Quadratic Graphs and Their Forms, SAT math lesson

Throw a ball and its path is a parabola. Different questions about that path need different numbers, such as where it starts, where it lands and how high it gets. The same quadratic can be written in three equivalent ways, and each one puts a different feature in plain view. Much of the work on SAT questions about quadratics is choosing the right form.

See also The Quadratic Function and Its Graph in Foundations for Calculus and Quadratic Functions and Vertex Form in Precalculus.

Key terms

Parabola

Definition The graph of a quadratic function $f(x) = ax^2 + bx + c$ with real constants $a$ , $b$ , $c$ and $a \ne 0$ is a parabola. It opens upward when $a > 0$ and downward when $a < 0$ . The larger $\lvert a \rvert$ is, the narrower the parabola.

In plain words A parabola is a U shape. A positive leading coefficient makes a cup that holds water, and a negative one turns it upside down.

Example Compare $y = x^2$ and $y = -2x^2 + 3$ .

$x$	$-2$	$-1$	$0$	$1$	$2$
$x^2$	$4$	$1$	$0$	$1$	$4$
$-2x^2 + 3$	$-5$	$1$	$3$	$1$	$-5$

The first opens upward from $(0, 0)$ . The second has $a = -2$ , so it opens downward from $(0, 3)$ and falls away faster.

Vertex

Definition The vertex of a parabola is its turning point. For $f(x) = ax^2 + bx + c$ with $a \ne 0$ its $x$ -coordinate is

h = -\frac{b}{2a},

and its $y$ -coordinate is $f(h)$ .

If $a > 0$ , the vertex gives the minimum value of $f$ over all real $x$ . If $a < 0$ , it gives the maximum. When $x$ is restricted to a closed interval, the minimum or maximum on that interval is at the vertex only if the vertex lies in the interval. Otherwise it is at an endpoint, so compare the values at the endpoints and at the vertex if it is inside.

In plain words The vertex is the bottom of the cup or the top of the arch. If a question limits the inputs, check whether that bottom or top is even allowed.

Example The function $f(x) = x^2 - 6x + 5$ has $a = 1$ and $b = -6$ , so the vertex is at $x = 3$ . Since $f(3) = 9 - 18 + 5 = -4$ , the vertex is $(3, -4)$ . Over all real numbers the minimum value is $-4$ . On the interval $4 \le x \le 7$ the vertex is not available, and $f$ is increasing there, so the minimum is $f(4) = -3$ and the maximum is $f(7) = 12$ .

Axis of symmetry

Definition The axis of symmetry of a parabola with vertex $(h, k)$ is the vertical line $x = h$ . For every real $d$ ,

f(h + d) = f(h - d),

so points at the same horizontal distance from the axis have the same height. If the parabola has two $x$ -intercepts $r$ and $s$ , the axis is halfway between them, at $x = \frac{r + s}{2}$ .

In plain words Fold the graph along the axis and the two halves land on each other.

Example For $f(x) = x^2 - 6x + 5$ the axis is $x = 3$ . Since $0$ and $6$ are both $3$ units from the axis, $f(0) = f(6)$ , and indeed both equal $5$ . The $x$ -intercepts $1$ and $5$ also have midpoint $3$ .

Intercepts

Definition The $y$ -intercept of $y = f(x)$ is the point $(0, f(0))$ , and for $f(x) = ax^2 + bx + c$ it is $(0, c)$ . An $x$ -intercept is a point $(r, 0)$ where $r$ is a real zero of $f$ , that is, a real solution of $f(x) = 0$ . A parabola has exactly one $y$ -intercept and two, one or no $x$ -intercepts.

In plain words The $y$ -intercept is where the graph crosses the vertical axis, so set $x = 0$ . The $x$ -intercepts are where it meets the horizontal axis, so set $y = 0$ .

Example For $y = x^2 - 6x + 5$ the $y$ -intercept is $(0, 5)$ . Setting $y = 0$ gives $(x - 1)(x - 5) = 0$ , so the $x$ -intercepts are $(1, 0)$ and $(5, 0)$ .

Standard form

Definition A quadratic function is in standard form when written as

f(x) = ax^2 + bx + c, \qquad a \ne 0.

This form shows the opening direction (the sign of $a$ ) and the $y$ -intercept $(0, c)$ directly, and it gives the vertex through $x = -\frac{b}{2a}$ .

In plain words Standard form is the expanded version. It is the easiest to evaluate at $x = 0$ and the one you need for the discriminant.

