Module 2: Algebra - Open School BCmedia.openschool.bc.ca/ocrmedia/pmath11/pdf/pma11_… · · 2015-06-22Module 2: Algebra Section 1: Quadratic Functions ... Module 2, Section 1

Module 2: Algebra

Section 1: Quadratic Functions

Introduction 3

Lesson 1 Relations and Functions 5

Lesson 2 Linear and Quadratic Functions 15

Lesson 3 Quadratic Functions: y = ax2 and y = ax2 + k 29

Lesson 4 Quadratic Functions: y = a(x – h)2 43

Lesson 5 Quadratic Functions: y = a(x – h)2 + k 51

Lesson 6 Completing the Square 59

Lesson 7 Special Features of the Quadratic Function 69

Lesson 8 Problems Involving Quadratic Functions 81

Review 87

Section Assignment 2.1 91

Section 2: Solving Equations

Introduction 103

Lesson 1 Solving Quadratic Equations Graphically 105

Lesson 2 Solving Quadratic Equations by Factoring 117

Lesson 3 Solving Quadratic Equations by Completingthe Square and by the Quadratic Formula 123

Lesson 4 The Nature of Roots 133

Lesson 5 Radical Equations 141

Lesson 6 Rational and Absolute Value Equations 147

Review 157

Section Assignment 2.2 161

Module 2 Answer Key 171

Principles of Mathematics 11 Contents 1

2 Contents Principles of Mathematics 11

Module 2, Section 1Quadratic Functions

IntroductionIn your previous mathematics course, you studied linearfunctions of the form y = mx + b. You learned about x- and y-intercepts of the function and the domain and range of thefunction.

After a brief review of these concepts, you will study, in detail,another important function that is quadratic in nature. You willlearn about its characteristics and how to use it to modelmathematical situations.

Section 1 — Outline

Lesson 1 Relations and Functions

Lesson 2 Linear and Quadratic Functions

Lesson 3 Quadratic Functions: y = ax2 and y = ax2 + k

Lesson 4 Quadratic Functions: y = a(x – h)2

Lesson 5 Quadratic Functions: y = a(x – h)2

+ k

Lesson 6 Completing the Square

Lesson 7 Special Features of the Quadratic Function

Lesson 8 Problems Involving Quadratic Functions

Review

Principles of Mathematics 11 Section 1, Introduction 3

Module 2

Notes

4 Section 1, Inroduction Principles of Mathematics 11

Module 2

Lesson 1

Relations and Functions

Outcomes

When you complete this lesson, you will be able to

• define a relation and a function

• state the domain and range of a given relation or function

• represent a relation or function ina) graph mode

b) rule mode

c) ordered pair mode

• determine whether a relation is or is not a function

• use functional notation to evaluate a function

Overview

In Principles of Mathematics 10 you studied the linear relationy = mx + b or f(x) = mx + b where m is the slope of the line andb is the y-intercept. The slope (m) of its graph shows you thatchanges in the x-value give proportional changes in the y-value.

The rate of change of y with respect to x is always constant. Inthe diagram, for every 1 unit of change in x, y changes 2 units.

In mathematics if two quantities are related in such a way thatthe value of the one quantity determines the value of the secondquantity, you have a mathematical model of a relation.

Principles of Mathematics 11 Section 1, Lesson 1 5

Module 2

12

1 2

12

y

x

•

•

•

Domain and RangeDe�nitionA relation is a set of ordered pairs (x, y).

The first components of the ordered pairs are called the elementsof the domain of the relation. The second components are theelements of the range of the relation.

Example 1Given the relation A = {(2, 1), (3, 2), (4, 3)}

Domain of A: {2, 3, 4}

Range of A: {1, 2, 3}

Example 2There are infinitely manypoints on the straightline. There are no limitson the x-coordinates orthe y-coordinates of thepoints on the line. SoDomain = Range = theset of all real numbers(written as R ).

Example 3

This relation also consists ofa finite number of points butthere are limits on the xand y coordinates. Thehorizontal span (Domain) isfrom -4 to 1, so we writeDomain

Similarly, Range =

The vertical line means “such that.”

A special type of relation is the function.

{ } { }− ≤ ≤ − ≤ ≤1 3 or | 1 3y y y

{ }{ }

= − ≤ ≤

= − ≤ ≤

4 1 or

| 4 1

x

x x

6 Section 1, Lesson 1 Principles of Mathematics 11

Module 2

4

3

y

x

3

-1-4

y

x

The Function

De�nition

A function is a set of ordered pairs (x, y) such that for eachvalue of x there is exactly one value of y.

Examples 1. Is the relation B = {(1, 4), (6, 2), (5, 3), (4, –4)} a function?

2. Is the relation C = {(1, 6), (4, 10), (1, 4), (2, –2)} a function?

3. Is the relation given by y2

= x a function?

4. Is the relation given by y = x2 a function?

5.

6.

Solution

1. Yes, the relation is a function because each of the firstcomponents is different: 1, 6, 5, 4 are all different.

2. No, the first component 1 is associated with 6 and 4 ratherthan with only one second component.

3. y2 = x can be rewritten as When relations aredesribed using an equation, it’s sometimes easier todetermine whether or not they’re functions by having a lookat some of the ordered pairs in the relation. We’ll do thatnow.

.y x= ±


Module 2

3

-3

-5

y

x5

3

-1-4

y

x

x –4 –1 0 1 4

y undefined undefined

Ordered NA NA (0,0) (1,1),(1,-1) (4,2),(4,–2)Pairs

We can see clearly that there is duplication in the x-coordinates,so this is not a function.

4. A quick look at some of the ordered pairs in this relation...

x –2 –1 0 1 2

y 4 –1 0 1 4

Ordered (–2,4) (–1,–1) (0,0) (1,1) (2,4)Pair

... shows no duplication in the x-values. This is a true function.

5. In this relation we can see that the ordered pairs (1,3) and(1,–1) are both elements of the relation. Hence, this is not afunction.

6. Most points on this figure (called an ellipse) have another pointabove or below them. Consequently, they have the same x-coordinate [for example (0,3) and (0,–3)]. Therefore this is not afunction.

Relations and functions can be represented in five main ways:

1. A set of ordered pairs, using “Set Notation.”

{(1,3), (2,5), (3,7), (4,9)}

2. Ordered pairs in a table of values

3. Mapping diagram

input x (domain) output y (range)

1234

3579

f

x 1 2 3 4

y 3 5 7 9

= −± 4 2, 2= −± 1 1, 1 =± 0 0

11 scitamehtaM fo selpicnirP1 nosseL ,1 noitceS8

Module 2

The arrowhead points to the range element associated withthe domain element of a function f.

4. Graph

5. Equation or rule

y = 2x + 1 where x ∈ {1, 2, 3, 4}

In functional notation, this rule would be written asf(x) = 2x + 1.

You can write f(1) = 3 which means if you substitute 1 intothe rule f, the answer is 3.

This notation is used for functions only. f(x) is pronounced “f of x.”

Function Notation

Examples1. Suppose f is defined using function notation as follows:

f(x) = 2x – 5. This, of course, means that y = 2x – 5 and ...

( ) ( )

( ) ( )( ) ( )

( ) ( )

− = − − = −

= − = −

= − = −

= − = −

= − = −2 2 2

2 2 2 5 9

1 12 5 4

2 2

0 2 0 5 5

3 2 3 5 6 5

2 5 2 5

f

f

f

f x x x

f x x x

1 2 3 4x

y

1

2

3

4

5

6

7

8

9

l

l

l

l


Module 2

2. If another function, g, is defined graphically as shown below,we can still determine functional values.

g(–2) = ?

Solution

Locate the point on the curve with x-coordinate –2. This pointis (–2,0). So g(–2) = 0.

Similarly g(0) = 3, g(2) = 0, g(4) = –1, and g(6) = 0.

3. Suppose h(x) = x2 – 2x + 1 and g(x) = 3 – x.

Determine a) h(g(–3)) and b) g(h(–3))

Solution

a) To determine h(g(–3)), first find (g(–3).

g(–3) = 3 – (–3) = 6 h(g(–3)) = h(6) = 62 – 2(6) + 1 = 25

b) To determine g(h(–3)), first find h(–3).

h(–3) = (–3)2 – 2(–3) + 1 = 9 + 6 + 1 = 16

g(h(–3)) = g(16) = 3 – 16 = –13

Now it’s your turn to try a few exercises on your own. Be sure tocorrect your answers using the Answer Key immediatelyfollowing this Lesson.


Module 2

⇒

⇒

1

–1–1

y = g(x)

x 1

Self-Marking Activity 1. Determine which of the following relations are functions:

a) A = {(2, 1) (2, 3) (2, 5) (2, 6)}

b) B = {(2, 1) (3, 1) (4, 1) (5, 1)}

c) y = 2x + 3

d) y = x2 + 2

2. It is possible to write an equation for a relation given a set ofordered pairs. For example,

From the first point it could be y = 3x or y = 2x + 1. Both ofthese would produce a y-value of 3 for an x-value of 1. Fromthe second point, y = 3x does not work but y = 2x + 1 doesbecause 2(2) + 1 = 5. Since y = 2x + 1 works for points threeand four as well, y = 2x + 1 is the relation represented bythose points. Now try writing an equation for each set ofordered pairs below.

a)

b)

c)x 2 4 6 8

y 1 2 3 4

x 1 2 3 4

y 1 4 9 16

x 1 2 3 4

y 3 6 9 12

x 1 2 3 4

y 3 5 7 9


Module 2

d)

3. If a relation is presented in the graphical mode, you candetermine whether it is, or is not, a function by using thevertical line test.

