Sketch a quadratic from standard form and identify key features

Ontario Grade 11 Mathematics

Quadratic Functions

Use the equation to find the shape, position, and key features of a parabola.

A quadratic graph is a U-shaped curve called a parabola. Its equation contains clues about where the graph sits, which way it opens, and where it crosses the axes. In this lesson, standard form means

y=ax^2+bx+c

, where

a

,

b

, and

c

are numbers and

a\ne 0

. You will use those clues to make a clear sketch and name its key features.

What you will learn

Recognize a quadratic in standard form and describe the role of its coefficients.
Find the vertex, axis of symmetry, and direction of opening from standard form.
Find intercepts when they can be read or calculated simply, then use key points to sketch the graph.
Identify the domain and range of a quadratic from its graph.

1. Prerequisite bridge: coordinates and intercepts

A point on a graph is written as an ordered pair, such as

(2,3)

. The first number is the

x

-coordinate, and the second is the

y

-coordinate. Substituting an

x

-value into an equation gives the matching

y

-value.

The

y

-intercept is where a graph crosses the

y

-axis. At that point,

x=0

. The

x

-intercepts are where the graph crosses the

x

-axis. At those points,

y=0

.

A quadratic has an

x^2

term and makes a curved graph. Its graph is symmetric: the two sides mirror each other across a vertical line. That line is called the axis of symmetry.

To find the $y$ -intercept, substitute $x=0$ .
To find $x$ -intercepts, set $y=0$ and solve when possible.
A quadratic graph is a parabola with a vertical axis of symmetry.

2. Read the shape and location from standard form

In standard form,

y=ax^2+bx+c

, the value of

a

controls the direction and width of the parabola. If

a>0

, the parabola opens upward and has a lowest point. If

a<0

, it opens downward and has a highest point. A larger value of |a| makes it narrower than the basic graph

y=x^2

; a value of |a|

between

0 and

1

makes it wider.

The constant

c

gives the

y

-intercept because setting

x=0

leaves

y=c

. The coefficient

b

helps determine the axis of symmetry and the vertex. The vertex is the turning point of the parabola: its lowest point when it opens upward, or its highest point when it opens downward.

For standard form, the

x

-coordinate of the vertex is

-b/(2a)

. Substitute that

x

-coordinate into the equation to find the vertex's

y

-coordinate. The axis of symmetry is the vertical line through the vertex. It is written

x=h

when the vertex is

(h,k)

.

To sketch, plot the vertex and

y

-intercept. If useful, find the

x

-intercepts by setting

y=0

. Use symmetry to add a point on the opposite side of the axis, at the same horizontal distance from it. Draw a smooth curve through the points.

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

The vertex is $(h,k)$ , where $h=-b/(2a)$ and $k$ is the value of $y$ at $x=h$ .
The axis of symmetry is $x=h$ .
The domain of any quadratic is all real numbers.
The range starts at the vertex's $y$ -value and extends upward or downward according to the opening direction.

3. Connect features to a sketch

A reliable sketch does not need many points. Start by marking the vertex and drawing the axis of symmetry as a light vertical guide. Then mark the

y

-intercept. The vertex and intercepts give the graph's main location and shape.

When the equation has easy factors, the

x

-intercepts can be found by setting the equation equal to zero and factoring. For example, if

y=(x-1)(x-5)

, then

y=0

when

x=1

or

x=5

. If the equation does not factor simply, a sketch can still be made from the vertex, the

y

-intercept, and additional substituted points.

The curve must mirror across its axis. For example, if one known point is two units to the left of the axis, its matching point is two units to the right and has the same

y

-value. This symmetry helps make the sketch balanced.

Check the sketch against the equation. The opening must match the sign of

a

, the graph must cross the

y

-axis at

(0,c)

, and the vertex must lie on the axis of symmetry.

Plot key points before drawing the curve.
Use equal distances from the axis to create matching points.
Label the vertex, axis of symmetry, and intercepts on the sketch.

4. Use the graph to describe its range

The domain describes which

x

-values are allowed. A vertical parabola continues left and right without stopping, so its domain is all real numbers.

The range describes which

y

-values appear on the graph. For an upward-opening parabola with vertex

(h,k)

, the smallest output is

k

, so the range is

y\ge k

. For a downward-opening parabola, the largest output is

k

, so the range is

y\le k

.

These features describe the graph, not just the equation. A complete answer to a sketching question should show the curve and clearly identify the vertex, axis, intercepts when found, opening direction, domain, and range.

Upward opening: the vertex gives the minimum $y$ -value.
Downward opening: the vertex gives the maximum $y$ -value.
Use the vertex and opening direction to state the range.

Feature clues in standard form

Feature	How to find or read it	What to show on the sketch
Opening direction	Check whether $a$ is positive or negative	Curve opens upward if $a>0$ ; downward if $a<0$
Vertex	Find $h=-b/(2a)$ , then calculate $y$ at $x=h$	Mark $(h,k)$
Axis of symmetry	Use the vertex's $x$ -coordinate	Draw or label $x=h$
$y$ -intercept	Set $x=0$ ; the result is $c$	Mark $(0,c)$
Range	Use the vertex height and opening direction	Upward: $y\ge k$ ; downward: $y\le k$

Worked example

Sketch a quadratic and identify its features

Sketch

y=x^2-4x+3

and identify its vertex, axis of symmetry, intercepts, opening direction, domain, and range.

