Vertex Form of a Quadratic

Standard form $y = ax^2 + bx + c$ is useful, but it hides the vertex behind a formula. Vertex form puts the vertex front and center so you can read it directly from the equation. If you need to know where the peak or valley of a parabola sits — for a bridge arch, a profit curve, or a projectile path — vertex form is the fastest way to get there.

This page covers what vertex form is, how to read it, how to convert to and from standard form, and why it matters in practice.

What Is Vertex Form?

The vertex form of a quadratic function is:

$y = a(x - h)^2 + k$

where:

• $(h, k)$ is the vertex of the parabola
• $a$ controls the direction and width (same as in standard form)

That is the entire point of vertex form: the vertex $(h, k)$ is visible right in the equation. No formula needed.

Reading the Vertex

Be careful with the sign of $h$. The form is $y = a(x - h)^2 + k$, with a minus sign before $h$. This means:

• $y = 2(x - 3)^2 + 5$ has vertex $(3, 5)$ — the $h$ is positive 3
• $y = 2(x + 4)^2 - 1$ has vertex $(-4, -1)$ — because $(x + 4) = (x - (-4))$, so $h = -4$
• $y = -(x - 0)^2 + 7 = -x^2 + 7$ has vertex $(0, 7)$

Example 1: Identify the vertex of $y = -3(x + 2)^2 + 9$

Rewrite the expression inside the parentheses:

$y = -3(x - (-2))^2 + 9$

So $h = -2$ and $k = 9$.

Vertex: $(-2, 9)$

Since $a = -3 < 0$, the parabola opens downward. The vertex is the maximum point.

Example 2: Identify the vertex of $y = \frac{1}{2}(x - 5)^2 - 4$

Reading directly: $h = 5$, $k = -4$.

Vertex: $(5, -4)$

Since $a = 1/2 > 0$, the parabola opens upward. The vertex is the minimum point. The parabola is also wider than $y = x^2$ because $|a| < 1$.

Converting from Standard Form to Vertex Form

The key technique is completing the square. If you have already studied that topic, this is a direct application. Here is the step-by-step process.

Given $y = ax^2 + bx + c$:

Step 1 — Factor $a$ out of the first two terms (leave $c$ outside):

$y = a\!\left(x^2 + \frac{b}{a}x\right) + c$

Step 2 — Complete the square inside the parentheses. Take half the coefficient of $x$, square it, and add/subtract it:

$y = a\!\left(x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2\right) + c$

Step 3 — Rewrite the perfect square trinomial and distribute the $-\left(\frac{b}{2a}\right)^2$ outside:

$y = a\!\left(x + \frac{b}{2a}\right)^2 + c - \frac{b^2}{4a}$

Now you have vertex form with $h = -\frac{b}{2a}$ and $k = c - \frac{b^2}{4a}$.

Let us see this with concrete numbers.

Example 3: Convert $y = 2x^2 - 12x + 14$ to vertex form

Step 1 — Factor out $a = 2$ from the $x$ terms:

$y = 2(x^2 - 6x) + 14$

Step 2 — Complete the square. Half of $-6$ is $-3$. Squaring: $(-3)^2 = 9$.

$y = 2(x^2 - 6x + 9 - 9) + 14$

Step 3 — Rewrite:

$y = 2\!\left((x - 3)^2 - 9\right) + 14 = 2(x - 3)^2 - 18 + 14$

$y = 2(x - 3)^2 - 4$

Vertex form: $y = 2(x - 3)^2 - 4$

Vertex: $(3, -4)$

Check: The vertex $x$-coordinate should be $-b/(2a) = 12/4 = 3$, and $y(3) = 2(9) - 36 + 14 = -4$. Confirmed.

Example 4: Convert $y = -x^2 + 8x - 10$ to vertex form

Step 1 — Factor out $a = -1$:

$y = -(x^2 - 8x) - 10$

Step 2 — Complete the square. Half of $-8$ is $-4$. Squaring: $(-4)^2 = 16$.

$y = -(x^2 - 8x + 16 - 16) - 10$

Step 3 — Rewrite:

$y = -\!\left((x - 4)^2 - 16\right) - 10 = -(x - 4)^2 + 16 - 10$

$y = -(x - 4)^2 + 6$

Vertex form: $y = -(x - 4)^2 + 6$

Vertex: $(4, 6)$

Example 5: Convert $y = 3x^2 + 6x + 1$ to vertex form

Step 1 — Factor out $a = 3$:

$y = 3(x^2 + 2x) + 1$

Step 2 — Half of $2$ is $1$. Squaring: $1^2 = 1$.

$y = 3(x^2 + 2x + 1 - 1) + 1$

Step 3 — Rewrite:

$y = 3(x + 1)^2 - 3 + 1 = 3(x + 1)^2 - 2$

Vertex form: $y = 3(x + 1)^2 - 2$

Vertex: $(-1, -2)$

Converting from Vertex Form to Standard Form

Going the other direction is straightforward — just expand the squared term and simplify.