Vertical Line Test

If any vertical line intersects the graph of a relation in morethan one point then the graph does not represent a function.

For example:

a)

For your information, the equation is x = y2 – 3. It is not afunction because you can draw a vertical line whichintersects the graph in two places. The line x = 1 (shownas a dotted line) is one such vertical line.

123

4

5

1 2 3 4 5 61234561

2

3

4

56

6

l

l

l

l

l

l

l

x 1 2 3 4

y 1 3 5 7


Module 2

b)

y = x2 – 3 is a function since no vertical line will intersectthe graph in more than one point.

3. Use the vertical line test to determine which relations arefunctions.

a) b)

c) d)

3

-3

3

3-3

3

123

45

1 2 3 4 5 612345612

345

67

ll

ll

ll

l


Module 2

4. Given the function f = {(–1, 3) (2, 6) (5, –3) (7, 7)}, find thevalue of

a) f(2) b) f( 5) c) f(–1) d) f(4)

5. Given the graph of the function h, find the value of

a) h(0) b) h( –1) c) h(2)

6. Given the function f(x) = 3 – x, find the value of

a) f(0) b) f( –1) c) f(3) d) f(t)

7. If f(x) = 2x2 – 3x – 1, find

a) f(–1) b) c) f(a)

d) f(2x) e) f(x – 1) f) f(f(1))

8. If f(x) = x2 – 2 and g(x) = –2x, find the value of

a) f(–3) b) g( –2) c) f(g(3)) d) g(f(3))

9. Find the domain and range of the relations in question 1.

10. Find the domain and range of the relations in question 3.

Check your answers in the Module 2 Answer Key.

f 2

x

y

1 21

11

11 scitamehtaM fo selpicnirP1 nosseL ,1 noitceS41

Module 2

( )

Module 2

Lesson 2Linear and Quadratic Functions

OutcomesWhen you complete this lesson, you will be able to

• identify linear functions

• find the slope and the intercepts of a linear function

• write the equation or rule of a linear function from its graph• recognize vertical and horizontal lines and write their

equations

• plot and describe data of quadratic form

• recognize functions of quadratic form

• find the vertex, domain and range, axis of symmetry,intercepts, zeros, and maximum/minimum values of aquadratic function

The Linear Function—a ReviewIn Principles of Mathematics 10, you studied linear functions.A linear function is a function whose graph is a straight line.Its form is f(x) = mx + b. Remember a straight line equation isof the form y = mx + b.

Remember on a line, equal changes in the x-value give equalchanges in the y-value. The rate of change of y with respect to xis a constant called the slope (m).

1 0

1 0

0

y

xover 1, up 1

and so on

over 1, up 1over 1, up 1


Module 2

Example 1

Graph the following linear function using the “slope-y intercept”method. f(x) = –3x + 4

Solution

Step 1: y = –3x + 4 has y intercept 4, so we plot the point (0,4).

Step 2: Since the slope is –3 (or –3/1), to get to the next point werun 1 unit to the right and 3 units down. That takes us to(1,1). We can repeat the process, obtaining the third point(2,–2).

Step 3: Join the points with a straight line and you have thedesired graph.

You are also asked to find the x- and y-intercepts of the function.

De�nitions

1. The x-intercept of a function is the x-coordinate(s) of thepoint(s) where the graph of the function intersects the x-axis.

2. The y-intercept of a function is the y-coordinate of the point(s)where the graph of the function intersects the y-axis.

Note: The y-intercept is symbolized using the letter b, wherey = mx + b.

Example 2

Find the x- and y-intercepts of the function f(x) = –2x + 3. Thenuse these to graph the function.


y

x

(0,4)

(2,–2)

1

3

3

1(1,1)

Solution

To find the x-intercept let f(x) = 0 since f(x) is always 0 on the x-axis. Then solve for x.

To find the y-intercept, let x = 0 since the x-coordinate is alwayszero on the y-axis. Then solve for y.

Its graph would be:

When you are given the graph of a linear function, you shouldbe able to write its rule or equation.

Example 3

Find the equation of the straight line that passes through thepoints (0, 2) and (–1, 0).Solution

Its slope (m) = and its y-intercept (b) is 2.2 0

0 12

−− −

=

(y-intercept)

y x

y

y

= − += − +=

2 3

2 0 3

3

(x-intercept)

− + =− = −

=

2 3 02 3

32

x

x

x

Senior 3 Pre-Calculus Mathematics Module 1, Lesson 2 17

Module 2

x

y

2

21

1

3

( )

( )

The equation of the linearfunction is f(x) = 2x + 2.

Sometimes the answer isalso written as {(x, 2x + 2)}.

Example 4A special case of the linear function results when the slope of thefunction is 0.

In this diagram, b = 2 and the slope is 0. Find its equation.

Solution

The equation is f(x) = 2 or y = 2.

Some points on the line are (1,2),(2,2), (–1,2).

Example 5

A vertical straight line is said to have a slope which is undefined.

The equation of the line is x = 2. Is this a function or a relation?

Solution

This is not a linear function but is a relation.

Some points on the line are (2,1),(2,3), (2,0), (2,–1).

The Quadratic FunctionNot every pattern in the world around you is linear. By studyingquadratic functions and relations, you greatly expand theapplications that you can model in science and engineering,business and industry, and in many other fields. Parabolas aregraphs of a family of functions known as quadratic functions. Aquadratic function has a squared term.

Remember that if you multiply two linear functions together youget a squared term.


Module 2

x

y

2

1

x

y

2

x

y

2

Example:x(x + 3) = x2 + 3x

DefinitionsA quadratic function is a function that can be written in theform f(x) = ax2 + bx + c, where a, b, and c are real number constantsand a ≠ 0.

A parabola is the graph of a quadratic function. Quadraticfunctions can be written in various forms

Example 1Here are six examples of quadratic functions. For each of the sixexamples,determine a, b, and c. Remember that a is the coefficientof x2, b is the coefficient of x, and c is the constant.

Solution

a b c

a) 2 1 1

b) 1 0 0

c) 3 0 –1

d) 0 2

e) –1 7 0

f) –1 1 0

12

a)

b)

c)

d)

e)

f)

f x x x

f x x

x x

f x x

g x x x

x x x

:

,

,

→ + +=

−

= +

= −

−

2 1

3 1

212

7

2

2

2

2

2

2

b gd io tb gb gd io t

f x ax bx c a

f x ax bx c a

Q x ax bx c a

b g

n s

= + + ≠

→ + + ≠

= + + ≠

2

2

2

0

0

0

:

( , )


Module 2

Example 2Explain whether the following are quadratic functions. In eachcase, x is the domain variable.

Solutiona) Yes, a = 3, b = 0, and c = 0.b) Yes. Re-arrange to get y = –2x2 + 4 or f(x) = –2x2 + 4 with

a = –2, b = 0, and c = 4.c) Since there is no squared term, it is not quadratic.d) The relation y2 = x is not a quadratic function because y is

squared rather than x.

In the following example, you will graph a quadratic functionwhere a = 1, b = 0, and c = 0.

Example 3Given the function f(x) = x2, draw the graph of f for –3 ≤ x ≤ 3.(That is, for values of x between –3 and 3.)

Solution

From the table of values,the points are plotted. Asmooth curve is drawnthrough the points.

1

2

3

4

5

1 2 3 4 5 6−5−6

6

7

8

9

10

x

y

−4 −3 −2 −1−2

−1

y f x= b gx f(x) = x2

–3 9–2 4–1 10 01 12 43 9

a)

b)

c)

d)

f x x

x y

x x

y x

b g

b gm r

=

+ =

+

=

3

12

2

2 5

2

2

2

,


Module 2

From the above graph of f(x) = x2, note the following:• The domain is –3 ≤ x ≤ 3.• The range is 0 ≤ y ≤ 9. If the domain of f was x ∈ R

(means x is a real number) then the range would be y ≥ 0,y ∈ R. The next diagram shows that scenario.

• The turning point of a parabola is called the vertex. In thisexample the vertex is (0, 0).

• The parabola opens upward. When the parabola opensupward, its minimum y-value is at its vertex. The functionis said to have a minimum value of 0. The minimumpoint is the vertex when the parabola opens up.

• A vertical line passing through the vertex is the axis ofsymmetry. This line divides a parabola into twosymmetrical parts, which are mirror images of each other.Every point on the parabola except its vertex has areflection image with respect to the line of symmetry. Inthe example, the y-axis is the axis of symmetry. Itsequation is x = 0. The points (–2,4) and (2,4) are reflectionsof each other, through the axis of symmetry x = 0.

• The zeros of a quadratic function are the value(s) of x whichmake the quadratic function zero. These values of x are calledthe x-intercepts of the graph and occur where the graphcrosses or touches the x-axis. In this example, the quadraticfunction is zero when x = 0.

