A1.4 Sketch quadratic relations in vertex form | Ontario Grade 11 Mathematics

Ontario Grade 11 Mathematics

Mathematical Models

Use the vertex, direction, and a few points to draw a clear parabola

A quadratic relation can be shown as a curved graph called a parabola. In vertex form, its equation gives you a helpful starting point: the vertex, or turning point, is built into the equation. In this lesson, you will use that information to sketch a graph by hand. You will also use symmetry to choose points efficiently and check that the sketch matches the relation.

What you will learn

Identify the vertex and axis of symmetry from vertex form.
Use the coefficient to decide whether a parabola opens up or down and whether it is wider or narrower.
Choose and calculate useful points, then sketch and label the curve.

1. Prerequisite bridge: coordinates and squares

A coordinate is written as an ordered pair, such as

(2,5)

. The first number tells you how far to move horizontally from the origin; the second tells you how far to move vertically. To plot a point, find its horizontal position first, then its vertical position.

You also need to evaluate simple squares. For example, if

x=3

, then

(x-1)^2=(3-1)^2=4

. Do the subtraction inside the brackets before squaring. The square is never negative: for example,

(-2)^2=4

.

When you sketch a relation, you do not need to plot every possible point. You plot a few correct points and draw a smooth curve that follows them. A sketch should show the important shape and features, not just a collection of dots.

An ordered pair gives a horizontal position and a vertical position.
Evaluate brackets before squaring.
A sketch uses selected points to show the curve’s shape.

2. Read the vertex form

The standard vertex form is

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

. The numbers

h

and

k

give the vertex

(h,k)

. The vertex is the turning point: the lowest point when the parabola opens up, or the highest point when it opens down.

The axis of symmetry is a vertical line through the vertex. It divides the parabola into two matching halves. Its equation is

x=h

. Points the same horizontal distance to the left and right of this line have the same

y

-value.

The coefficient

a

controls the opening and the steepness. If

a

is positive, the parabola opens up. If

a

is negative, it opens down. If the size of

a

is greater than

1

, the graph is narrower than

y=x^2

. If the size of

a

is between

0

and

1

, it is wider than

y=x^2

. If

a=1

or

a=-1

, it has the same width as

y=x^2

.

Take care with signs in the brackets. In

y=2(x-3)^2+1

, the vertex is

(3,1)

. In

y=2(x+3)^2+1

, rewrite the bracket as

x-(-3)

: the vertex is

(-3,1)

. The sign inside the bracket is opposite the horizontal coordinate of the vertex.

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

Read the vertex as $(h,k)$ from $y=a(x-h)^2+k$ .
The axis of symmetry is $x=h$ .
The sign of $a$ gives the opening; its size compared with $1$ gives the width.

3. Build a sketch with points

Start by marking the vertex and drawing its axis of symmetry lightly. Then choose horizontal positions on both sides of the vertex. Equal distances from the axis make matching pairs of points, which helps you check your arithmetic and your sketch.

Substitute each chosen

x

-value into the relation to calculate

y

. Begin with nearby values, often one and two units from the vertex. If the coefficient is a fraction or the graph is especially wide, choose distances that make the calculations manageable. Plot the points, then draw a smooth curve through them and the vertex.

The curve should turn at the vertex, match the opening direction, and have two sides that mirror each other across the axis. Label the vertex and axis. A rough sketch does not need a perfectly scaled curve, but it must agree with the values and features you have found.

Use equal horizontal distances on either side of the axis.
Calculate and plot points before drawing the curve.
Check the vertex, opening, width, and symmetry.

4. Apply the method to a context

A quadratic relation in vertex form can describe how a quantity changes as the input changes. For example, the height of an object over time may rise to a highest point and then fall. The vertex identifies that highest point when the graph opens down. The same sketching steps apply: identify the vertex and opening, select useful input values, calculate outputs, and draw a smooth curve.

The graph shows the relation across the values you choose to display. A sketch is not a claim that every possible input is practical in a real situation. Focus on the relation’s shape and the part that makes sense for the context, when a context is given.

The vertex can describe a maximum or minimum value in a practical relation.
Use the same graphing steps whether the relation is abstract or contextual.

Points for Example 1

Input $x$	Output $y$	Point
$1$	$-3$	$(1,-3)$
$0$	$-1$	$(0,-1)$
$2$	$-1$	$(2,-1)$
$-1$	$5$	$(-1,5)$
$3$	$5$	$(3,5)$

Worked example

Example 1: An upward-opening parabola

Sketch

y=2(x-1)^2-3

. Identify its vertex, axis of symmetry, opening, and relative width, then plot enough points to support the sketch.

Read the key features
Compare the relation with vertex form. The vertex is $(1,-3)$ , so the axis of symmetry is $x=1$ . Since $a=2$ is positive, the parabola opens up. Its width is narrower than $y=x^2$ because the size of $a$ is greater than $1$ .
$y=2(x-1)^2-3$
Choose matching inputs
Use inputs one and two units from $x=1$ . Matching distances on the two sides should give matching outputs because the graph is symmetric about its axis.
$x=0,2,-1,3$
Calculate the outputs
Substitute each input into the relation. For example, at $x=0$ , the bracket is $-1$ , whose square is $1$ , so $y=2(1)-3=-1$ . The other values give the points shown below.
$y(0)=-1,\quad y(2)=-1,\quad y(-1)=5,\quad y(3)=5$
Plot and sketch
Plot the vertex $(1,-3)$ and the four calculated points. Draw a smooth, upward-opening curve through them. The pairs $(0,-1)$ and $(2,-1)$ , and $(-1,5)$ and $(3,5)$ , are equally spaced from the axis and have matching heights.
$(1,-3),\ (0,-1),\ (2,-1),\ (-1,5),\ (3,5)$

Answer: The sketch has vertex

(1,-3)

, axis

x=1

, and opens up. It is narrower than

y=x^2

.

