Math Core

Standard form $y = ax^2 + bx + c$ hides the vertex: you need a formula and some arithmetic to find it. Vertex form puts the vertex right in the equation, so you can read the turning point, the axis of symmetry and the maximum or minimum at a glance, and you can write the equation of a parabola straight from its graph.

Sliding the parent function

Start with $y = x^2$ , whose vertex is $(0, 0)$ . Now compare it with $y = (x - 3)^2 + 1$ .

The expression $(x - 3)^2$ is smallest, namely $0$ , when $x = 3$ . So the lowest point has moved from $x = 0$ to $x = 3$ .
At that point, $y = 0 + 1 = 1$ . Every output is $1$ more than it would have been.

So the whole graph has slid $3$ units right and $1$ unit up, and the new vertex is $(3, 1)$ . The shape is unchanged.

y = (x - 3)² + 1 is the parent parabola shifted 3 units right and 1 unit up.Open in grapher →

Vertex form

A quadratic function in vertex form is

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

The vertex is $(h, k)$ and the axis of symmetry is $x = h$ .
$a$ plays the same role as in standard form: $a > 0$ opens up, $a < 0$ opens down, and a larger $|a|$ is narrower.
The minimum (if $a > 0$ ) or maximum (if $a < 0$ ) value of the function is $k$ .

Reading the vertex

The form has a minus sign built in: $x - h$ . When the equation shows a plus sign, rewrite it as subtracting a negative. For example,

y = (x + 2)^2 - 5 = \big(x - (-2)\big)^2 + (-5),

so $h = -2$ , $k = -5$ , and the vertex is $(-2, -5)$ .

Common mistake

The sign of $h$ is the most common mistake with vertex form. In $y = (x + 2)^2 - 5$ the vertex is $(-2, -5)$ , not $(2, -5)$ . Ask yourself: what value of $x$ makes the squared part zero? Here $x + 2 = 0$ when $x = -2$ . The $k$ value keeps its sign as written.

Worked example: Reading and sketching

Describe the graph of $y = 2(x + 1)^2 - 8$ and find its intercepts.

The vertex is $(-1, -8)$ , and $a = 2 > 0$ , so the parabola opens up, is narrower than $y = x^2$ , and has a minimum value of $-8$ . The axis of symmetry is $x = -1$ .

For the y-intercept, set $x = 0$ : $y = 2(1)^2 - 8 = -6$ .

For the x-intercepts, set $y = 0$ :

2(x + 1)^2 = 8 \quad\Longrightarrow\quad (x + 1)^2 = 4.

The numbers whose square is $4$ are $2$ and $-2$ , so $x + 1 = 2$ or $x + 1 = -2$ , giving $x = 1$ or $x = -3$ . They sit $2$ units on either side of the axis, as symmetry says they should.

y = 2(x + 1)² - 8: vertex (-1, -8), x-intercepts -3 and 1, y-intercept -6.Open in grapher →

Writing an equation from a graph

To write the equation of a parabola, you need its vertex and one other point. The vertex gives $h$ and $k$ ; the other point lets you solve for $a$ .

Worked example: Vertex plus one point

A parabola has vertex $(2, -3)$ and passes through $(4, 5)$ . Write its equation in vertex form.

Substitute the vertex: $y = a(x - 2)^2 - 3$ . Now use the point $(4, 5)$ :

\begin{aligned} 5 &= a(4 - 2)^2 - 3 \\ 5 &= 4a - 3 \\ 8 &= 4a \\ a &= 2 \end{aligned}

The equation is $y = 2(x - 2)^2 - 3$ .

From vertex form to standard form

To convert to standard form, expand the square and simplify. You practiced squaring binomials in the special products lesson: $(x - h)^2 = x^2 - 2hx + h^2$ .

Worked example: Expanding

Write $y = -(x - 4)^2 + 9$ in standard form.

\begin{aligned} y &= -(x^2 - 8x + 16) + 9 \\ &= -x^2 + 8x - 16 + 9 \\ &= -x^2 + 8x - 7 \end{aligned}

Check with the vertex formula: $x = -\dfrac{8}{2(-1)} = 4$ , which matches $h = 4$ .

From standard form to vertex form

Going the other way, the value of $a$ is the same in both forms. So you only need the vertex, and you already know how to find it: $h = -\dfrac{b}{2a}$ , and $k$ is the function's value at $h$ . (The next lessons show a second method, completing the square.)

Worked example: Using the vertex formula

Write $y = 3x^2 - 12x + 7$ in vertex form.

Here $a = 3$ . The vertex has

h = -\frac{-12}{2(3)} = 2, \qquad k = 3(2)^2 - 12(2) + 7 = 12 - 24 + 7 = -5.

So $y = 3(x - 2)^2 - 5$ .

Tip

To check a conversion, expand your vertex form and compare it with the original. Here $3(x - 2)^2 - 5 = 3(x^2 - 4x + 4) - 5 = 3x^2 - 12x + 7$ . It matches. A quicker spot check is to substitute one value, such as $x = 0$ , into both forms.

Practice

Practice 1

What is the vertex of $y = (x - 5)^2 + 2$ ?

Practice 2

Find the vertex of $y = -3(x + 4)^2 + 1$ .

Enter a point like (2, -3)

Practice 3

What is the minimum value of $f(x) = 2(x - 1)^2 + 6$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 4

Which equation matches the graph?

y = -(x - 2)^2 + 3(2, 3)Open in grapher →

Practice 5

A parabola has vertex $(-1, 4)$ and passes through $(1, -4)$ . Its equation is $y = a(x + 1)^2 + 4$ . Find $a$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 6

Write $y = (x + 3)^2 - 10$ in standard form $y = ax^2 + bx + c$ . What is $c$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

When $y = 2x^2 + 12x + 11$ is written in vertex form $y = a(x - h)^2 + k$ , what are $h$ and $k$ ? Enter them as $(h, k)$ .

Enter a point like (2, -3)

Practice 8

The graph of $y = x^2$ is shifted $4$ units left and $6$ units down. What is the y-intercept of the new parabola?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.