�

�

�

x

y

axis of

symmetry

vertex

reflection

image of

(–2, 4)

reflection

image of

(2, 4)

(–2, 4) (2, 4)

(0, 0)


Module 2

Example 4Draw the graph of g(x) = –x2 + 6x – 5, –1 ≤ x ≤ 7

Write the domain, range, coordinate of the vertex, the equationof the axis of symmetry, x-intercept(s), y-intercept(s), and zeros.

SolutionTable of values

Domain: –1 ≤ x ≤ 7, x ∈ RRange: –12 ≤ y ≤ 4, y ∈ RVertex: (3, 4) You can see the vertex in the graph or the tableof values.Equation of axis of symmetry: x = 3x-intercepts: 1, 5y-intercept: –5 Zeros: 1, 5

This function is said to have a maximum value when y = 4. The graph has a maximum point when the graph opensdownward.

1

2

3

4

5

1 2 3 4 5 612341

2

6

7

8

9

10

x

y

7 8 9

3

4

1112

x –1 0 1 2 3 4 5 6 7

g(x) –12 –5 0 3 4 3 0 –5 –12


Module 2

Self-Marking Activity 1. Draw the graph of the following functions. (Use the slope-

intercept method to graph in parts a and b.)

a) y = 3x + 2

b) 2x – y + 1 = 0

c) Join these points intoa smooth curve.

2. In question 1 (c), what would you expect the x-value to bewhen y = 2?

3. Which of the following describe a quadratic function?

e)5

5 5

5

b)

d)

g x x

f x xx

: → −

= −

2 5

31

3

b ga)

c)

f x x

x x x

:

,

→

− +

3

2 1

2

2d io t

x –3 –2 0 2 3

y 10 5 1 5 10


Module 2

f)

4. A graph of a quadratic function is shown below. Each tick onthe axes represents 1 unit.

a) What is the domain of the graph?b) What is the range of the graph?c) What are the coordinates of the vertex?d) What is the equation of the axis of symmetry?e) What are the zeros of the function?f) What are the x-intercepts?g) Does the curve have a maximum?h) Does the curve have a minimum?

5

5

5

5

5 5

5


Module 2

5. For each of the following write the domain, range,approximate coordinates of the vertex, equation of the axis ofsymmetry, zeros, y-intercept, and maximum or minimumvalue.a)

b)

1

2

3

4

5

1 2 3 4 5 61234561

2

3

4

5

6

x

y7

1

2

3

4

5

1 2 3 4 5 61234561

2

3

4

5

6

6

x

y


Module 2

c)

6. Sketch the graphs of the following quadratic function

a) f(x) = x2 – 4x – 5

Hint: In your table of values, choose values of x from –1 to 5.

b) {(x, –x2 – 2x + 8)}

Hint: In your table of values, choose values of x from –4 to 2.

Find the domain, range, coordinates of vertex, equation ofaxis of symmetry, zeros, x-intercept(s), y-intercept(s), andthe maximum value or minimum value.

7. Match the equation with its graph.a) y = x2 – 2 b) y = x2 + 4x – 1 c) y = –x2 – 4x – 3 d) y = x2 + 2 e) y = –x2 + 4x + 1 f) y = –x2 + 2

1

2

3

4

5

1 2 3 4 5 612341

2

3

4

5

6

6

7 8


Module 2

i) ii)

iii) iv)

v) vi)

8. a) The point (4, 3) is on the graph of f(x) = 2x2 – 8x + 3. Usesymmetry to find the coordinates of another point on theparabola. Its axis of symmetry is x = 2.

b) A parabola has a vertex at (2, 3) and passes through thepoint P(–1, 6). Find the coordinates of a third point on theparabola.

c) Given certain reflection image points (0, 2) and (4, 2) on aparabola, find the equation of the line of symmetry.


2

4

224 x

y

2

4

6

2

2 424 x

y

2

4

6

2

4

2 424 x

y

2

4

2

4

6

2 424 x

y

4

2

4

2 424 x

y

2

4

2

4

6

2 424 x

y


Module 2

Module 2

Notes


Lesson 3Quadratic Functions: y = ax2 and y = ax2 + k

OutcomesWhen you complete this lesson, you will be able to• determine basic characteristics of the quadratic function

y = ax2

• use translations and reflections with respect to the quadraticfunction of the form y = ax2 + k

OverviewYou have learned characteristics of the graphs of quadraticfunctions that were written in the form f(x) = ax2 + bx + c, wherea ≠ 0.The simplest and most basic quadratic function is f(x) = x2.

Observe that for f(x) = x2 if (x, y) is a point on the graph, (–x, y)is also. The y-axis is the line of symmetry. The point (0, 0) is theminimum or lowest point of the graph and it is called thevertex. You will see how the terms of the quadratic functiondetermines the size of the corresponding parabola and itsplacement on the coordinate plane. These changes that occurare called geometric transformations.

1

2

3

4

5

6

7

8

9

10

1

2

1 2 3 4 512345 x

y


Module 2

Example 1Use a table of values or a graphing calculator, if you have one, tograph the following functions on the same coordinate plane.

State the similarities and differences thatoccur.

SolutionYour table of values for each of the functions is listed below.

Notice thatg(–3) = 18 which is equal to 2 • f(–3)g(–2) = 8 which is equal to 2 • f(–2), and so on.

and

DefinitionsIf a > 1, the graph of y = ax2 can be obtained from the graph ofy = x2 with a vertical stretch of the graph of y = x2 performed bymultiplying by a factor of a.

If 0 < a < 1 (any real number between 0 and 1), the graph of y = ax2 can be obtained from the graph of y = x2 with a verticalshrink of the graph of y = x2 by multiplying by a factor of a.

k f

k f

− = −

− = −

392

12

3

2 212

2

b g b g

b g b g

is equal to

is equal to and so on.,

x –3 –2 –1 0 1 2 3

g(x) 18 8 2 0 2 8 18

f(x) 9 4 1 0 1 4 9

k(x) 2 0 29

21

2

1

29

2

g x x

f x x

k x x

b gb gb g

=

=

=

2

12

2

2

2


Module 2

For comparison, the graph of f is dashed.

Notice the similarities and differences:1. The graphs open upward.2. The graphs have the same axis of symmetry, x = 0. 3. The graphs share the same vertex (0, 0) and each have a

minimum y-value of 0.4. The greater the coefficient a, when a > 1, the narrower the

corresponding parabola, e.g., the graph of j(x) = 3x2 isnarrower than the graph of f(x) = x2.

5. The smaller the coefficient a, when 0 < a < 1, the wider the

graph, e.g., the graph of is wider than the graph

of f(x) = x2.6. The domain is x ∈ R for all functions. 7. The range is y ≥ 0, y ∈ R for all functions.8. The zeros and the x-intercept are 0. 9. The y-intercept is 0.

( )= 213

j x x

5

x0 5

g f k

y


Module 2

Example 2Graph the following functions on the same co-ordinate systemusing a table of values or a graphing calculator

Note: –x2 means –(x2), not (–x)2

SolutionA table of values for each of the functions is listed.

The diagram on the next page contains all the graphs fromExample 1 (two pages back) and Example 2 (above). Notice thatin general, the graph of y = –x2 is the mirror reflection in the x-axis of the graph of y = x2.

x –3 –2 –1 0 1 2 3

m(x) –18 –8 –2 0 –2 –8 –18

F(x) –9 –4 –1 0 –1 –4 –9

n(x) –2 0 –2− 92

− 12

− 12

− 92

( )( )

( )

= −

= −

= −

2

2

2

2

12

m x x

F x x

n x x


Module 2

Notice also the following characteristics of graphs with negativea values.

1. The graphs opendownward.

2. The graphs share theline of symmetry, x = 0.

3. The graphs share acommon vertex (0, 0)and have a maximumy-value of 0.

4. The graphs have thesame shapes as thosefor positive values of a,but now they opendownward.

5. The domain is x ∈ R forall functions.

6. The range is y ≤ 0, y ∈R for all functions.

7. The zeros and the x-intercept are 0.

8. The y-intercept is 0.

DefinitionAny graph of the form y = –ax2 can be obtained by reflectingthe graph of y = ax2 over the x-axis.

Example 3Describe how a complete graph of g(x) = –5x2 can be obtainedfrom the graph of f(x) = x2.

SolutionThere are two solutions. You can vertically stretch the graph off(x) = x2 by a factor of 5 and then reflect in the x-axis. Or, youcan reflect the graph of f(x) = x2 in the x-axis and then stretchthe graph of y = –x2 by a factor of 5.

Example 4Write an equation in the form y = ax2, of the function thatcontains the ordered pair (–2,8).

5

x5

g f k

m F n

5

y


Module 2

SolutionSubstitute x = –2 and y = 8 into y = ax2

8 = a(–2)2

8 = a(4)2 = aThe equation is y = 2x2.In general, the following statements are true about eachquadratic function of the type y = ax2.1. Its axis of symmetry is the y-axis.2. The coordinates of its vertex are (0, 0).3. Its graph opens upward if a > 0, or downward if a < 0.4. Its graph is narrower than the graph of {(x, y)|y = x2} if

|a|> 1 and wider if |a| < 1. (Recall that |a| is the absolutevalue of a.

5. It has a minimum value of 0 if a > 0, or a maximum value of 0if a < 0.

You will now examine graphs of quadratic functions of the typey = ax2 + k or f(x) = ax2 + k. What is the effect of adding a realnumber k to ax2 in the defining equation y = ax2?

