Vertex form is y = a(x − h)² + k, where (h, k) is the vertex. From standard form y = ax² + bx + c, h = −b ÷ (2a) and k = c − b² ÷ (4a) — for example, y = x² + 6x + 5 becomes y = (x + 3)² − 4 with vertex (−3, −4). This vertex form calculator converts between standard, vertex and factored form, shows the completing-the-square steps with exact fractions, and gives the vertex, axis of symmetry, discriminant, roots, y-intercept and a graph.
How to use the vertex form calculator
- Choose the form you already have: standard, vertex or factored.
- Type the three numbers. Fractions (1/2), decimals (0.75) and negatives all work.
- Read the vertex form at the top and the other two forms below it.
- Open the step-by-step panel to see how the square was completed, line by line.
- Copy the link to share the exact equation.
The three forms of a quadratic
| Form | Equation | Shows directly |
|---|---|---|
| Standard (general) | y = ax² + bx + c | y-intercept c |
| Vertex | y = a(x − h)² + k | vertex (h, k), axis x = h, max/min value k |
| Factored (intercept) | y = a(x − r₁)(x − r₂) | x-intercepts r₁ and r₂ |
All three describe the same parabola and share the same a. The OpenStax College Algebra vertex formulas link them: h = −b/(2a), k = f(h) = c − b²/(4a), and the roots are h ± √(−k/a), which is the quadratic formula (−b ± √(b² − 4ac))/(2a) rearranged.
Worked example: completing the square
Convert y = 2x² − 8x + 3 to vertex form. The calculator's own steps:
- Start: y = 2x² − 8x + 3
- Factor 2 out of the x terms: y = 2(x² − 4x) + 3
- Half of the x coefficient −4 is −2; its square is 4.
- Add and subtract 4 inside the brackets: y = 2(x² − 4x + 4) − 2 × 4 + 3
- The bracket is a perfect square: x² − 4x + 4 = (x − 2)².
- Combine the constants: 3 − 8 = −5.
- Vertex form: y = 2(x − 2)² − 5, so h = 2 and k = −5.
The vertex is (2, −5); because a = 2 is positive, the parabola opens upward and −5 is its minimum value. The discriminant is 40, positive but not a perfect square, so the roots are irrational: x = 2 − √10/2 and 2 + √10/2.
Worked example: the shortcut formula
For y = 3x² + 5x − 2: h = −5 ÷ (2 × 3) = −5/6 and k = −2 − 25 ÷ 12 = −49/12, so y = 3(x + 5/6)² − 49/12. The discriminant 49 is a perfect square, so the roots are rational (−2 and 1/3) and the factored form is y = 3(x + 2)(x − 1/3).
Vertex form to standard form and factored form
Going the other way is just expanding. y = 2(x − 3)² − 5 expands to y = 2x² − 12x + 13. Starting from factored form, y = (1/2)(x + 1)(x − 3) expands to y = (1/2)x² − x − 3/2, and its vertex form is y = (1/2)(x − 1)² − 2 — the vertex sits halfway between the roots, at x = (−1 + 3) ÷ 2 = 1.
Examples table
| Standard form | Vertex form | Vertex | Discriminant | Roots |
|---|---|---|---|---|
| y = x² + 6x + 5 | y = (x + 3)² − 4 | (−3, −4) | 16 | −5 and −1 |
| y = 2x² − 8x + 3 | y = 2(x − 2)² − 5 | (2, −5) | 40 | 2 − √10/2 and 2 + √10/2 |
| y = −x² + 4x + 1 | y = −(x − 2)² + 5 | (2, 5) | 20 | 2 − √5 and 2 + √5 |
| y = x² + 2x + 5 | y = (x + 1)² + 4 | (−1, 4) | −16 | −1 − 2i and −1 + 2i |
| y = 3x² + 5x − 2 | y = 3(x + 5/6)² − 49/12 | (−5/6, −49/12) | 49 | −2 and 1/3 |
| y = x² − 4x + 4 | y = (x − 2)² | (2, 0) | 0 | 2 |
| y = (1/2)x² + 3x | y = (1/2)(x + 3)² − 9/2 | (−3, −9/2) | 9 | −6 and 0 |
Row four, y = x² + 2x + 5, has a negative discriminant: the vertex (−1, 4) is above the x-axis and the parabola opens upward, so it never crosses, and the roots are the complex pair −1 − 2i and −1 + 2i.
