Vertex form puts the turning point of a parabola right in the equation. Here is how to read it, how to convert to and from standard form, and how to find the vertex.
What is vertex form?
Vertex form is a way of writing a quadratic as y = a(x − h)² + k, where (h, k) is the vertex of the parabola. The number a is the same a as in standard form and controls how wide the parabola is and which way it opens.
Standard form, y = ax² + bx + c, is good for reading the y-intercept. Vertex form is better for graphing, because the turning point and the axis of symmetry x = h are right there in the equation.
The value of a works the same in both forms. If a is bigger than 1 (or less than −1), the parabola is narrower than y = x². If a is between −1 and 1, it's wider. A negative a flips it upside down.
How do you read the vertex from vertex form?
The vertex is (h, k), but watch the sign on h. The form has x minus h, so y = 3(x − 4)² + 1 has vertex (4, 1), and y = (x + 2)² − 3 has vertex (−2, −3).
A quick way to think of it: h is the x value that makes the bracket zero. For (x + 2), that's x = −2. The value of k is what's left of y at that point.
Sometimes h or k is hidden because it's zero. In y = x² − 7 there's no bracket, so h = 0 and the vertex is (0, −7). In y = −2(x − 5)² there's no number added at the end, so k = 0 and the vertex is (5, 0), sitting right on the x-axis.
How do you convert standard form to vertex form?
Factor a out of the x terms, complete the square inside, and move the extra constant out (multiplied by a). Here is y = 2x² − 8x + 3.
- Factor 2 out of the x terms: y = 2(x² − 4x) + 3.
- Half of −4 is −2, and (−2)² = 4. Add and subtract 4 inside: y = 2(x² − 4x + 4 − 4) + 3.
- Bring the −4 out, multiplied by 2: y = 2(x² − 4x + 4) − 8 + 3.
- Write the square and combine: y = 2(x − 2)² − 5.
Is there a faster way to get vertex form?
Yes. Find h with h = −b/2a, plug h into the original equation to get k, and drop both into a(x − h)² + k. The a value doesn't change.
For y = 2x² − 8x + 3, h = −(−8)/(2 × 2) = 8/4 = 2. Then k = 2(2)² − 8(2) + 3 = 8 − 16 + 3 = −5. So y = 2(x − 2)² − 5, the same answer as completing the square.
How do you convert vertex form to standard form?
Expand the squared binomial, multiply by a, and add k. Take y = 3(x + 1)² − 4.
How do you find the vertex of a parabola?
If the equation is in vertex form, read (h, k) directly. If it's in standard form, use x = −b/2a for the x coordinate and substitute to get y, or complete the square.
There's a third route when the roots are easy. The vertex sits halfway between the two x-intercepts. For y = −x² + 6x − 5, the roots are x = 1 and x = 5, so the vertex is at x = 3, and y = −9 + 18 − 5 = 4. In vertex form that's y = −(x − 3)² + 4.
Is the vertex a maximum or a minimum?
If a is positive, the parabola opens up and the vertex is the minimum point. If a is negative, it opens down and the vertex is the maximum.
In y = 2(x − 2)² − 5, the lowest y value is −5, reached at x = 2. In y = −(x − 3)² + 4, the highest y value is 4, at x = 3. That's why vertex form is the go-to form for "maximum height" and "minimum cost" word problems.
The question usually asks for one of two things. If it wants the largest or smallest value, give k. If it wants the input that produces that value, give h. Reading the question twice before answering saves a lot of lost marks here.
How do you write vertex form from a vertex and a point?
Put h and k into y = a(x − h)² + k, substitute the other point for x and y, and solve for a. Say the vertex is (1, −2) and the parabola passes through (3, 6).
How do you graph a parabola from vertex form?
Plot the vertex, draw the axis of symmetry through it, then add a few points on one side and mirror them across the axis. Vertex form gives you the first two for free.
- Start with y = (x − 1)² − 4. The vertex is (1, −4).
- Draw the axis of symmetry, the vertical line x = 1.
- Find the y-intercept by setting x = 0: (0 − 1)² − 4 = −3, so (0, −3).
- Mirror it across x = 1 to get (2, −3).
- Find the x-intercepts by setting y = 0: (x − 1)² = 4, so x − 1 = ±2, giving x = 3 and x = −1.
- Sketch a smooth U through the points. It opens up because a = 1 is positive.
What are the most common vertex form mistakes?
Almost all of them are sign or distribution errors. These are the ones worth checking each time.
- Reading h with the wrong sign. In (x + 2)², h is −2.
- Forgetting to multiply the subtracted constant by a when completing the square.
- Changing a during the conversion. The leading coefficient stays the same in both forms.
- Expanding (x − h)² as x² − h² instead of x² − 2hx + h².
Frequently asked questions
What is the vertex form equation?
The vertex form of a quadratic is y = a(x − h)² + k. The point (h, k) is the vertex, x = h is the axis of symmetry, and a is the same leading coefficient as in standard form.
How do you find h and k from standard form?
Use h = −b/2a, then substitute h into the equation to find k. For y = 2x² − 8x + 3, h = 2 and k = −5, so the vertex is (2, −5).
Why is it x − h and not x + h?
Shifting a graph right by h replaces x with x − h. So a vertex at x = 4 gives (x − 4)², and a vertex at x = −2 gives (x − (−2))², which simplifies to (x + 2)².
Can you find the x-intercepts from vertex form?
Yes. Set y = 0 and solve for x. For y = (x + 2)² − 3, (x + 2)² = 3, so x = −2 ± √3.
How do you convert vertex form to standard form quickly?
Square the binomial, multiply by a, then add k. For y = −2(x − 5)², that gives −2(x² − 10x + 25) = −2x² + 20x − 50.
What is the axis of symmetry in vertex form?
It is the vertical line x = h. For y = 2(x − 2)² − 5, the axis of symmetry is x = 2, and the parabola is a mirror image on either side of that line.