Example 5Using a table of values or a graphing calculator, graph thefollowing functions on the same coordinate system.

SolutionYour table of values should look like the following:

x –2 –1 0 1 2

F: y 4 1 0 1 4

G: y 6 3 2 3 6

H: y 1 –2 –3 –2 1

F x x

G x x

H x x

=

= +

= −

,

,

,

2

2

2

2

3

d io td io td io t


Module 2

Notice that G(–2) = F(–2) + 2 and H(–2) = F(–2) – 3

In general, if f(x) = ax2 and g(x) = ax2 + k, then g(x) = f(x) + k forall x.

DefinitionThe graph of y = ax2 + k is obtained from the graph of y = ax2 bya vertical shift.• If k > 0, the shift is up k units.• If k < 0, the shift is down |k| units.

For any k, the graph has a vertex at (0, k) and the y-axis is theline of symmetry.

From the graph, F is the basic function y = x2.

Notice that if the graph of y = x2 is shifted up 2 units,you obtain the graph ofy = x2 + 2. If the graph ofy = x2 is shifted down3 units, you obtain thegraph of y = x2 – 3.

Observe the graphs of Gand H: • The axis of symmetry is

the y-axis; i.e., x = 0• The vertices for G and H

are (0, 2) and (0, –3),respectively.

• The minimum values of G and H are 2 and –3, respectively.• The size and shape of the resulting graph is same as for

y = x2.

Example 6Using your graphing calculator or a table of values graph

Predict the shape of the graphs.

K y x

L y x

M y x

:

:

:

= −= − += − −

2

2

2

2

1

5

x5

G

F

y

3

H


Module 2

Solution

Notice that L(–2) = K(–2) + 2 and M (–2) = K(–2) – 1.

Notice that it follows the same idea as Example 5.

Graph of L is the graph of Kshifted up 2 units and the graph ofM is the graph of K shifted down1 unit.

In general, the followingstatements are true about eachquadratic function of the type y = ax2 + k.

1. Axis of symmetry is the y-axiswith equation x = 0.

2. Coordinates of vertex: (0, k).3. y = ax2 + k is the graph of y = ax2 shifted |k| units vertically.

If k > 0, the parabola is shifted upward.If k < 0, the parabola is shifted downward.

4. If a > 0, the minimum value of y is k.5. If a < 0, the maximum value of y is k.

In the assignment you will be expected to do the stretch or shrinkand the vertical shift rather than use tables of values.

x

5 K

y

3

L

M

x –2 –1 0 1 2

K: y –4 –1 0 –1 –4

L: y –2 +1 +2 +1 –2

M: y –5 –2 –1 –2 –5


Module 2

Self-Marking Activity1. State whether each statement is true or false. If the

statement is false, rewrite it so that it is true.a) x2 + y = 1 is a quadratic function.b) t(x) = 1 – x is a quadratic function.c) {(x, 4x2)} is a quadratic function.d) The graph of r(x) = x2 + 1 is symmetric with respect to the

vertical or y-coordinate axis.e) The vertex of the parabola

pictured at the right is(1, 2).

f) An equation of the axis ofsymmetry of the parabolapictured at the right is y =2.

g) The graph of y = x2 is a“wider”parabola than the

graph of

h) The graph of y = 5x2 is a “narrower” parabola than thegraph of y = x2.

i) The graph of

is a parabola that opensdownward.

j) The graph of y = –2x2 is aparabola that opensdownward.

k) The maximum value ofthe function pictured atthe right is 4.

l) The minimum value ofthe function pictured atthe right is 0.

m) The domain of the function pictured at the right is{y|y ≤ 4}.

5

x0 4

yy x= 1

42

y x= 14

2.

4

x0 4

y


Module 2

2. Complete the following table.

3. Describe how to arrive at the graph of each of the followingusing y = x2 as the reference curve.

4. Find the following for the graph given by y = ax2: a) What is the direction of the opening of the graph if a > 0?b) What is the direction of the opening of the graph if a < 0?c) If a > 0, does y = ax2 have a minimum point or a maximum

point?d) If a < 0, does y = ax2 have a maximum point or a minimum

point?

a)

b)

c)

y x

y x

y x

=

=

= −

4127

2

2

2

Vertex

Equation of axis ofsymmetry

Domain

Range

x-intercepts or zeros

Direction of opening

Maximum y-value

Minimum y-value

y-intercept

y x= 1

32 y x= 2 y x= −3 2


Module 2

5. Match the letter corresponding to each function with thenumber corresponding to its graph.

6. Write an equation in the form y = ax2 if (1, –4) lies on thecurve.

7. Write the equation of the graph. Then write the equation ofthe reflection in the x-axis. Note each mark represents 1unit.

8. State whether each statement is true or false. If thestatement is false, rewrite it so it is true.a) The graph of y = 3x2 + 4 is the graph of y = 3x2 shifted

downward 4 units.b) The graph of {(x, –x2 – 3)} is the graph of {(x, –x2)} shifted

downward 3 units.c) The vertex of the graph of y = 2x2 is the origin.

d) The vertex of the graph of is (0, –2).

e) An equation of the axis of symmetry of the graph of y = x2 – 1 is x = –1.

y x= −12

22

x

y

1 2

3 4

a)

b)

c)

d)

y x

y x

y x

y x

= −== −

=

2

1 6

2

6

2

2

2

2

.

π


Module 2

f) The graph of y = x2 – 1 opens downward.g) The graph of y = –x2 – 2 opens downward.h) The maximum value of {(x, y)|y = –2x2 + 3} is –2.i) The minimum value of {(x, y)|y = 3x2 – 4} is –4.j) If f(x) = 2x2 + 3, then f(x) = f(–x), for all x.k) The graph of y = x2 + 4 is symmetric with respect to the

y-axis.l) The graphs of y = x2 + 2 and y = –x2 + 2 are mirror

reflections of each other in the x-axis.m) The graph of y = x2 is a narrower parabola than the graph

of

n) The range of {(x, –x2 + 2)} is {y|y ≤ 2}.


10. Describe how to arrive at the graph of each of the followingif y = x2 is the reference curve:

( ){ }( )

→ +

− +

= −

− + =

2

2

2

2

a) : 5 4

b) , 2 1

1c) 12

d) 3 2

f x x

x x

f x x

y x

Vertex


Domain

Range

x-intercepts

y-intercepts


Maximum value

Minimum value

y x= +2 4 y x= 2 y x= −2 1 y x= − +2 1

= +213.

2y x


Module 2

11. A quadratic function f is defined by f:x → ax2 + k, a ≠ 0.Describe the graph of f ifa) a > 0 and k > 0b) a > 0 and k < 0c) a < 0 and k > 0d) a < 0 and k < 0

12. Write an equation in the form y = ax2 + k given the vertex isat (0, –3) and the parabola passes through the point (1, –1).



Module 2

Notes


Module 2

Lesson 4Quadratic Functions: y = a(x – h)2

OutcomeWhen you complete this lesson, you will be able to• identify the characteristics of the graph of y = a(x – h)2 which

is a horizontal shift of y = ax2

OverviewCertain quadratic functions can be written in the form y = a(x – h)2 + k.

Example 1Use a table of values or a graphing calculator to graph thefollowing functions on the same coordinate plane.

Let a have a fixed value of 1 to consider the effect of changes inthe term of y = a(x – h)2.

SolutionThe following tables of values give some ordered pairs thatsatisfy the above functions.

x –2 –1 0 1 2

y 4 1 0 1 4

x 4 5 6 7 8

y 4 1 0 1 4

x –8 –7 –6 –5 –4

y 4 1 0 1 4

f(x)

g(x)

e(x)

( )( ) ( )( ) ( )

=

= −

= +

2

2

2

6

6

f x x

g x x

e x x


Module 2

Notice the following pattern:

In general, if f(x) = x2 and g(x) = (x – h)2 then g(x) = f(x – h) for all x.

DefinitionThe graph of y = (x – h)2 is said to be obtained from the graph of y= x2 by a horizontal shift.• If h > 0, the shift is h units to the right.• If h < 0, the shift is |h| units to the left.

For any h, the graph has a vertex (h, 0) and the line of symmetryx = h.

You can visualize the graph of y = (x – 6)2 as being the graph of y = x2 shifted 6 units to the right and the graph of y = (x + 6)2 asbeing the graph of y = x2 shifted 6 units to the left.

Note: y = (x + 6)2 is by the definition y = (x – (–6))2

Notice the following:a) The vertex of g(x) is (6, 0) and the vertex of e(x) is (–6, 0).b) The equation of the axis of symmetry of g(x) is x = 6 and of e(x)

is x = –6.c) The minimum values of g(x) and e(x) are 0.d) The size and shape of the resulting graphs are the same as

y = x2.

x

y

5

50

5

e gf

( ) ( ) ( ) ( )( ) ( ) ( ) ( )2 2 6 4 2 2 6 4

1 1 6 1 1 1 6 1

f g f e

f g f e

− = − + = − = − − =

− = − + = − = − − =


Module 2

Example 2Continue your investigation by giving “a” different values.Using a combination of shifts and stretches or shrinks, graphthe following functions on a coordinate plane.

