−6) ÷ (2 × 2) = 3/2; k = 7 − 36 ÷ 8 = 5/2. Vertex (3/2, 5/2), axis x = 3/2, a > 0 so it opens up with a minimum of 5/2. Vertex form y = 2(x − 3/2)² + 5/2. q = −(5/2) ÷ 2 < 0: no real roots.

y = x² − 4x + 3. h = 4 ÷ 2 = 2; k = 3 − 16 ÷ 4 = −1. q = 1 ÷ 1 = 1, so x = 2 ± 1: roots 1 and 3.

y = −0.5(x + 2)² + 8. a = −1/2, h = −2, k = 8. b = −2 × (−1/2) × (−2) = −2; c = −1/2 × 4 + 8 = 6: y = −(1/2)x² − 2x + 6. Opens down with a maximum of 8. q = −8 ÷ (−1/2) = 16, so x = −2 ± 4: roots −6 and 2.

y = x² + 2x − 1. h = −1; k = −1 − 4 ÷ 4 = −2. q = 2 is not a square, so x = −1 ± √2 = −2.414213562 and 0.4142135624.

Other questions people ask

How do I find the vertex of a parabola?

For y = ax² + bx + c, the x of the vertex is h = −b ÷ (2a). Put h back into the equation for k. For y = 2x² − 6x + 7, h = 6 ÷ 4 = 3/2 and k = 2(9/4) − 9 + 7 = 5/2, so the vertex is (3/2, 5/2).

Is there a formula for k?

Yes: k = c − b² ÷ (4a). For y = 2x² − 6x + 7, k = 7 − 36 ÷ 8 = 5/2, the same as putting h into the equation.

What is the axis of symmetry?

The vertical line through the vertex, x = h. The parabola is a mirror image on each side of it.

How do I tell a minimum from a maximum?

Look at the sign of a. When a > 0 the parabola opens up and the vertex is its lowest point (a minimum). When a < 0 it opens down and the vertex is its highest point (a maximum).

How do I convert vertex form to standard form?

Expand a(x − h)² + k: b = −2ah and c = ah² + k. For y = −0.5(x + 2)² + 8, b = −2 × (−0.5) × (−2) = −2 and c = −0.5 × 4 + 8 = 6, so y = −0.5x² − 2x + 6.

How are the roots found from the vertex?

Set y = 0: a(x − h)² = −k, so x = h ± √(−k ÷ a). When −k ÷ a is negative there is no real root; when it is 0, the vertex is the only root.