Math Core

Every quadratic function can be written in several equivalent ways, and each way answers a different question at a glance. In Algebra 1 you met standard form and vertex form. Here you'll add factored form, move freely between all three, and learn to pick the form that fits the job.

Three forms, one parabola

The three equations below all describe the same function. You can check by expanding the second and third.

f(x) = x^2 - 6x + 5 \qquad f(x) = (x - 3)^2 - 4 \qquad f(x) = (x - 1)(x - 5)

Each one puts a different feature of the graph on display.

Standard form $f(x) = x^2 - 6x + 5$ shows the y-intercept: $f(0) = 5$ .
Vertex form $f(x) = (x - 3)^2 - 4$ shows the vertex $(3, -4)$ and so the minimum value $-4$ .
Factored form $f(x) = (x - 1)(x - 5)$ shows the zeros: $f(x) = 0$ exactly when $x = 1$ or $x = 5$ .

One parabola, three readings: y-intercept 5 (standard form), vertex (3, -4) (vertex form), zeros 1 and 5 (factored form).Open in grapher →

The three forms of a quadratic function

form	equation	what you can read directly
standard	$f(x) = ax^2 + bx + c$	y-intercept $c$ ; axis $x = -\dfrac{b}{2a}$
vertex	$f(x) = a(x - h)^2 + k$	vertex $(h, k)$ ; max or min value $k$
factored	$f(x) = a(x - r)(x - s)$	zeros $r$ and $s$ ; axis $x = \dfrac{r + s}{2}$

In all three, the leading coefficient $a$ is the same number. It controls the direction ( $a > 0$ opens up, $a < 0$ opens down) and the width.

Because $a$ never changes from form to form, converting really means finding the other two pieces of information (the vertex, or the zeros).

Standard form to vertex form: completing the square

In Algebra 1 you completed the square when the leading coefficient was $1$ . When $a \ne 1$ , factor $a$ out of the $x$ -terms first, so the expression in parentheses starts with $x^2$ .

Take $f(x) = 2x^2 - 12x + 13$ .

\begin{aligned} f(x) &= 2(x^2 - 6x) + 13 \\ &= 2(x^2 - 6x + 9 - 9) + 13 \\ &= 2(x - 3)^2 - 18 + 13 \\ &= 2(x - 3)^2 - 5 \end{aligned}

Half of $-6$ is $-3$ , and $(-3)^2 = 9$ . Adding and subtracting $9$ inside the parentheses doesn't change the value. When the $-9$ comes out of the parentheses, it gets multiplied by the $2$ in front, which is where the $-18$ comes from.

Common mistake

The most common error is forgetting that the number you subtract is still inside the factored-out $a$ . In the example above, the $-9$ becomes $-18$ when it leaves the parentheses, not $-9$ . A quick check: substitute $x = 0$ into both forms. The original gives $13$ , and $2(0 - 3)^2 - 5 = 18 - 5 = 13$ . If the two numbers disagree, look at that step.

You can also skip completing the square and use the vertex formula: $h = -\dfrac{b}{2a}$ and $k = f(h)$ . For $2x^2 - 12x + 13$ , $h = \dfrac{12}{4} = 3$ and $k = 2(9) - 36 + 13 = -5$ , the same answer. Completing the square is worth knowing anyway, because it is exactly how the quadratic formula is built, and you'll use it again for circles.

Factored form and its symmetry

A parabola is symmetric about its axis, so its two zeros sit at equal distances on either side of it. That means the axis passes through the midpoint of the zeros.

Worked example: Vertex from factored form

Find the vertex of $g(x) = -(x + 1)(x - 7)$ .

The zeros are $x = -1$ and $x = 7$ . The axis is halfway between them:

h = \frac{-1 + 7}{2} = 3.

The vertex has $x = 3$ , so $k = g(3) = -(3 + 1)(3 - 7) = -(4)(-4) = 16$ . The vertex is $(3, 16)$ , and since $a = -1 < 0$ it is a maximum.

Not every quadratic has a factored form with real numbers. A parabola that never crosses the x-axis has no real zeros, so there are no real $r$ and $s$ to write down. (Later in this unit you'll see that such quadratics do factor if you allow complex numbers.)

Writing a quadratic from given information

Pick the form that uses the information you have, then solve for $a$ with one more point.

Given the vertex: start from $y = a(x - h)^2 + k$ .
Given the zeros: start from $y = a(x - r)(x - s)$ .
Given three random points: substitute each into $y = ax^2 + bx + c$ and solve the system for $a$ , $b$ , $c$ .

Worked example: From zeros and a point

A parabola has x-intercepts $-2$ and $4$ and a y-intercept of $-16$ . Write its equation in standard form.

The zeros give $y = a(x + 2)(x - 4)$ . The point $(0, -16)$ gives

-16 = a(0 + 2)(0 - 4) = -8a, \qquad\text{so}\qquad a = 2.

Then $y = 2(x + 2)(x - 4) = 2(x^2 - 2x - 8) = 2x^2 - 4x - 16$ .

Choosing the right form

In an application, the question tells you which form helps most. "When is it highest?" or "What is the minimum cost?" is a vertex question. "When does it hit the ground?" or "Where does the profit equal zero?" is a zeros question. "What was the starting value?" is a y-intercept question.

Worked example: A launched object

A ball is thrown upward from a platform. Its height in feet after $t$ seconds is $h(t) = -16t^2 + 64t + 5$ . Find the maximum height.

This is a vertex question. Complete the square:

\begin{aligned} h(t) &= -16(t^2 - 4t) + 5 \\ &= -16(t^2 - 4t + 4 - 4) + 5 \\ &= -16(t - 2)^2 + 64 + 5 \\ &= -16(t - 2)^2 + 69 \end{aligned}

The ball reaches its maximum height of $69$ feet after $2$ seconds. Notice the sign: $-16 \times (-4) = +64$ .

Tip

To check that two forms match, compare them at two or three convenient inputs such as $x = 0$ and $x = 1$ . Two different quadratics can agree at two points, but if they also have the same leading coefficient $a$ , agreeing at two points is enough to prove they are identical.

Practice

Practice 1

Which form of a quadratic function shows its zeros most directly?

Practice 2

Find the vertex of $f(x) = -2(x + 3)^2 + 7$ .

Enter a point like (2, -3)

Practice 3

Write $f(x) = x^2 + 10x + 18$ in vertex form $f(x) = (x - h)^2 + k$ . What is $k$ ?

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

Practice 4

Write $f(x) = 3x^2 - 18x + 20$ in vertex form. Enter the vertex $(h, k)$ .

Enter a point like (2, -3)

Practice 5

Find the vertex of $f(x) = 2(x + 1)(x - 7)$ .

Enter a point like (2, -3)

Practice 6

A parabola has zeros $-1$ and $5$ and passes through $(2, 18)$ . Write its equation in standard form $y = ax^2 + bx + c$ . What is $b$ ?

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

Practice 7

Which of these is not equivalent to $f(x) = -x^2 + 4x + 5$ ?

Practice 8

Find the maximum value of $f(x) = -3x^2 + 12x - 1$ .

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