SolutionEach graph begins with the basic graph y = x2.1. Shift y = x2 five units to the right to get y = (x – 5)2

[the dashed curve on the grid].2. a) For k(x), stretch vertically by a factor of 4 to get

y = 4(x – 5)2.b) For l(x), shrink vertically by a factor of 4 to get

y = ¼(x – 5)2.

c) For m(x), flip over the x-axis to get m(x) = –(x – 5)2.

Note: k(x) = 4(x – 5)2 is narrower than the graph of y = (x – 5)2

In the figure below, the graphs of k(x), l(x), and m(x) arecontrasted with the graph of y = (x – 5)2 [dashed curve].

x

y

5

30

l(x)

9

m(x)

k(x)

k x x

l x x

m x x

b g b gb g b gb g b g

= −

= −

= − −

4 5

14

5

5

2

2

2


Module 2

x y y x, |b g b g{ }= − 5 2

Example 3The vertex of a given parabola is at (2, 0). Another point on thegraph is (–2, 8). Write its equation in the form y = a(x – h)2.

SolutionSince the vertex of y = x2 is shifted 2 units to the right, h = 2. Theother ordered pair yields y = 8 when x = –2.

The equation is

The following generalizations can be made for functions of the typey = a(x – h)2:1. Axis of symmetry is the line x = h.

2. Coordinates of the vertex are (h, 0).

3. y = a(x – h)2 is the graph of y = ax2 shifted |h| unitshorizontally.If h > 0, the parabola is shifted to the right.If h < 0, the parabola is shifted to the left.

4. If a > 0, the minimum value of y is 0.If a < 0, the maximum value of y is 0.

y x= −12

2 2b g .

∴ = −

= − −

= −

=

=

y a x h

a

aa

a

b gb gb g

2

2

2

8 2 2

8 4

8 1612


Module 2

Self-Marking Activity 1. State whether each statement is true or false. If the

statement is false, rewrite it so that it is true.a) The graph of y = (x + 5)2 is the graph of y = x2 shifted 5

units to the left.b) The graph of y = (x – 7)2 is the graph of y = x2 shifted 7

units to the left.c) The vertex of the graph of y = (x – 1)2 is (1, 0).d) The vertex of the graph of y = 2(x – 1)2 is (2, 0).e) An equation of the axis of symmetry of the graph of

f) An equation of the axis of symmetry of the graph of

g) The graph of is a narrower parabola

than the graph of y = –3(x – 2)2.

h) The graph of opens downward.

i) The quadratic function has 0 for

its minimum value.

j) If g(x) = (x + 4)2, then g(x – 4) = g(–x – 4) for all x.

x y y x, |b g b g= −RSTUVW

13

2 2

y x= − +14

4 2b g

y x= − −13

2 2b g

y x y= − + = −12

3 32b g is .

y x x= − =12

312

2b g is .


Module 2

2. Match the equation with its graph.a) y = (x – 4)2

b) y = x2 + 4c) y = –(x + 2)2

d) y = –(x – 3)2

i) ii)

iii) iv)

3. Without using tables of values, sketch the graphs of thefollowing. Provide a written explanation of the steps involvedin the shift.

a)

b)

c)

f x x

y x

x x

:

,

→ −

= − −

+

12

4

2 1

2 1

2

2

2

b gb gb g{ }

2

4

6

24 x

y

8

2 4

2

4

6

24 x

y

8

6 2

2

4

6

2 4 x

y

8

6 8

2

4

6

2 4 x

y

8

6 8


Module 2



Vertex


Domain

Range

x-intercepts

y- intercepts


Maximum value

Minimum value

y x= + 2 2b g y x= −1 2b g y x= −2 32b g y x= +4

36

2b g


Module 2


Module 2

Notes

Lesson 5Quadratic Functions: y = a(x – h)2 + k

OutcomeWhen you complete this lesson, you will be able to• investigate graphs of quadratic functions of the form

f(x) = a(x – h)2 + k

OverviewIn the previous lessons, you explored the effects of the constantsa, h, and k on the graphs of quadratic functions. In this lesson,you will investigate graphs of quadratic functions of the form f(x) = a(x – h)2 + k.

DefinitionThe standard form of a quadratic function y = f(x) is f(x) = a(x – h)2 + k, where a, h, and k are real number constants.

Transforming y = x2 to f(x) = a(x – h)2 + k.

The graph of a quadratic function in standard form f(x) = a(x – h)2 + k can be obtained from the graph of y = x2

through the following sequence of transformations, performedin the indicated order:1. Horizontal shift h units: Shift right if h > 0 and left if h < 0.2. Vertical stretch/shrink by factor |a|. Stretch if |a| > 1 and

shrink if 0 < |a| < 1.3. Reflect in x-axis only if a < 0.4. Vertical shift k units: Shift up if k > 0 and down if k < 0.

The vertex of the graph of f is the point (h, k), and its line ofsymmetry is x = h.


Module 2

Example 1Sketch the graph given by y = (x + 2)2 – 3.

Solution1. Sketch the graph of y = x2. No

doubt, you already observedthat the formation of the graphof y = x2 is as shown here.

2. To obtain the graph of y = (x + 2)2, apply a horizontalshift of 2 units to the left to thegraph in step 1.

3. To obtain the graph of y = (x + 2)2 – 3, apply a verticalshift of 3 units downward to thegraph in step 2.

Therefore, the graph of y = (x +2)2 – 3 is obtained.

x

y

1

1

2

3

4

5

1 213 24 056

2

3

x

y

1

1

2

3

4

5

1 213 24 056

x

y

1

1

2

3

4

5

1 2 3 413 24 0


Module 2

Note: The point(2,4) on y = x2

shifts “left two”to (0,4) on y = (x + 2)2 .

Note: The point(0,4) on y = (x + 2)2– 3shifts “threedown” to (0,1) ony = (x + 2)2 – 3.

In Example 1, notice that a = 1 had no affect on the graph of y = ax2. In the next example, a = –1 and a reflection will beapplied.

Example 2Sketch the graph given by y = –(x – 1)2 + 2.

Solution1. Sketch the graph of y = x2,

using the standardformation.

2. To obtain the graph of y = (x – 1)2, apply ahorizontal shift of 1 unit tothe right to the graph instep 1.

3. To obtain the graph of y = –(x – 1)2, apply areflection to the graph instep 2 in the x-axis.

When a graph is reflected,the x-value stays the sameand y-value becomesopposite: (3,4) becomes(3,–4).

x

y

1

2

3

4

5

1 2 3 413 2 0

1

5

x

y

1

1

2

3

4

5

1 2 3 413 24 0

x

y

1

1

2

3

4

5

1 2 3 413 24 0


Module 2

4. To obtain the graph of y = –(x – 1)2 + 2, apply avertical shift of 2 units up tothe graph in step 3.

Example 3Write a sequence of transformations that will produce the graphof y = –(x – 7)2 – 3 from y = x2.

SolutionReflect y = x2 over the x-axis, shift horizontally 7 units to theright and 3 units vertically downward.

Below is a summary of observations concerning the graphs of

1. The graph is a parabola.2. The coordinates of the vertex are (h, k).3. An equation defining the axis of symmetry is x = h.4. The graph opens upward if a > 0 or downward if a < 0.5. The graph is narrower than the graph of y = x2 if |a| > 1 and

wider if |a| < 1.6. If a > 0, the minimum value of y is at y = k.7. If a < 0, the maximum value of y is at y = k.

( ) ( ){ } [ ]= − + ≠2, | 0x y y a x h k a

x

y

1

2

3

4

1 2 3 413 2 0

1

5

2

3


Module 2

Self-Marking Activity 1. State whether each statement is true or false. If the

statement is false, rewrite it so it is true.a) The vertex of the graph of y = (x – 4)2 + 2 is (–4, 2). b) The vertex of the graph of y = 3(x – 5)2 – 2 is (5, 2).c) The vertex of the graph of y = 2x2 + 7 is (2, 7).d) The vertex of the graph of y = –3x2 + 4 is (0, 4).e) The vertex of the graph of y = –2(x – 3)2 is (–2, 3).f) An equation of the axis of symmetry of the graph of

y = –3(x – 6)2 + 1 is x = 1.g) An equation of the axis of symmetry of the graph of

h) An equation of the axis of symmetry of the graph ofy = (x + 2)2 + 1 is x = –2.

i) The graph of y = 4(x – 3)2 – 1 opens downward.j) The graph of y = 4(x + 2)2 – 3 opens upward.k) The range of {(x, y)|y = –2(x + 3)2 + 4} is {y|y ≤ 4}.l) The maximum value of {(x, y)|y = 3(x – 5)2 – 2} is – 2.

2. Match each equation with its graph.

( )

( )

( ) ( )

( ) ( )

= − = +

−= + = − −

= + = − −

+ = + − = − −

22

22

2 2

2 2

1a) 3 e) 1 23

1b) f) 1 3 23

1c) 3 2 g) 23

1d) 1 3 2 h) 1 23

y x y x

y x y x

y x y x

y x y x

= − + =

21 13 4 is .2 2

y x x


Module 2


x

y

5

55

i)

x

y

5

55

iii)

x

y

5

5

10

5

iv)

x

y

5

5

10

5

v)

x

y

5

5

10

5

vi)

x

y

5

5

10

5

vii)

x

y

5

55

viii)

x

y

5

55

ii)

Module 2


4.. For each of the following parabolas statea) the direction of openingb) the coordinates of the vertexc) the equation of the axis of symmetryd) whether the given graph is narrower or wider than y = x2

5. Without making a table of values, sketch the graph of eachof the following functions. Explain the various steps. Statethe vertex and the equation of the axis of symmetry.

