Vertex form writes a quadratic as y = a(x − h)² + k, and the whole point of it is that the turning point sits there in plain sight at (h, k). Standard form contains exactly the same curve and hands you a different fact — the y-intercept — while making you work for the vertex. Completing the square i...
(x − 3)² has its vertex at +3, not −3. The bracket subtracts h, so a plus sign printed inside means h is negative. Reading the sign straight off the bracket reflects the answer about the y-axis every time. The other half of the trouble is the leading coefficient: a has to be taken outside the x terms before b is halved, or the vertex comes out wrong by a factor of a — and when a is 1 both orders agree, which is why the habit survives until it does not. This page shows every line of the completion, and expands the result back to check it returns the coefficients it started from.
For y = ax² + bx + c. Whole numbers, decimals and fractions all work.
A
B
C
y = x² − 6x + 5
VERTEX FORM
y = (x − 3)² − 4
vertex (3, -4)
The vertex is at (3, -4), and expanding this form returns the coefficients it came from.
The bracket and the vertex have opposite signs. This one reads y = (x − 3)² − 4, and the vertex is at x = 3. Reading the sign printed inside the bracket would give -3 — the same curve reflected about the y-axis.
THE VERTEX
(3, -4)
both coordinates whole
AXIS OF SYMMETRY
x = 3
the mirror line through the turning point
MINIMUM VALUE
-4
a is positive, so it opens upward
EXPANDS BACK
yes
to the coefficients it started from
WHAT IS HAPPENING
y = x² − 6x + 5
Vertex form says the same thing rearranged, so nothing is added or thrown away. What changes is what can be read off at a glance — the turning point rather than the y-intercept.
half of -6 is -3, and -3² = 9
That square is what is missing from x² + px to make it a perfect square. Adding it and taking it away again in the same line changes nothing, which is the whole trick.
(x − 3)² − 4
The bracket closes into a perfect square, and what is left over is the constant. Note the sign: the bracket holds -3, so the vertex sits at x = 3 — the opposite sign to the one printed inside it.
vertex (3, -4), axis of symmetry x = 3
Because a is positive the parabola opens upward, so -4 is its minimum value and it is reached at x = 3.
Vertex form is a set of instructions. Start from y = x², stretch it vertically by a, move it h to the right and k up, and you have the curve. Standard form contains exactly the same information and hands you a different piece of it — the y-intercept, which vertex form makes you work for. Neither is more correct; they answer different questions, and completing the square is the translation between them.
Exact fractions throughout · the vertex form is expanded back to check it returns the coefficients
Choose a direction. Standard to vertex completes the square; vertex to standard expands it. Both show every line, because the intermediate steps are where the mistakes live.
Enter the three values. For standard form they are a, b and c; for vertex form they are a, h and k — where h is the vertex position, not the number printed inside the bracket.
Check the sign warning. The page always states what the bracket reads and where the vertex actually is, so the two can be compared rather than confused.
Look at the shift drawn against y = ax². The dashed curve is the base parabola and the solid one is yours, with h and k marked as the movement between them.
Take y = x² − 6x + 5. Half of −6 is −3, and (−3)² is 9. Add nine and take nine away in the same line: x² − 6x + 9 − 9 + 5. The first three terms close into (x − 3)², and what is left is −4. So y = (x − 3)² − 4, and the vertex is at (3, −4). Note the sign. The bracket holds −3 and the vertex is at +3. Everything about vertex form that trips people up is in that one line. Check it by expanding: (x − 3)² is x² − 6x + 9, and 9 − 4 is 5, so y = x² − 6x + 5. Back where we started, which is the only check worth doing. Now y = 2x² − 8x + 3, where the leading coefficient changes the order of operations. Take the 2 outside the x terms first: 2(x² − 4x) + 3. Now half of −4 is −2 and (−2)² is 4, so inside the bracket we have 2(x² − 4x + 4 − 4) + 3, which is 2(x − 2)² − 8 + 3, or 2(x − 2)² − 5. The vertex is at (2, −5). Try it the other way round to see the failure. Halving b straight off gives −8 ÷ 2 = −4, suggesting a vertex at x = 4. The real vertex is at x = 2. The two differ by exactly the factor of 2 that was never taken out, and on a quadratic with a = 1 the same shortcut would have been right — which is precisely why it survives. Now y = 3x² + 5x + 1, where the answer is not tidy. h = −5 ÷ 6 and k = 1 − 25 ÷ 12 = −13 ÷ 12, so y = 3(x + 5/6)² − 13/12. Nothing has gone wrong. Only 14.3% of quadratics with integer coefficients between −10 and 10 have a whole-number vertex, and completing the square produces sixths and twelfths by construction. Keeping these as fractions rather than 0.8333 and 1.0833 is the difference between an exact answer and a nearly-exact one. One last thing worth knowing: k is the discriminant divided by −4a. A vertex sitting exactly on the x-axis is precisely a discriminant of zero — the two facts are the same fact seen from different sides.
| Rule | What it says | Why |
|---|---|---|
| Vertex form | y = a(x − h)² + k | The turning point is at (h, k), readable without any working. |
| The sign of h | (x − 3)² has vertex at +3 | The bracket subtracts h, so a plus inside means a negative h. The commonest error by far. |
| Standard form | y = ax² + bx + c | Same curve. This one hands you the y-intercept instead of the vertex. |
| Finding h | h = −b ÷ 2a | Divide by 2a, not by 2. The a matters and is easy to drop. |
| Finding k | k = c − b² ÷ 4a | Equivalently, substitute h back in. Both give the same number. |
| Completing the square | factor a out first | Halving b before dividing by a puts the vertex out by a factor of a. |
| The square to complete | (half of p)², where p = b ÷ a | Add it and subtract it in the same line, so nothing changes. |
| Axis of symmetry | x = h | The mirror line through the turning point. Same number as the vertex x. |
| Maximum or minimum | k, decided by the sign of a | a positive opens upward so k is the minimum; a negative opens downward so it is the maximum. |
| a is the same in both forms | unchanged by the rearranging | Only b and c are absorbed into h and k. |
| k against the discriminant | k = −D ÷ 4a | The two are the same fact. k = 0 is exactly a zero discriminant. |
| The y-intercept | c, or a·h² + k | What the curve does at x = 0, which vertex form does not show directly. |
| Transformations | shift h right, k up, stretch by a | Vertex form reads as instructions applied to y = x². |
| Whole vertices are rare | 14.3% of small integer cases | Completing the square produces halves and quarters by construction. |
It is y = a(x − h)² + k, and its value is that the turning point is (h, k) with no working required. Standard form gives the y-intercept just as directly and hides the vertex. Neither is more correct — they answer different questions about the same curve. Vertex form also reads as a set of instructions: take y = x², stretch it vertically by a, move it h to the right and k up.
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Last updated: August 5, 2026 · Exact fractions throughout, never floating point · The bracket subtracts h, so (x − 3)² has its vertex at +3.