What is the vertex form of a quadratic?

y = a(x − h)² + k. The vertex is (h, k), the axis of symmetry is x = h, and a sets the direction and the width.

How do you find the vertex from standard form?

For y = ax² + bx + c, the x-coordinate is h = −b ÷ 2a. Put that x into the rule to get k. The vertex is (h, k).

Why is the vertex (h, k) when the bracket says x − h?

The bracket (x − h) is zero when x = h, which is where the parabola reaches its turning point. The + k then lifts that point to height k.

What does the sign of a tell you?

If a is positive the parabola is a bowl and the vertex is the lowest point. If a is negative it is an arch and the vertex is the highest point.

Part 27 of 30 · QuadraticsSee the whole course

Vertex form of a quadratic: y = a(x − h)² + k

By The Pip team · October 24, 2026 · 7 min read

Short answer

In vertex form, y = a(x − h)² + k, the vertex is the point (h, k) and the axis of symmetry is the line x = h. The number a opens it up (positive) or down (negative) and sets how narrow it is.

Why it works: it is x² moved

Vertex form is just the transformations you already know, applied to y = x²:

(x − h) slides it h to the right (see horizontal shifts).
+ k lifts it k up (see vertical shifts).
a in front stretches it and flips it if negative (see stretch and compression).

The vertex of x² is (0, 0). After the moves it is at (h, k).

Example 1: y = (x − 2)² − 1

a = 1, so a bowl. h = 2 and k = −1, so the vertex is (2, −1) and the axis is x = 2.
This is the same parabola as y = x² − 4x + 3 from the last post. Expand to check: (x − 2)² − 1 = x² − 4x + 4 − 1 = x² − 4x + 3 ✓.

Example 2: y = −2(x + 1)² + 8

a = −2: an arch, narrower than x². The bracket (x + 1) is x − (−1), so h = −1. k = 8.
The vertex is (−1, 8), the highest point.
It lands where −2(x + 1)² + 8 = 0, so (x + 1)² = 4, so x + 1 = 2 or −2: x = 1 or x = −3.

This is the figure’s starting point. Slide a, h and k and watch the gold vertex and the dashed mirror line.

y = −2(x + 1)² + 8

a, in front −2

h, sideways −1

k, up and down 8

vertex (h, k) = (−1, 8): from x + 1 and + 8
axis of symmetry x = −1; highest value 8
expanded: y = −2x² − 4x + 6

It is an arch with its top at (−1, 8). It lands on the x-axis at x = −3 and x = 1.

From standard form to vertex form

To rewrite y = x² + 6x + 5, find the vertex first. The x-coordinate is h = −b ÷ 2a = −6 ÷ 2 = −3. Then k = the height at x = −3: 9 − 18 + 5 = −4. So:

y = (x + 3)² − 4
Check: (x + 3)² − 4 = x² + 6x + 9 − 4 = x² + 6x + 5 ✓

A ball: the highest point

A ball’s height after t seconds is h(t) = −5(t − 2)² + 20. The vertex is (2, 20): the ball is highest, 20, at t = 2. It is back on the ground when −5(t − 2)² + 20 = 0, so (t − 2)² = 4, t = 0 or t = 4.

Watch it explained in under a minute

The same idea, narrated step by step. Square version for sharing: download the 1:1 video.

The four mistakes to avoid

Reading the sign of h the wrong way. (x + 1) means h = −1, so the vertex is at x = −1.
Reading k as the y-intercept. k is the height of the vertex. They match only when h = 0.
Forgetting that a negative a makes an arch, so the vertex is the highest point, not the lowest.
Dropping a when finding the landing spots. Move k across first, then divide by a.

Try it yourself

Tap a question to see the answer.

1. y = 3(x − 4)² + 5: find the vertex and the lowest value.

The vertex is (4, 5). Since a = 3 is positive it opens up, so the lowest value is 5.

2. y = −(x + 2)² + 9: find the vertex and the x-intercepts.

The vertex is (−2, 9). Set it to 0: (x + 2)² = 9, so x + 2 = 3 or −3: x = 1 or x = −5.

3. Write y = x² − 8x + 10 in vertex form.

h = 8 ÷ 2 = 4. k = 16 − 32 + 10 = −6. So y = (x − 4)² − 6. Check: x² − 8x + 16 − 6 = x² − 8x + 10 ✓.

Where this fits in

Vertex form shows the top. The next form shows where the parabola lands on the x-axis: roots from the factored form.