( ) ( )

( ) ( )

= + = − − +

= + − = − +

2 2

2 2

1a) 2 1 b) 1 62

c) 2 6 10 d) 6 1 8

y x y x

y x y x

( ) ( )

( ) ( )

= + = − − +

= + − = − +

2 2

2 2

1i) 2 1 ii) 1 62

iii) 2 6 10 iv) 6 1 8

y x y x

y x y x

Vertex

Axis of symmetryequation

Domain

Range

y-intercept


Maximum value

Minimum value

( )= − +21 2y x ( )= − − −2

1 2y x ( )= + +21 2y x ( )= − + −2

1 2y x


Module 2

6. For each of the following quadratic functions,• specify whether the function has a minimum or a maximum

value and what that value is• specify the range

7. If is a point on the quadratic function represented by

y = (x – 2)2 + k, what real number is the minimum value of thefunction?


31,

2

( ) ( )

( ) ( ) ( )

= − + + = + −

= − − − = + +

22

2 2

1a) 2 b) 2 4 2

21

c) 3 2 1 d) 1 42

f x x y x

g x x y x


Module 2

Lesson 6Completing the Square

OutcomesWhen you complete this lesson, you will be able to• apply a geometric approach to the process called completing

the square• transform f(x) = ax2 + bx + c into the form y = a(x – h)2 + k, by

completing the square, in order to analyze the features of itsgraph.

OverviewYou discovered how to find the maximum or minimum value ofa quadratic function when it is written in the form f(x) = a(x –h)2 + k. You will now concentrate on writing any generalquadratic function f(x) = ax2 + bx + c into the formf(x) = a(x – h)2 + k. This process is called completing thesquare.

In algebra there are many times when you can complete thesquare to help solve a problem. Completing the square inquadratic expressions can be modelled with geometric shapescalled tiles.

Example 1* Solutiona) Model x2 with a square tile that is

x units long on each side.

b) Model x with a rectangular tile that isx units long and 1 unit wide.

c) Model a constant with a square tile that is 1 unit long on each side.

d) Model a square that has 1 x2-tile, 4 x-tiles,and 4 unit tiles.

1

1

1

x

x-tile

x

x

x 2 tile


Module 2

*You don’t haveto draw or use

tiles inMathematics 11.

However, somestudents don’t

“catch on” tocompleting the

square until theysee it done in

tiles—then theyfind it easy.

Examples 1 and 2are for those

students.

The area of this square is the sum of the tiles x2 + 4x + 4.

Since the length of a side of this squareis x + 2, the area can also be written as(x + 2)2.

This model of x2 + 4x + 4 forms acomplete square. The quadraticexpression is a complete square and iscalled a perfect-square trinomial.

A perfect-square trinomial such as x2 + 4x + 4 can also be writtenas (x + 2)2 and is called the square of a binomial.

Example 2a) Write the quadratic expression modelled by these tiles.

b) Find the minimum number of additional unit tiles needed toform a square.

c) Write a perfect trinomial modelled by the completed square.

Solutiona) There are 1 x2 tile and 6 x-tiles and 0 unit tiles. The

expression is x2 + 6x.

b) Take of the x-tiles and move them to the adjacent side of x2.

There are 32 or 9 unit tiles needed to fill in the missing corner or tocomplete the square.

c) By adding the tiles, the new squaremodels the perfect square trinomialx2 + 6x + 9.

12

x

2

2


Module 2

In a perfect square trinomial such as x2 + 4x + 4, the square of

half the coefficient of x or is the constant term 4.

In a perfect square trinomial such as x2 + 6x + 9, the square of

half the coefficient of x or is the constant term 9.

If x2 + bx + c is a perfect square trinomial, then

To complete the square in the expression

You can use the technique of completing the square totransform a general quadratic function f(x) = ax2 + bx + c intothe standard form y = a(x – h)2 + k.

Example 3Complete the square, without using tiles, to find the vertex andthe equation of the axis of symmetry for the quadratic functiony = x2 + 6x.

SolutionThe number required to complete the square is the square ofone-half the coefficient of the x-term.

Equation of axis of symmetry x = –3.

Add 9 to each side

Factor the perfect square trinomial

Subtract 9 from each side

62

3 92

2FHGIKJ = =y x x

y x x

y x

y x

V

= +

+ = + +

+ = +

= + −

= − −

2

2

2

2

6

9 6 9

9 3

3 9

3 9

b gb gb g,

x bx cb2

2

2+ = FHG

IKJ, .add

bc

2

2FHGIKJ = .

62

2FHGIKJ

42

2FHGIKJ


Module 2

Example 4Express y = x2 – 4x + 3 in the standard form and find themaximum or minimum value.

Solution

Since a = 1 and is positive, the parabola opens upward and thefunction has a minimum value of –1.

Example 5Find the coordinates of the vertex and the equation of the axis ofsymmetry of f(x) = 3x2 + 12x +16.

SolutionExpress the equation in the form y = a(x – h)2 + k.

The vertex is at (–2, 4) and the equation of the axis of symmetryis x = –2.

Factor the trinomialsimplify

42

42F

HGIKJ =Complete the square

Isolate the x-termsy x x

y x x

y x x

y x

y x

= − +

− = −

− + = − +

+ = −

= − −

2

2

2

2

2

4 3

3 4

3 4 4 4

1 2

2 1

b gb g


Module 2

f x x x

f x x x

f x x x

f x x

f x x

b g

b g d ib g d i

b g b gb g b g

= + +

− = +

− + = + +

− = +

= + +

3 12 16

16 3 4

16 12 3 4 4

4 3 2

3 2 4

2

2

2

2

2

Isolate the x-terms and factor out the 3

Complete the square

42

42F

HGIKJ = but it is really 4

multiplied by 3 or 12that has to be added to each side

Factor

and add 4 to each side

Example 6Express f(x) = –2x2 + 8x + 5 in standard form and find thedomain and range of the function.

Solution

Domain: x ∈ RRange: {y|y ≤ 13}

Example 7

Express in standard form. Determine whether

the function has a maximum or minimum value. Determine thevalue of x that is paired with the maximum or minimum value.

Solution

Since the parabola opens downward, the function has amaximum value at x = –1. The x-value that is paired with themaximum is –1.


Module 2

f x x x

f x x x

f x x x

f x x

f x x

b gb g d ib g d ib g b gb g b g

= − + +

− = − −

− − = − − +

− = − −

= − − +

2 8 5

5 2 4

5 8 2 4 4

13 2 2

2 2 13

2

2

2

2

2

Isolate the x-terms and factor out –2

Note the sign changes

Complete the square and simplify

y x x= − − +12

42

y x x

y x x

y x x

y x

y x

y x

= − − +

− = − +

− − = − + +

− − = − +

− = − +

= − + +

12

4

412

2

412

12

2 1

412

12

1

92

12

1

12

192

2

2

2

2

2

2

d i

d i

b g

b g

b g

Complete the square and add to each side

− 1Isolate the terms and factor out

2x

− 12

Simplify

Note: to factor1 from 12

1try 12

211

2

− −

− ÷ −

= − × −=

Example 8Find the values of a, h, and k that will make the statement true.

a) x2 – 6x = (x – h)2 + k

b) 3x2 – 12x + 5 = a(x – h)2 + k

Solutiona) a = 1

You have to complete the square to get the right-hand side.

b)

One of the more important roles of quadratic functions ismodelling the path of a falling object. You can write a quadraticfunction that models the height of an object influenced only bygravity near the Earth’s surface.

After t seconds, the height of an object with an initial upwardvelocity of v0 metres per second and an initial height of h0 metresis

h t t v t hb g = − + +4 9 20 0.

a h k= − = − =3 2 17, ,

Rearrange andcomplete thesquare

− + = − + −

− + + FHGIKJ

FHG

IKJ = − + − −

− + = − + −

− + + = − +

3 4 5

3 442

5 3 2

3 2 17

3 2 17

2 2

22

2 2

2 2

2 2

x x a x h k

x x a x h k

x a x h k

x a x h k

d i b g

b g b g

b g b gb g b g

∴ = = = −a h k1 3 9, ,

x x x h k

x x x h k

x x h k

22

22

2 2

2 2

662

62

6 9 9

3 9

− + FHGIKJ = − + + FHG

IKJ

− + = − + +

− − = − +

b gb g

b g b g


Module 2

In this equation, the force of gravity is represented by thesquared term in the negative direction. As the time increases,the t2 term overpowers the t term and the ball falls.

Example 9A tennis ball at an initial height of 1.5 metres is hit upwardwith an initial velocity of 25 metres per second.a) Write a quadratic function g(t) for the height in metres after

t seconds. (This is a physics formula which you may or maynot have seen before.)

b) Use your TI-83 graphing calculator to find the maximumheight reached by the ball. (This is optional. See thecalculator’s manual about using the CALC function.)

Solutiona) Substitute the values into h(t) = –4.9t2 + v0t + h0 where v0 is

the initial velocity in metres per second and h0 is the initialheight.

b) Input y1 = –4.9x2 + 25x + 1.5Set your windows as followsXmin = 0Xmax = 10Xscl = 1Ymin = 0Ymax = 35Yscl = 5Xres = 0

Press 2nd (CALC) #4 to find the maximum, which willgive the coordinates of the vertex. V(2.55, 33.39).Therefore, the maximum height is 33.4 m.

h t t tb g = − + +4 9 25 152. .


