Vertex form writes a quadratic as y = a(x − h)² + k, and the vertex of the parabola is the point (h, k). If your quadratic is in the form y = ax² + bx + c, find the vertex with h = −b / 2a, then substitute h back in to get k. Completing the square gives the same point and rewrites the whole equation in vertex form at the same time.
Below, one parabola is solved by all three routes (the −b / 2a formula, completing the square, the midpoint of the roots) so you can see they agree. Then come conversions both ways, an equation from a vertex and one point, and a thrown ball whose vertex is its highest point.
What Is Vertex Form?
y = a(x − h)² + k, with a ≠ 0. The vertex is (h, k) and the axis of symmetry is the vertical line x = h. If a > 0 the parabola opens up and k is its minimum value; if a < 0 it opens down and k is its maximum value.Why does the vertex sit at (h, k)? Look at the squared bracket. (x − h)² can never be negative, and it equals zero only when x = h. So when a is positive, a(x − h)² is at least zero, and y is smallest exactly when that term vanishes, at x = h, where y = k. When a is negative the same term is at most zero, so y is largest there instead.
You can also see vertex form as the basic parabola y = x² after function transformations: shifted right by h, stretched by a (and flipped if a is negative), then shifted up by k. The point (0, 0) on y = x² ends up at (h, k).
This guide calls ax² + bx + c "standard form" and a(x − h)² + k "vertex form". Textbooks don't all agree: OpenStax College Algebra calls ax² + bx + c the general form and a(x − h)² + k the standard form, noting that it is also called vertex form. If a question says "standard form", check which one your course means.
How to Read the Vertex from Vertex Form
The only tricky part is the sign of h. The formula has a minus sign built in, so whatever you see inside the bracket, h is its opposite: (x − 4) means h = 4, and (x + 3) means x − (−3), so h = −3. The k outside the bracket keeps its sign.
Find the vertex, the axis of symmetry and the maximum or minimum of y = −2(x + 3)² + 5.
Rewrite the bracket to match the template: x + 3 = x − (−3), so h = −3. The constant outside is k = 5.
Here a = −2 is negative, so the parabola opens down and the vertex is the highest point.
Axis of symmetry: x = −3. Maximum value: y = 5, reached at x = −3.
vertex (−3, 5), axis x = −3, maximum y = 5Two special cases come up often. y = (x − 4)² has no k written, so k = 0 and the vertex is (4, 0), sitting on the x-axis. y = 3x² − 7 has no bracket shift, so h = 0 and the vertex is (0, −7), on the y-axis.
Three Ways to Find the Vertex from Standard Form
Most problems hand you y = ax² + bx + c. Running all three routes on one parabola shows they are really one idea, and gives you a second method to check with.
Find the vertex of y = 2x² − 12x + 10.
Route 1, the formula. Here a = 2, b = −12, c = 10. So h = −b / 2a = −(−12) / (2 × 2) = 12 / 4 = 3. Substitute: k = 2(3)² − 12(3) + 10 = 18 − 36 + 10 = −8.
Route 2, completing the square. Factor 2 out of the x terms: y = 2(x² − 6x) + 10. Half of −6 is −3, and (−3)² = 9. Add and subtract 9 inside the bracket: y = 2(x² − 6x + 9 − 9) + 10 = 2(x − 3)² − 18 + 10 = 2(x − 3)² − 8.
Route 3, the roots. Factor: y = 2(x² − 6x + 5) = 2(x − 1)(x − 5), so the roots are x = 1 and x = 5. The vertex lies halfway between them: h = (1 + 5) / 2 = 3. Then k = 2(3 − 1)(3 − 5) = 2(2)(−2) = −8.
vertex (3, −8); vertex form y = 2(x − 3)² − 8Route 3 works because a parabola is symmetric about its axis, so its two x-intercepts sit at equal distances from x = h. If you need a refresher on getting the roots, see how to factor quadratic equations.
