How to Write a Quadratic Function in Standard Form (With Examples)

The standard form of a quadratic function is f(x) = ax² + bx + c, where a, b, and c are real numbers and a ≠ 0. That’s the direct answer to how to write a quadratic function in standard form: rearrange or expand whatever you’re given until the terms appear in descending order of degree, with the constant last. Most problems start from either vertex form, factored form, or a graph, and each requires a slightly different move.

Converting From Vertex Form to Standard Form

Vertex form, f(x) = a(x − h)² + k, tells you the vertex (h, k) immediately, but converting to standard form is pure algebra: expand the squared binomial and combine like terms.

Take f(x) = 2(x − 3)² + 4.

• Square the binomial: (x − 3)² = x² − 6x + 9.
• Distribute the 2: 2x² − 12x + 18.
• Add the constant: 2x² − 12x + 18 + 4 = 2x² − 12x + 22.

Here a = 2, b = −12, and c = 22. The vertex still sits at (3, 4) — you haven’t changed the function, only its appearance. As Khan Academy notes, these forms are equivalent representations, so converting back and forth is a good way to check your work.

One caution: watch the sign inside the parentheses. In f(x) = (x + 5)² − 1, the vertex is (−5, −1), not (5, −1). Expanding gives x² + 10x + 25 − 1 = x² + 10x + 24.

A shortcut for the coefficients

You don’t always have to expand step by step. Starting from a(x − h)² + k:

• a stays the same.
• b = −2ah
• c = ah² + k

With a = 2, h = 3, k = 4: b = −2(2)(3) = −12, and c = 2(9) + 4 = 22. Same answer, fewer lines.

Converting From Factored Form

If you’re given f(x) = a(x − r₁)(x − r₂), multiply the binomials first using FOIL, then distribute a. The roots r₁ and r₂ are where the graph crosses the x-axis, but they don’t appear directly in standard form.

Example: f(x) = 3(x − 1)(x + 4).

• FOIL the binomials: x² + 4x − x − 4 = x² + 3x − 4.
• Distribute the 3: 3x² + 9x − 12.

Check with the shortcut: since the roots are 1 and −4, the vertex’s x-coordinate is their midpoint, −1.5, which gives h = −1.5 and k = 3(−1.5 − 1)(−1.5 + 4) = 3(−2.5)(2.5) = −18.75. Using b = −2ah: −2(3)(−1.5) = 9. ✓

Writing the Function From a Graph or Data Points

When you’re handed a graph or a table of values, your job is to find a, b, and c. Standard form gives you a system of equations: each point (x, y) on the curve satisfies y = ax² + bx + c.

Three points method

Pick three points with distinct x-values. Substitute each into y = ax² + bx + c to get three equations, then solve.

Suppose the points are (0, 3), (1, 4), and (2, 9).

• (0, 3): c = 3.
• (1, 4): a + b + 3 = 4, so a + b = 1.
• (2, 9): 4a + 2b + 3 = 9, so 4a + 2b = 6, or 2a + b = 3.

Subtracting the second equation from the third: a = 2. Then b = 1 − 2 = −1. The function is f(x) = 2x² − x + 3.

The point (0, 3) gave away c for free. That’s not a coincidence — in standard form, c is always the y-intercept, because f(0) = c. If your graph or table includes the y-intercept, you’ve already solved for one unknown.

When the vertex is known

If the graph shows the vertex, vertex form is faster to write first, then convert. A parabola with vertex (−2, 5) passing through (0, 1):

• Start with f(x) = a(x + 2)² + 5.
• Plug in (0, 1): 1 = a(4) + 5, so 4a = −4 and a = −1.
• Expand: −(x² + 4x + 4) + 5 = −x² − 4x + 1.

As Purplemath points out, the sign of a is visible either way — a negative a opens the parabola downward, and here the vertex sits above the y-intercept, which confirms it.

Checking Your Work

A few habits catch most errors:

• Confirm a ≠ 0. If your expansion cancels the x² term, you didn’t have a quadratic to begin with.
• Test one point. Plug an original point back into your standard-form result. For f(x) = 2x² − x + 3, testing x = 2 gives 8 − 2 + 3 = 9. ✓
• Verify the vertex. The x-coordinate of the vertex is −b/(2a). For the same function, that’s −(−1)/(4) = 0.25, and 2(0.25)² − 0.25 + 3 = 2.875. Since the three original points didn’t include the vertex, this is just a sanity check that the shape is reasonable.

Paul’s Online Math Notes frames this well: standard form is the most useful version when you need the y-intercept, the axis of symmetry, or the vertex formula, while vertex form and factored form each expose different features.

One last reminder on order: standard form means descending powers of x, ending with the constant. Writing 3 + 2x − x² is not standard form, even though it describes the same parabola — always rewrite it as −x² + 2x + 3.

Key Takeaways

• Standard form is f(x) = ax² + bx + c, with a ≠ 0 and terms in descending order of degree.
• From vertex form a(x − h)² + k: expand, or use b = −2ah and c = ah² + k.
• From factored form a(x − r₁)(x − r₂): FOIL the binomials, then distribute a.
• From three points: substitute each into the standard equation and solve the resulting system; the y-intercept gives c directly.
• The coefficient c is always the y-intercept, and the vertex’s x-coordinate is −b/(2a).
• Always plug one original point back in to confirm your answer.