Module 2

Self-Marking Activity 1. Write a quadratic expression modelled by this set of tiles.

2. What value of k makes the expression a perfect squaretrinomial? Write each expression as the square of a binomial.

3. Find the values of a, r, and t for each of the following:Note r is representing h, and t is representing k.

4. For each of the following equations, answer the questionsbelow:

i) Express in the form f(x) = a(x – h)2 + k.ii) Find the coordinate of vertex and equation of the axis of

symmetry.

a)

b)

c)

d)

e)

f x x x

f x x x

x x x

f x x x

y x x

b g

d io tb g

= − +

→ + −

− −

= − +

− + = −

2

2

2

2

2

6

2 5 3

4

2 12 19

3 3 3

:

,

a)

b)

c)

x x a x r t

x x a x r t

x x a x r t

2 2

2 2

2 2

8

5

12

2 4

− = − +

+ = − +

+ + = − +

b gb gb g

a) b)

c) d)

e) f)

x x k x x k

x x k x x k

x x k x x k

2 2

2 2

2 2

8 8

20 2

5 7

+ + − ++ + − +− + + +


Module 2

iii) Describe in words how you would use y = x2 to sketch thecurve.

iv) Find the range.v) Find the y-intercept.vi) Determine the maximum or minimum value.

5. Given y = ax2 + bx + c, a ≠ 0, express it in the form y = a(x – h)2 + k. Write the coordinates of the vertex.

Reasons:


Module 2

( )( )

= + +

− = +

− = + − + = + +

− + = + = + + −

− = + + −−

2

2

2

2 22

22

2 2

2 2

2

2 2

4 2

2 4

42 4

4,

2 4

y ax bx c

y c ax bx

by c a x x

a

b b by c a a x x

a a a

b by c a x

a a

b by a x c

a a

b ac by a x

a a

b ac bV

a a

a)

b)

c)

d)

e)

f)

g)

You can use for finding the x-coordinate at the

vertex if the given equation is in the form y = ax2 + bx + c.You can substitute that value into the quadratic equation tofind the corresponding y-coordinates.

xba

= −2

Note: to makethis questioneasier, we havedone the mathfor you. All youdo is supply thereasons..

6. a) A quadratic function has its vertex at (2, 6). If the samecurve passes through (1, 7), find the value of a if theequation is in the form y = a(x – h)2 + k.

b) Does the function have a maximum or minimum value?c) Write the equation of the parabola.

7. A projectile is shot straight up from a height of 6 metreswith an initial velocity of 80 metres per second. Its height inmetres above the ground after t seconds is given by theequation h = 6 + 80t – 5t2. After how many seconds does theprojectile reach its maximum height, and what is thisheight?



Module 2

Lesson 7Special Features of the Quadratic Function

OutcomesWhen you complete this lesson, you will be able to• find the coordinates of the vertex of a quadratic function if

you know its x-intercepts• find the coordinates of the vertex of a quadratic function of

the form y = ax2 + bx + c• use interval notation to express the domain and range of

relations and functionsOverviewThe zeros of a function f are the values of x that make f(x) = 0.They can be used to find the x-intercepts. In order to find thex-intercepts of a quadratic function that is written as theproduct of linear functions, you use the zero product property.

DefinitionIf ab = 0, then either a = 0 or b = 0 must be true.

Example 1Consider the function y = x2 – 2x – 8.

It can be factored into y = (x – 4) (x + 2), which is a product oftwo linear functions.

Its zeros can be found by letting y = 0.

∴ (x – 4) (x + 2) = 0

Using the Zero Product PropertyEither

The zeros are 4 and –2.

The x-intercepts are at (4, 0) and (–2, 0). Remember that thesepoints are on the x-axis, and the y-coordinate of these points isalways 0.

x xx x

− = + == = −

4 0 2 04 2

oror


Module 2

In Principles of Mathematics 10, you used the midpoint formulato find the midpoint of a line segment given its endpoints atP1(x1, y1) and P2(x2, y2). This method is averaging the two x-coordinates and the two y-coordinates.

midpoint of x-coordinate

midpoint of y-coordinate

To find the x-coordinate of the vertex, find the midpoint of the x-coordinates of the x-intercepts.

midpoint of x-coordinates

To find the y-coordinate of the vertex, substitute the x-coordinateat the vertex into the original function.

The coordinates of the vertex are at (1, –9). Verify using agraphing calculator.

The axis of symmetry is the vertical line passing through thevertex, that is, x = 1.

This method can be used to find the vertex of a quadratic functionif it factors readily.

A second method of determining the vertex quickly is by using aformular which we will develop here. (This was also derived inthe last Self-Marking Activity.

y x x

y

y

= − +

= − +

= − = −

4 2

1 4 1 2

3 3 9

b gb gb gb gb gb g

becomes

=+ −

= =4 2

222

1b g

= +y y1 2

2

x x= +1 2

2


Module 2

Module 2

= + +

− = +

− = + − = + + • − •

− = + −

+ − = +

+ − = +

− = + +

2

2

2

2 22

2 2

22

22

2

If , then:

1 1 Completing the square2 2

2 4

4 2

44 4 2

42

y ax bx c

y c ax bx

by c a x x

a

b b by c a x x

a a a

b by c a x

a a

b by c a x

a a

b ac by a x

a a a

b ac by a x

a

2

4a

The Vertex of the parabola which is the graph of this function is:

The x-coordinate for the vertex can be derived quickly as –b/2awhen the function is given as f(x) = ax2 + bx + c and does not factorreadily.

Example 2Find the coordinates of the vertex and equation of axis ofsymmetry of f(x) = x2 – 6x + 4.

This function does not factor readily so to find the vertex do thefollowing:

a) Find the x-coordinate using

a b

x

= = −

=− −

= =

1 6

62 1

62

3b gb g

2– 4 –,

2 4b ac b

Va a

2– 4 –,

2 4b ac b

Va a


Module 2

b) Find y by substituting x = 3 into the original function.

The vertex is at (3, –5).

(Alternately, you can use the formula to find the y-coordinate of the vertex.)

An alternate method for expressing the domain and range offunctions and relations is called interval notation. The notationuses square brackets [ ] to indicate that all the numbersbetween two specified numbers and the numbers themselvesare included, whereas round parentheses ( ) are used to excludethe end numbers.

The symbol, ∞, is used to indicate that the interval of numbershas no upper limit: –∞ means the numbers decrease withoutlimit and ∞ means they increase without limit.

The following charts provide you with the link between theinequality notation that you have used and interval notation.

− 244

ac ba

y x x

y

yy

= − +

= − +

= − += −

2

2

6 4

3 6 3 4

9 18 45

b g b g


Module 2


a b

InequalityNotation

a ≤ x ≤ b

IntervalNotation

[a, b]a and b are included so squarebrackets are used

a b

a ≤ x < b[a, b)

the open circle indicates b isnot included and so a roundparenthesis is used with b

a ba < x ≤ b

(a, b]the open circle at a indicates ais not included and so a roundparenthesis is used with a

a ba < x < b

(a, b)the open circles at a and bmean that a and b are notincluded and hence the roundparentheses

a b

x ≥ b

[b, ∞)the line designated by asquare bracket includes b andgoes infinitely to the right,designated by a parenthesis

a b

x > b(b, ∞)

means b is excluded but thatthe line goes infinitely to theright

CorrespondingLine Graph

a b

a b

(–∞, a]a is included but the linecontinues infinitely to the leftof a

(–∞, a)a is excluded but the linecontinues infinitely to the leftof a

x ≤ a

x < a

Example 3Describe the line illustrated using interval notation.


Module 2

3012

Answer

[–2, 3]

The square brackets are usedto indicate that all numbersbetween and including –2 and3 equivalent to {x|–2 ≤ x ≤ 3}in inequality notation.

3012 1 2

Answer

[–1, ∞)

–1 is included and numbersbecome infinitely largeequivalent to {x|x ≥ –1}.

3012 1 2

Answer

(–∞, 1)

1 is excluded on the right butnumbers go to negativeinfinity on the left equivalentto {x|x < 1}.

a)

b)

c)

3012 1 2

Answer

(–∞, 2]

Includes 2 on the right andgoes left to negative infinity.Equivalent to {x|x ≤ 2}.

d)

3012 1 2

Answer

[–2, 1)

Includes –2, and excludes 1.Equivalent to {x|–2 ≤ x < 1}.

e)

3012 1 2

Answer

(–∞, ∞)

Goes infinitely in bothdirections; a square bracket isnever used with infinity.equivalent to {x|x ∈ R}.

f)

5 8

Answer

(–5, 8]

Equivalent to {x|–5 < x ≤ 8}. g)

6 80

Answer

(–∞, –6) ∪[0, 8]

(∪ means or)

Equivalent to {x|x < –6 or 0 ≤ x ≤ 8}.

h)

Example 4Find the domain and range of the following functions or relationsusing interval notation.