Example 6: Convert $y = 4(x - 1)^2 + 3$ to standard form

Step 1 — Expand $(x - 1)^2$:

$(x - 1)^2 = x^2 - 2x + 1$

Step 2 — Multiply by $a = 4$:

$4(x^2 - 2x + 1) = 4x^2 - 8x + 4$

Step 3 — Add $k = 3$:

$y = 4x^2 - 8x + 4 + 3 = 4x^2 - 8x + 7$

Standard form: $y = 4x^2 - 8x + 7$

Example 7: Convert $y = -2(x + 3)^2 + 5$ to standard form

Step 1 — Expand $(x + 3)^2$:

$(x + 3)^2 = x^2 + 6x + 9$

Step 2 — Multiply by $a = -2$:

$-2(x^2 + 6x + 9) = -2x^2 - 12x - 18$

Step 3 — Add $k = 5$:

$y = -2x^2 - 12x - 18 + 5 = -2x^2 - 12x - 13$

Standard form: $y = -2x^2 - 12x - 13$

Why Use Vertex Form?

Vertex form is especially useful when you need to:

1. Read the vertex instantly — no calculation required
2. Graph quickly — you know the vertex and direction immediately, then plot a few symmetric points
3. Find the maximum or minimum — the $k$-value is the answer, and $h$ tells you where it occurs
4. Write an equation from a graph — if you can identify the vertex and one other point, you can determine $a$

Example 8: Write the equation from a graph

A parabola has its vertex at $(2, -1)$ and passes through the point $(4, 7)$. Find the equation in vertex form.

Step 1 — Start with vertex form: $y = a(x - 2)^2 + (-1) = a(x - 2)^2 - 1$

Step 2 — Substitute the known point $(4, 7)$:

$7 = a(4 - 2)^2 - 1 = 4a - 1$

$8 = 4a \implies a = 2$

Equation: $y = 2(x - 2)^2 - 1$

Check: At $x = 2$: $y = 2(0) - 1 = -1$. At $x = 4$: $y = 2(4) - 1 = 7$. Confirmed.

Real-World Application: Carpentry — Designing a Parabolic Arch

A carpenter is building a decorative arch over a doorway. The doorway is 3 feet wide (from $x = 0$ to $x = 3$ feet), and the arch should peak at 1.5 feet above the top of the door frame, centered over the doorway.

The vertex is at the center of the doorway at the peak height: $(1.5, 1.5)$. The arch must touch the top of the frame at both sides, meaning $y = 0$ at $x = 0$ and $x = 3$.

Step 1 — Use vertex form with vertex $(1.5, 1.5)$:

$y = a(x - 1.5)^2 + 1.5$

Step 2 — Substitute the point $(0, 0)$:

$0 = a(0 - 1.5)^2 + 1.5 = 2.25a + 1.5$

$-1.5 = 2.25a \implies a = -\frac{1.5}{2.25} = -\frac{2}{3}$

Equation:

$y = -\frac{2}{3}(x - 1.5)^2 + 1.5$

Verify at $x = 3$:

$y = -\frac{2}{3}(3 - 1.5)^2 + 1.5 = -\frac{2}{3}(2.25) + 1.5 = -1.5 + 1.5 = 0 \checkmark$

Now the carpenter can calculate the arch height at any point. For instance, at $x = 0.5$ feet from the left edge:

$y = -\frac{2}{3}(0.5 - 1.5)^2 + 1.5 = -\frac{2}{3}(1) + 1.5 = -0.667 + 1.5 = 0.833 \text{ feet}$

The arch is about 10 inches high at that point. These measurements let the carpenter cut a precise template for the arch.

Comparing the Three Forms

Quadratic functions have three common forms. Each reveals different information at a glance:

Being able to convert between forms is a core algebra skill. Completing the square takes you from standard to vertex form. Expanding takes you from vertex to standard form. Factoring (when possible) takes you from standard to factored form.

Common Mistakes to Avoid

1. Sign error on $h$. In $y = 2(x + 3)^2 - 1$, the vertex is $(-3, -1)$, not $(3, -1)$. The form has a minus sign: $(x - h)$. If the equation shows $(x + 3)$, then $h = -3$.
2. Forgetting to distribute $a$ after completing the square. When you add and subtract the completing-the-square constant inside parentheses that are multiplied by $a$, the subtracted part must be multiplied by $a$ when brought outside. In Example 3, the $-9$ inside $2(\ldots)$ became $-18$ outside.
3. Mixing up $h$ and $k$. The vertex is $(h, k)$ where $h$ is horizontal and $k$ is vertical. Double-check which value goes with $x$ and which is the constant added at the end.
4. Expanding incorrectly. When going from vertex to standard form, remember that $(x - h)^2 = x^2 - 2hx + h^2$. The middle term has a factor of $2$.
5. Forgetting the $+k$ when converting. After completing the square and distributing $a$, do not forget to add the original constant $c$ (or equivalently, the $k$ value at the end).

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Key Takeaways

• Vertex form $y = a(x - h)^2 + k$ shows the vertex $(h, k)$ directly — no formula needed
• Watch the sign: $(x + 3)$ means $h = -3$, not $h = 3$
• Convert from standard form by completing the square: factor out $a$, add/subtract the perfect square constant, rewrite
• Convert to standard form by expanding the squared binomial and simplifying
• Vertex form is ideal for graphing, optimization, and writing equations from graphs
• All three forms (standard, vertex, factored) describe the same parabola — each form just makes different features easy to read

Return to Algebra 1 for more topics in this section.