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Characteristics of Quadratic Functions

Read a quadratic graph from its leading coefficient, vertex, symmetry axis, intercepts, range, and worked graphing examples.

Shape of Quadratic Function Graphs

The graph of a quadratic function is a parabola. The leading coefficient $a$ controls its opening direction and how narrow or wide it appears.

Influence of the Leading Coefficient on Graph Shape

When the Leading Coefficient Is Positive

If $a > 0$ , the graph of the quadratic function will open upward. This means the graph has a minimum point.

Examples of functions with $a > 0$ :

$f(x) = x^2$ , the simplest example with $a = 1$
$a$ , an example with $a = 2$

The sign of $a$ controls the opening direction. Its magnitude controls the width: compared with $y=x^2$ , the graph is narrower when $|a|>1$ and wider when $0<|a|<1$ .

Quadratic Function Graph with

a > 0

Graph opens upward and has a minimum point.

When the Leading Coefficient Is Negative

If $a < 0$ , the graph of the quadratic function will open downward. This means the graph has a maximum point.

Examples of functions with $a < 0$ :

$f(x) = -x^2$ with $a$
$f(x) = -3x^2 - 12x - 15$ with $a = -3$

Quadratic Function Graph with

a < 0

Graph opens downward and has a maximum point.

Why the Leading Coefficient Cannot Be Zero

When $a = 0$ , the function form becomes $f(x) = bx + c$ . This is no longer a quadratic function, but a linear function. A quadratic function must have $a \neq 0$ so that the highest power of the variable $x$ is $2$ .

Key Points and Lines on the Parabola

Vertex

The vertex is the highest point (if $a < 0$ ) or the lowest point (if $a > 0$ ) on the graph. The coordinates of the vertex are expressed as $(-\frac{b}{2a}, f(-\frac{b}{2a}))$ .

Axis of Symmetry

The axis of symmetry is a vertical line that divides the parabola into two symmetrical parts. The equation of the axis of symmetry is $x = -\frac{b}{2a}$ .

Vertical Axis Intercept

The $y$ -intercept is obtained when $x = 0$ . Its value is $f(0) = c$ .

Horizontal Axis Intercepts

The intercepts on the $x$ -axis are obtained when $f(x) = 0$ , that is, when $ax^2 + bx + c = 0$ . The solutions can be found using the formula:

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Steps to Graph a Quadratic Function

Determine whether the parabola opens upward ( $a > 0$ ) or downward ( $a < 0$ ).
Calculate the coordinates of the vertex $(-\frac{b}{2a}, f(-\frac{b}{2a}))$ .
Calculate the $y$ -intercept: $(0, c)$ .
Calculate the intercepts on the $x$ -axis, if any.
Choose several other $x$ values and calculate their corresponding $y$ values.
Plot all points in the coordinate system.
Connect the points with a parabolic curve.

Drawing Quadratic Function Graphs

Upward Opening Parabola Example

Graph the function $f(x) = x^2 - 2x - 3$ :

Coefficient $(-\frac{b}{2a}, f(-\frac{b}{2a}))$ , so the parabola opens upward.
Vertex:
$x = -\frac{b}{2a} = -\frac{-2}{2 \cdot 1} = 1$
So the vertex is at $(1, -4)$ .
$y$ -intercept:

$x = -\frac{b}{2a} = -\frac{-2}{2 \cdot 1} = 1$
Intercepts on the $x$ -axis: $f(x) = 0$ or $x^2 - 2x - 3 = 0$
Calculate some additional points:
$f(-2) = (-2)^2 - 2 \cdot (-2) - 3 = 4 + 4 - 3 = 5$

Graph of

f(x) = x^2 - 2x - 3

Parabola opens upward with vertex at

x^2 - 2x - 3 = 0

and intercepts on the

x

-axis at

(-1, 0)

and

(3, 0)

.

Downward Opening Parabola Example

Graph the function $f(x) = -x^2$ :

Coefficient $a = -1 < 0$ , so the parabola opens downward.
Vertex:
$x = -\frac{b}{2a} = -\frac{0}{2 \cdot (-1)} = 0$
So the vertex is at $(0, 0)$ .
$y$ -intercept:

$f(0) = 0$

So the $y$ -intercept is at $(0, 0)$ .
Intercepts on the $x$ -axis: $f(x) = 0$ or $-x^2 = 0$
Calculate some additional points:
$f(-2) = -((-2)^2) = -4$

Graph of

f(x) = -x^2

Parabola opens downward with vertex at

(0, 0)

.

Opening Direction Vertex and Range

Write the vertex as $(x_v,y_v)$ , where $y_v=f(x_v)$ .

Condition	Opening	Vertex value	Range
$a > 0$	Upward	Minimum	$[y_v,\infty)$
$a < 0$	Downward	Maximum	$(-\infty,y_v]$

Practice Problem

Analyze the graph of $g(x)=-2(x+1)^2+8$ . Find its opening direction, vertex, symmetry axis, intercepts, and range.

Worked Solution

The function is already in vertex form $a(x-h)^2+k$ . Here $a=-2$ , $h=-1$ , and $k=8$ .

Because $a<0$ , the parabola opens downward.
The vertex is $(-1,8)$ , so its maximum value is $8$ .
The symmetry axis is $x=-1$ .

For the $y$ -axis intercept, set $x=0$ :

g(0)=-2(0+1)^2+8=6

The graph crosses the $y$ -axis at $(0,6)$ . For the $x$ -axis intercepts, solve $g(x)=0$ :

-2(x+1)^2+8=0

The intercepts are $(-3,0)$ and $(1,0)$ . Since the parabola opens downward from the maximum $8$ , its range is $(-\infty,8]$ .

Types of Quadratic Equation Roots

Constructing Quadratic Functions

Characteristics of Quadratic FunctionsRead a quadratic graph from its leading coefficient, vertex, symmetry axis, intercepts, range, and worked graphing examples.

On this page

Shape of Quadratic Function Graphs
Influence of the Leading Coefficient on Graph Shape
Key Points and Lines on the Parabola
Steps to Graph a Quadratic Function
Drawing Quadratic Function Graphs
- Upward Opening Parabola Example
- Downward Opening Parabola Example
Opening Direction Vertex and Range
Practice Problem
- Worked Solution

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Characteristics of Quadratic Functions

Published: Apr 19, 2025. Updated: Sep 8, 2026.

So the $y$ -intercept is at $(0, -3)$ .

Using the quadratic formula:

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{2 \pm \sqrt{4 + 12}}{2} = \frac{2 \pm \sqrt{16}}{2} = \frac{2 \pm 4}{2}

So the intercepts on the $x$ -axis are $(-1, 0)$ and $(3, 0)$ .

f(2) = 2^2 - 2 \cdot 2 - 3 = 4 - 4 - 3 = -3

x^2 = 0

x = 0

So the $x$ -intercept is at $(0, 0)$ .

f(1) = -(1^2) = -1

f(2) = -(2^2) = -4