Why h = −b / 2a
The formula is just Route 2 done once with letters. Factor a out of the x terms and complete the square:
y = a(x² + (b/a)x) + c = a(x + b/2a)² + c − b²/4a
Matching this with a(x − h)² + k gives h = −b / 2a and k = c − b² / 4a. You can use that second formula for k, but substituting h into the original equation is usually faster and harder to get wrong. For Example 2, k = 10 − 144 / 8 = 10 − 18 = −8, the same answer.
| Route | Best when | Watch out for |
|---|---|---|
h = −b / 2a, then substitute | You only need the vertex, especially with awkward numbers | Dropping the minus sign when b is negative |
| Completing the square | The question asks for the equation in vertex form | Forgetting to multiply by a when you balance the added number |
| Midpoint of the roots | The quadratic is already factored, or factors easily | Only works with real roots: y = x² + 2x + 5 has none, yet its vertex is (−1, 4) |
How to Convert Standard Form to Vertex Form
When the answer has to be an equation, complete the square. The full method, including why "half of b, squared" makes a perfect square, is in how to complete the square. Here is the vertex-form version when a is not 1.
- Factor a out of the x terms only
Leave the constant
coutside the bracket. Inside, thex²term now has coefficient 1. - Add and subtract the square of half the x coefficient
Take the coefficient of
xinside the bracket, halve it, square it, and add and subtract it inside. - Move the subtracted number out, multiplied by a
The extra number sits inside the bracket, so it gets multiplied by
aon the way out. Combine it withc. - Write the perfect square
The bracket is now
(x − h)². Read off(h, k).
Write y = −3x² + 12x − 7 in vertex form.
Factor −3 out of the x terms: y = −3(x² − 4x) − 7. Check by expanding: −3 × −4x = 12x, so the sign is right.
Half of −4 is −2, and (−2)² = 4: y = −3(x² − 4x + 4 − 4) − 7.
Move the −4 out. It gets multiplied by −3, so it becomes +12: y = −3(x² − 4x + 4) + 12 − 7.
So y = −3(x − 2)² + 5. Cross-check with the formula: h = −12 / (2 × −3) = 2 and −3(4) + 24 − 7 = 5.
y = −3(x − 2)² + 5, vertex (2, 5), a maximumIn Example 3, the most common wrong answer is y = −3(x − 2)² − 11. It comes from adding 4 inside the bracket and then subtracting 4 outside, as if a were 1. But the 4 inside is multiplied by −3, so you actually added −12 to the right side. To undo that you add +12, not subtract 4. The habit that prevents it: after moving the number out, expand your answer once and check you get the original c back.
Odd values of b give fractions, and the method does not change. For y = x² + 5x + 1, half of 5 is 5/2 and its square is 25/4, so y = (x + 5/2)² − 25/4 + 1 = (x + 5/2)² − 21/4. The vertex is (−5/2, −21/4), or (−2.5, −5.25). Keep the fractions exact until the end; rounding 25/4 early is a common source of a wrong k.
How to Convert Vertex Form to Standard Form
Going the other way is just expanding. Square the bracket first, then multiply by a, then add k. Take Example 1:
y = −2(x + 3)² + 5 = −2(x² + 6x + 9) + 5 = −2x² − 12x − 18 + 5 = −2x² − 12x − 13
A quick check: −b / 2a = 12 / (−4) = −3, which matches h = −3. And the y-intercept of the expanded form, −13, should equal the vertex form at x = 0: −2(3)² + 5 = −18 + 5 = −13. Squaring the bracket before multiplying matters. Writing −2(x + 3)² as (−2x − 6)² squares the −2 as well and gives the wrong parabola.
How to Write the Equation from the Vertex and a Point
Vertex form has three numbers, a, h and k. The vertex gives you two of them, so one more point on the parabola is enough to find a.
A parabola has its vertex at (1, −4) and passes through (3, 4). Find its equation.
Start from the template with h = 1 and k = −4: y = a(x − 1)² − 4.