Reading the graph from vertex form
- Vertex: (h, k). Note the sign flip: y = (x − 2)² has h = +2, and y = (x + 3)² has h = −3.
- Direction: a > 0 opens up (k is the minimum); a < 0 opens down (k is the maximum), as in y = −(x − 2)² + 5.
- Width: |a| greater than 1 makes the parabola narrower than y = x²; between 0 and 1, wider.
- Axis of symmetry: the line x = h; points the same distance either side have equal y-values.
- Shift: y = a(x − h)² + k is y = ax² moved h units right and k units up.
Where vertex form is used
Any problem asking for a maximum or minimum is a vertex problem: the highest point of a thrown ball, the price that maximises revenue, the dimensions that give the largest area for a fixed fence. Write the quadratic in vertex form and k is the answer, reached at x = h. OpenStax works a fencing example this way: a fixed length of fence gives an area that is a downward-opening quadratic in one side length, and its vertex gives the best dimensions.
Common mistakes
- Getting the sign of h wrong. h = −b/(2a); in vertex form the bracket shows x − h, so a positive h appears as a minus sign.
- Forgetting to factor out a. Completing the square only works directly when the x² coefficient is 1.
- Not multiplying the subtracted square by a. When a ≠ 1, the square you add inside the bracket is worth a × square outside it.
- Rounding too early. Keep fractions exact; −5/6 is not −0.83 when you square it.
- Assuming real roots. A negative discriminant means no x-intercepts.
More algebra tools: matrix inverse calculator, fraction calculator, linear interpolation and the 30-60-90 triangle calculator.
Frequently asked questions
What is vertex form?
Vertex form writes a quadratic as y = a(x − h)² + k. The point (h, k) is the vertex — the lowest point of the parabola if a is positive, the highest if a is negative. OpenStax calls it the standard form of a quadratic function and the ax² + bx + c version the general form; many classes say vertex form and standard form instead.
How do you convert standard form to vertex form?
Either complete the square or use the shortcut h = −b ÷ (2a) and k = c − b² ÷ (4a). For y = x² + 6x + 5: h = −6 ÷ 2 = −3 and k = 5 − 36 ÷ 4 = −4, so y = (x + 3)² − 4.
How do you find the vertex from standard form?
The x-coordinate is h = −b/(2a). Substitute it back into the equation to get k, the y-coordinate. That is the vertex formula OpenStax gives, and the axis of symmetry is the vertical line x = h.
How do you convert vertex form to standard form?
Expand the square and collect terms. y = 2(x − 3)² − 5 becomes 2(x² − 6x + 9) − 5 = 2x² − 12x + 13. Choose "vertex form" in the calculator to do it automatically.
Is a the same in standard form and vertex form?
Yes. The leading coefficient a is the same in all three forms — standard, vertex and factored — and it alone decides whether the parabola opens up or down and how narrow it is.
How do you complete the square?
Factor a out of the x² and x terms, take half of the new x coefficient, square it, add and subtract that square inside the bracket, then write the first three terms as a perfect square. The calculator shows every step with exact fractions.
What does the discriminant tell you?
b² − 4ac positive means two real roots (the parabola crosses the x-axis twice), zero means one repeated root (the vertex sits on the axis) and negative means two complex roots (no x-intercepts).
How do you find the roots from vertex form?
Set y = 0: a(x − h)² + k = 0, so (x − h)² = −k/a and x = h ± √(−k/a). If −k/a is negative there are no real roots.
What is factored (intercept) form?
y = a(x − r₁)(x − r₂), where r₁ and r₂ are the x-intercepts. It exists over the real numbers only when the discriminant is zero or positive; the calculator gives the factored form whenever the roots are rational.
Can the calculator handle fractions and decimals?
Yes. Enter 1/2, −3/4 or 0.25 and everything is computed exactly with fractions, so a vertex like (−5/6, −49/12) is shown exactly instead of as a rounded decimal.
What if a = 0?
Then there is no x² term and the equation is a straight line, not a parabola, so it has no vertex form.
Sources & method
- OpenStax College Algebra 2e §5.1 — Quadratic functions (general and standard/vertex form, h = −b/2a, k = f(h))
- OpenStax Intermediate Algebra 2e §9.2 — Solve quadratic equations by completing the square
- OpenStax Intermediate Algebra 2e §9.3 — Solve quadratic equations using the quadratic formula (discriminant)
Results are estimates for general information. Found an error? It helps everyone — see our methodology.