Check: The two pairs of points have equal heights and equal distances from

x=1

, as symmetry requires.

Worked example

Example 2: A downward-opening parabola

Sketch

y=-\frac{1}{2}(x+2)^2+4

. Identify its vertex, axis of symmetry, opening, and relative width, then plot useful points.

Read the key features
The bracket $x+2$ can be written as $x-(-2)$ , so the vertex is $(-2,4)$ . The axis is $x=-2$ . The coefficient is negative, so the graph opens down. Its size is less than $1$ , so it is wider than $y=x^2$ .
$y=-\frac{1}{2}(x+2)^2+4$
Choose matching inputs
Choose values one and two units to either side of the vertex’s horizontal coordinate, $-2$ . This gives two pairs of inputs that should produce matching heights.
$x=-1,-3,0,-4$
Calculate the outputs
At $x=-1$ , the bracket is $1$ , so the output is $4-\frac{1}{2}=3.5$ . At $x=0$ , the bracket is $2$ , whose square is $4$ , so the output is $4-2=2$ . The matching inputs on the other side give the same outputs.
$y(-1)=3.5,\quad y(-3)=3.5,\quad y(0)=2,\quad y(-4)=2$
Plot and sketch
Plot the vertex and four calculated points. Draw a smooth curve that turns at $(-2,4)$ and opens down. Check that the paired points sit at equal heights on opposite sides of $x=-2$ .
$(-2,4),\ (-1,3.5),\ (-3,3.5),\ (0,2),\ (-4,2)$

Answer: The sketch has vertex

(-2,4)

, axis

x=-2

, and opens down. It is wider than

y=x^2

.

Check: The matching pairs are equally spaced from

x=-2

and have equal outputs. The vertex is the highest plotted point.

Common mistakes and how to avoid them

Reading the sign inside the brackets as the vertex’s horizontal coordinate.

Correction: Rewrite the bracket as

x-h

. For example,

x+2=x-(-2)

, so the horizontal coordinate is

-2

.

Treating a negative coefficient as a negative width rather than a downward opening.

Correction: The sign tells the opening direction. The size of the coefficient, ignoring its sign, tells whether the graph is wider or narrower than

y=x^2

.

Plotting only one point on each side and drawing an uneven curve.

Correction: Use matching distances from the axis of symmetry. Check that each pair has equal outputs before drawing.

Lesson summary

In $y=a(x-h)^2+k$ , the vertex is $(h,k)$ .
The axis of symmetry is $x=h$ .
A positive $a$ opens the graph up; a negative $a$ opens it down.
Compare the size of $a$ with $1$ to judge width.
Plot the vertex and symmetric points, then draw a smooth curve.

Check your understanding

Question 1

For

y=3(x+1)^2-2

, which description is correct?

Vertex $(-1,-2)$ ; axis $x=-1$ ; opens up and is narrower than $y=x^2$ .
Vertex $(1,-2)$ ; axis $x=1$ ; opens up and is narrower than $y=x^2$ .
Vertex $(-1,-2)$ ; axis $x=-1$ ; opens down and is wider than $y=x^2$ .
Vertex $(-1,2)$ ; axis $x=-1$ ; opens up and is narrower than $y=x^2$ .

Show answer and explanation

Vertex

(-1,-2)

; axis

x=-1

; opens up and is narrower than

y=x^2

.

Since

x+1=x-(-1)

, the vertex is

(-1,-2)

. The positive coefficient opens up, and its size,

3

, is greater than

1

, so the graph is narrower.

Question 2

For

y=-\frac{1}{4}(x-2)^2+5

, what output matches the input

x=4

?

$4$
$5$
$6$
$-4$

Show answer and explanation

4

Substitute

4

: the bracket is

2

, and its square is

4

. Then

y=-\frac{1}{4}(4)+5=-1+5=4

.

Question 3

A sketch of

y=-2(x-3)^2+1

should have which feature?

A vertex at $(3,1)$ and a downward opening.
A vertex at $(-3,1)$ and an upward opening.
A vertex at $(3,-1)$ and a downward opening.
A vertex at $(3,1)$ and an upward opening.

Show answer and explanation

A vertex at

(3,1)

and a downward opening.

The vertex is

(3,1)

. The coefficient

-2

is negative, so the graph opens down.

Key terms

Quadratic relation: A relation whose graph is a parabola and whose variable is squared.
Vertex: The turning point of a parabola; its highest or lowest point.
Axis of symmetry: The line that divides a parabola into two matching halves.
Sketch: A drawn representation that shows a graph’s important features and shape, without needing exact scale.

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About this lesson

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Mathematics (MBF3C), expectation A1.4. It is a study resource, not an official curriculum publication.

Official curriculum reference

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A1.4 · Sketch quadratic relations in vertex form

Ontario Grade 11 Mathematics

Use the vertex, direction, and a few points to draw a clear parabola

What you will learn

1. Prerequisite bridge: coordinates and squares

2. Read the vertex form

3. Build a sketch with points

4. Apply the method to a context

Points for Example 1

Example 1: An upward-opening parabola

Example 2: A downward-opening parabola

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through MBF3C

About this lesson