Substitute the point (3, 4): 4 = a(3 − 1)² − 4, so 4 = 4a − 4.
Solve: 4a = 8, so a = 2.
Check with symmetry: the mirror image of (3, 4) across x = 1 is (−1, 4), and 2(−1 − 1)² − 4 = 8 − 4 = 4.
y = 2(x − 1)² − 4, or y = 2x² − 4x − 2Vertex Form in a Word Problem
Any question that asks for a maximum height, a minimum cost or a largest area of a quadratic is a vertex question. The vertex gives both parts of the answer: h says when or where, and k says how much.
A ball is thrown straight up from a height of 1.5 m at 19.6 m/s. Its height after t seconds is H(t) = −4.9t² + 19.6t + 1.5 meters, using g = 9.8 m/s² (the −4.9 is −½g; the exact standard value is 9.80665 m/s² from NIST). When does it reach its highest point, and how high is it?
Time of the vertex: t = −b / 2a = −19.6 m/s ÷ (2 × −4.9 m/s²) = 2.00 s. The units cancel to seconds, as they should.
Height there: H(2) = −4.9(4) + 19.6(2) + 1.5 = −19.6 + 39.2 + 1.5 = 21.1 m.
In vertex form: H(t) = −4.9(t − 2)² + 21.1. Expanding gives back −4.9t² + 19.6t − 19.6 + 21.1, and −19.6 + 21.1 = 1.5.
Physics cross-check: at the top the velocity is zero, and v = 19.6 − 9.8t = 0 at t = 2 s, the same time. The heights are given to 0.1 m, so the answer keeps one decimal place.
highest point 21.1 m, reached at t = 2.00 sThe kinematic equations guide explains where a height formula like this comes from. Maximum-area and minimum-cost problems use the same vertex step, and in calculus the same answers come from setting a derivative to zero, as in optimization problems.
How to Check Your Vertex
Three checks catch nearly every error, and each takes under a minute.
- Symmetry test. Pick two x-values the same distance either side of
h; they must give the same y. For Example 2,f(2) = 8 − 24 + 10 = −6andf(4) = 32 − 48 + 10 = −6. Equal, sox = 3really is the axis. Both values are also above−8, which fits a parabola opening up. - Expand back. Multiply your vertex form out and compare it with the original
ax² + bx + c. All three coefficients must match. This is the check that catches the Common Mistake above. - Second route. If you completed the square, confirm
hwith−b / 2a, or with the midpoint of the roots if they're easy to find.
The free quadratic formula calculator shows the vertex, the roots and a graph of the parabola with the steps, so you can compare it with your own working once you've finished. If a scanned homework problem has you stuck, Solver AI shows the conversion step by step, which makes it easier to spot the line where a sign flipped.
Practice Problems
Work each one by hand first, then check it with a second route.
- Find the vertex and the minimum value of
y = 4(x − 1)² − 9. - Write
y = x² + 8x + 11in vertex form and give the vertex. - Write
y = 2x² + 6x − 1in vertex form. Keep fractions exact. - A farmer has 60 m of fencing to make a rectangular pen against a barn wall, so only three sides need fence. With width
xmeters, the area isA = x(60 − 2x). Find the dimensions that give the largest area, and that area. - A parabola has its vertex at
(−2, 3)and passes through(0, −5). Find its equation in vertex and standard form, then find its x-intercepts exactly.
1. Vertex (1, −9); minimum y = −9 at x = 1. 2. y = (x + 4)² − 5, vertex (−4, −5). 3. y = 2(x + 3/2)² − 11/2, vertex (−3/2, −11/2). 4. A = −2(x − 15)² + 450, so the pen is 15 m by 30 m with area 450 m². 5. a = −2, so y = −2(x + 2)² + 3 = −2x² − 8x − 5; setting y = 0 gives (x + 2)² = 3/2, so x = −2 ± (√6)/2, about −0.775 and −3.225.