Example For $f(x) = -2x^2 + 8x - 3$ the parabola opens downward because $a = -2 < 0$ , and it crosses the $y$ -axis at $(0, -3)$ . The vertex is at $x = -\frac{8}{2(-2)} = 2$ , with $f(2) = -8 + 16 - 3 = 5$ , so the maximum value is $5$ .

Factored form

Definition A quadratic function with real zeros $r$ and $s$ (possibly equal) can be written in factored form

f(x) = a(x - r)(x - s), \qquad a \ne 0.

By the zero product property its zeros are exactly $r$ and $s$ , so the $x$ -intercepts are $(r, 0)$ and $(s, 0)$ . A quadratic with no real zeros has no factored form of this kind with real numbers.

In plain words Factored form puts the $x$ -intercepts on display. Just flip the sign of each number inside the parentheses.

Example The function $f(x) = x^2 - 6x + 5$ factors as $(x - 1)(x - 5)$ , so its zeros are $1$ and $5$ . The vertex is midway, at $x = 3$ , where $f(3) = (2)(-2) = -4$ . The vertex did not move when the form changed, only the way it is found.

Vertex form

Definition A quadratic function is in vertex form when written as

f(x) = a(x - h)^2 + k, \qquad a \ne 0.

Its vertex is $(h, k)$ and its axis of symmetry is $x = h$ . Watch the sign inside the square. The form $(x + 4)^2$ means $(x - (-4))^2$ , so $h = -4$ . The value $k$ is the minimum of $f$ when $a > 0$ and the maximum when $a < 0$ . Completing the square turns standard form into vertex form.

In plain words Vertex form tells you where the parabola turns and how high or low it gets, which is what most “maximum” and “minimum” questions want.

Example A ball is tossed so its height after $t$ seconds is $H(t) = -5t^2 + 20t + 1$ meters. Completing the square,

\begin{aligned} H(t) &= -5(t^2 - 4t) + 1 \\[0.4em] &= -5(t - 2)^2 + 20 + 1 \\[0.4em] &= -5(t - 2)^2 + 21. \end{aligned}

So the ball reaches its greatest height of $21$ meters at $t = 2$ seconds. If the question only asked about the first second, $0 \le t \le 1$ , the vertex would be outside that interval. The height is increasing on that interval, so the greatest height there would be $H(1) = 16$ meters.

Check yourself

Which statement about the graph of $y = -3(x + 2)^2 + 7$ in the $xy$ -plane is true? (A) It opens upward and has vertex $(-2, 7)$ . (B) It opens downward and has vertex $(-2, 7)$ . (C) It opens downward and has vertex $(2, 7)$ . (D) It opens downward and has vertex $(-2, -7)$ .

Show the answer

Choice (B) is correct. The leading coefficient $-3$ is negative, so the parabola opens downward, which rules out (A). Writing $(x + 2)^2$ as $(x - (-2))^2$ shows $h = -2$ , and the constant outside is $k = 7$ . Choice (C) misreads the sign inside the square, and choice (D) changes the sign of $k$ for no reason.

The function $f$ is defined by $f(x) = (x + 1)(x - 7)$ . What is the minimum value of $f(x)$ ?

Show the answer

The answer is $-16$ . The zeros are $-1$ and $7$ , so the axis of symmetry is at their midpoint, $x = 3$ . The parabola opens upward, so the minimum is $f(3) = (4)(-4) = -16$ . A common slip is to answer $3$ , which is where the minimum occurs, not its value.

The function $f(x) = x^2 - 8x + 12$ is written in four equivalent forms below. Which form displays the minimum value of $f$ as a constant or coefficient? (A) $f(x) = (x - 2)(x - 6)$ , (B) $f(x) = x(x - 8) + 12$ , (C) $f(x) = (x - 4)^2 - 4$ , (D) $f(x) = (x - 6)^2 + 4x - 24$

Show the answer

Choice (C) is correct. It is vertex form with vertex $(4, -4)$ , and since the parabola opens upward the minimum value is $-4$ . Choice (A) displays the zeros $2$ and $6$ . Choice (B) displays the $y$ -intercept $12$ , and choice (D) shows no feature of the graph directly. All four expand to $x^2 - 8x + 12$ .

Quadratic Graphs and Their Forms

Theory, part 53

Practice, part 54

Key terms

Parabola

Vertex

Axis of symmetry

Intercepts

Standard form

Factored form

Vertex form

Check yourself

How this lesson connects

Builds on

Used later in

Resources for this lesson

Function transformation explorer