Read the coefficients
Compare the equation with $y=ax^2+bx+c$ . Here, $a=1$ , $b=-4$ , and $c=3$ . Since $a$ is positive, the parabola opens upward.
$a=1,\quad b=-4,\quad c=3$
Find the vertex and axis
The vertex's horizontal coordinate is $-b/(2a)$ . Substituting the coefficients gives $x=2$ . Put $x=2$ into the equation to find $y=-1$ , so the vertex is $(2,-1)$ . The axis of symmetry is the vertical line through the vertex.
$h=-\frac{-4}{2(1)}=2,\quad y(2)=2^2-4(2)+3=-1$
Find the intercepts
At the $y$ -intercept, $x=0$ , so $y=3$ . For the $x$ -intercepts, set $y=0$ and factor. The graph crosses the $x$ -axis at $x=1$ and $x=3$ .
$y(0)=3,\quad x^2-4x+3=(x-1)(x-3)$
Plot points and sketch
Plot the vertex $(2,-1)$ and the intercepts $(0,3)$ , $(1,0)$ , and $(3,0)$ . The axis is $x=2$ . The point $(0,3)$ is two units left of the axis, so its matching point is $(4,3)$ . Draw a smooth, upward-opening parabola through the points.
$(0,3),\quad (1,0),\quad (2,-1),\quad (3,0),\quad (4,3)$
State the domain and range
The parabola continues in both horizontal directions, so its domain is all real numbers. It opens upward and has a lowest point at $y=-1$ , so its range is $y\ge -1$ . x∈ $\mathbb{R}$ , y≥ -1

Answer: The graph opens upward, has vertex

(2,-1)

, and has axis of symmetry

x=2

. Its

x

-intercepts are

(1,0)

and

(3,0)

, and its

y

-intercept is

(0,3)

. Its domain is all real numbers and its range is

y\ge -1

.

Check: The two

x

-intercepts are equally spaced around

x=2

, and the points

(0,3)

and

(4,3)

have equal heights on opposite sides of the axis. These checks agree with the symmetry of the parabola.

Common mistakes and how to avoid them

Using

b/(2a)

as the vertex's

x

-coordinate.

Correction: Include the negative sign: use

-b/(2a)

.

Calling

c

the

x

-intercept.

Correction: The value

c

gives the

y

-intercept

(0,c)

. Find

x

-intercepts by setting

y=0

.

Drawing the two sides at different heights for points equally far from the axis.

Correction: A parabola is symmetric. Matching points on opposite sides of its axis have the same

y

-value.

Giving the range without checking the opening direction.

Correction: For an upward-opening graph, the vertex is the minimum. For a downward-opening graph, it is the maximum.

Lesson summary

Write the equation as $y=ax^2+bx+c$ and use the sign of $a$ to determine its opening direction.
Find the vertex with $h=-b/(2a)$ and substitute $h$ into the equation to find $k$ .
The axis of symmetry is $x=h$ , and the $y$ -intercept is $(0,c)$ .
Find $x$ -intercepts by setting $y=0$ when possible. Plot key points and use symmetry to draw the parabola.
The domain is all real numbers. The vertex and opening direction determine the range.

Check your understanding

Question 1

For

y=-2x^2+8x-5

, which statement gives the vertex and opening direction?

Vertex $(2,3)$ ; opens downward
Vertex $(-2,3)$ ; opens downward
Vertex $(2,-5)$ ; opens upward
Vertex $(4,-5)$ ; opens downward

Show answer and explanation

Vertex

(2,3)

; opens downward

Here

a=-2

and

b=8

, so the vertex's

x

-coordinate is

-8/(2\cdot -2)=2

. Substitution gives

y=3

. Since

a<0

, the graph opens downward.

Question 2

What is the

y

-intercept of

y=3x^2+2x-7

?

$(0,3)$
$(0,2)$
$(0,-7)$
$(-7,0)$

Show answer and explanation

(0,-7)

Set

x=0

. Then

y=-7

, so the

y

-intercept is

(0,-7)

.

Question 3

A parabola opens downward and has vertex

(1,4)

. Which statement gives its range?

$y\ge 4$
$y\le 4$
$x\le 4$
All real $y$ -values

Show answer and explanation

y\le 4

A downward-opening parabola has its highest point at the vertex. Its

y

-values are therefore

4

or less.

Key terms

Quadratic: An equation whose highest power of the variable is $2$ , such as $y=ax^2+bx+c$ with $a\ne 0$ .
Parabola: The U-shaped curve made by the graph of a quadratic.
Vertex: The turning point of a parabola; its minimum or maximum point.
Axis of symmetry: The vertical line that divides a parabola into matching left and right sides.
Intercept: A point where a graph crosses an axis.
Domain: The set of allowed $x$ -values for a graph.
Range: The set of $y$ -values a graph reaches.

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