Module 2

a)

3

2x

y SolutionD: (–∞, ∞)

R: (–∞, ∞)

No restrictions are on the valuesthat the line contains.

b)

2

x


R: {2}

The x-values are not restrictedfor this line; this notationindicates the range is constantat 2.

c)

2 x

ySolutionD: {2}

R: (–∞, ∞)

The x-values are the same on avertical line; y-values are notrestricted.


Module 2

d)

2

x2

4


R: [0, ∞)

The x-values are not restrictedfor this curve. The y-valuesinclude 0 to positive infinity.

e)

x

y

4321

SolutionD: (–∞, ∞)

R: [0, ∞)

Notice that the domain andrange are the same as in (d).

f)

2x

2

4


R: (–∞, 0]

Notice that the domain is thesame as in (d); y-values include0 but go to negative infinity.


Module 2

g)

2x

4

ySolutionD: (–∞, ∞)

R: [–4, ∞)

The x-values are not limited; y-values include –4 andcontinue to infinite largenumbers.

h)

2

x1

y

4

SolutionD: (–∞, ∞)

R: (–∞, –1]

Because the parabola opensdownward, –1 is included aswell as an infinite number ofy-values approachingnegative infinity.

i)

x

y

2

2

2

2

SolutionD: [–2, 2]

R: [–2, 2]

Example 5For the function y = x2 + 6x + 5, state thea) the coordinates of the vertexb) equation of axis of symmetryc) the x-interceptsd) the y-intercepte) domain and range

Solutionsa) The x-coordinate of the vertex is given by

Substitute the x-value into

b) The axis of symmetry is the vertical line passing through thevertex, ∴ x = –3.

c) Because the equation factors readily y = (x + 5)(x + 1).Let y = 0 and use the zero product property: (x + 5)(x + 1) = 0.Either x + 5 = 0 or x + 1 = 0;x = –5 or –1.The zeros are –5 and –1.The x-intercepts are at the points (–5, 0) and (–1, 0).

y x x

y

yy

V

= + +

= − + − +

= − += −− −

2

2

6 5

3 6 3 5

9 18 54

3 4

b g b g

b g,

xba

a b

x

= −

= =

= − = −

21 6

62 1

3

,

b g


Module 2

d) The y-intercept is found by letting x = 0 in

∴ the y-intercept is at the point (0, 5)e) D: (–∞, ∞)

R: [–4, ∞)

Self-Marking Activity1. Using interval notation, describe the indicated solution to

the inequality.

2. Using interval notation, describe the indicated solution tothe inequality.

a)

b)

c)

d)

0 6

3 10

2 7

2 5

a) or

b)

c)

d) or

e) or

x x x

x x

y y

y y y

x x x

|

|

|

|

|

< − ≥

− < ≤

> −

> < −

− < ≤ − ≤

3 2

10 5

8

5 7

5 2 2

l ql ql ql ql q

y x x

y

y

= + +∴ = + +

=

2

2

6 5

0 6 0 5

5b g


Module 2

e)

3. For each of the following quadratic equationsa) y = x2 – 2x – 3b) y = –x2 – 6x – 8find the following and then sketch the graph:i) x-interceptsii) vertexiii) equation of axis of symmetryiv) y-interceptv) direction of openingvi) maximum valuevii) minimum valueviii) domain and range (state in both notations)ix) sketch the graph

4. For each of the following quadratic functions

find

i) coordinates of the vertex using

ii) equation of axis of symmetryiii) x-interceptsiv) y-interceptv) direction of openingvi) maximum or minimum valuevii) domain and range in interval notation


xba

= −2

a) b)

c) d)

e) f)

g) h)

y x x y x x

y x x y x x

y x x y x x

y x x y x x

= + − = − −

= + − = + +

= + + = − + +

= − − − = − + +

2 2

2 2

2 2

2 2

6 7 4 60

2 8 10 3 24 21

5 6 3 4

2 5 2 3 2 1

6 823


Module 2

Lesson 8Problems Involving Quadratic Functions

OutcomeWhen you complete this lesson, you will be able to• find a function to model problem situations that are

quadratic in nature

OverviewThe area of a rectangular space can be modelled by a quadraticfunction and its graph can be used to find its maximum area.

Example 1John has 50 m of fencing to enclose a rectangular area. Heneeds to know the maximum area that can be enclosed. Whatare the dimensions that will produce this maximum area?

SolutionDraw a table showing the possibilities

You can generalize by lettingx = width of the rectangle25 – x = length

Remember, the perimeter = x + x + 25 – x + 25 – x = 50 m.

25 x

x

Width (m)

1

2

3

x

…

Length (m)

24

23

22

25 – x

…

Area (m2)

24

46

66

x(25 – x)

…


Module 2

If you represent the area by y = x (25 – x), the equationy = 25x – x2

y = –x2 + 25xis a quadratic function (representing the area of the space).

Since a is negative, this function gives you a maximum value:y = –x2 + 25x

By completing the square of the above function, you can get thewidth giving you the maximum area.

This can be interpreted that a width of m gives a maximum

area of

Example 2An apple orchard has 20 trees per acre. The average yield is 300apples per tree. It is estimated that for each additional tree peracre, the average yield per tree is reduced by 10 apples. Howmany trees per acre yields the largest number of apples.

SolutionLet x = number of new trees planted per acre20 + x = number of trees per acre300 – 10x = average yield per treef(x) = total number of apples

2625m .

4

252

( )

( )

2

2 22

2

2

25

25 251 25

2 2

625 254 2

25 6252 4

y x x

y x x

y x

y x

= − − +

+ − = − − +

− = − −

= − − +


Module 2

Remember, the total number of apples = (number of trees peracre)(average yield per tree)

From the equation, you can determine that the vertex of theparabola is at (5, 6250). This means that when the originalnumber of trees per acre has 5 added to it, (20 + 5 additional)for a total of 25 trees, the maximum yield will be 6250 apples.

An alternate solution would be:f(x) = –10x2 + 100 x + 6000

The maximum value occurs at the x-coordinate of the vertex. Inthis case

Example 3A car dealership can sell 20 cars per week at a profit of $2400each. The dealership has determined that for every $300increase in profit, one less car per week is sold. What is themaximum profit the dealership can make and how many carswould then be sold per week?

SolutionMaximum profit = (number of cars sold)(profit per car)x = the decrease in the number of cars soldThen, 20 – x = number of cars sold2400 + 300x = profit per carf(x) = maximum profit

xba

= − = −−

=2

10020

5

( ) ( )( )

( ) ( )( ) ( )

( ) ( )

( ) ( )

2

2

2 22

2

2

20 300 10

6000 100 10

6000 10 10

10 106000 10 10 102 2

10006000 10 54

10 5 6250

f x x x

x x

f x x x

f x x x

f x x

f x x

= + −

= + −

− = − −

− − − + − = − − +

− − = − −

= − − +


Module 2

Note: Thisequation is the

second line inthe above

solution. Thenuse the vertex

formula to get xand substitute

to get y.

This means a decrease of 6 cars will yield a maximum profit of$58 800.

∴ number of cars sold: 20 – 6 = 14 cars per week

Example 4The price of a radio is $50, and 40 are sold each day. For each$1.00 the price is raised, the store sells one less radio. If eachradio costs $18 to make, how much should the price be set at tomaximize profits?

SolutionMaximum profit = (profit/radio) × (no. of radios sold)

Profit/radio = $50 – $18 = $32Let x = the decrease in the number of radios 40 – x = total number of radios sold(32 + x) = profit per radiof(x) = maximum profit

f x x x

x x x

x x

f x x x

f x x

f x x

b g b gb g

b g d ib g b gb g b g

= + −

= − + −

= − + +

− − = − − +

− = − −

= − − +

32 40

1280 32 40

8 1280

1280 16 8 16

1296 4

4 1296

2

2

2

2

2

f x x x

x x x

f x x x

f x x x

f x x x

f x x x

f x x x

f x x

x

b g b gb g

b gb gb g d i

b g b g

b g d ib g b g

= − +

= + − −= + −

− = − +

− = − −

− + − −FHGIKJ = − − + −F

HGIKJ

FHG

IKJ

− − = − − +

= − − +

=

20 2400 300

48000 6000 2400 300

48000 3600 300

48000 300 3600

48000 300 12

48000 300122

300 12122

48000 10800 300 12 36

300 6 58800

6

2

2

2

2

22

2

2

2

cars


Module 2

When four fewer radios are sold, a maximum profit of $1296is achieved.

Profit/radio = 32 + x= 32 + 4= 36

New price = $18 + $36= $54

The new price of the radio would be $54.

Check: Profit now: 54 – 18 = 36Number of radios sold: 40 – 4 = 36Maximum profit: 36 x 36 = $1296

Self-Marking Activity1. The height, h, in metres, after the launching of a rocket at

any time, t, in seconds, is defined by the equation

Find the maximum height reached by the rocket and the timeit takes to reach this height.

2. At the local beach, the lifeguard has 620 m of marker buoys torope off a safe swimming area. Calculate the dimensions ofthe rectangular swimming area to create maximumswimming room if one side of the area is to be the beach.

3. If 65 apple trees are planted in an orchard, the average yieldper tree will be 1500 apples per year. For each additional treeplanted in the orchard, the annual yield per tree drops by 20apples. How many trees should be planted in order to producea maximum yield?

4. The difference between two numbers is 14. Find the twonumbers so that their product is a minimum.

5. Honest John’s used car lot sells an average of 20 cars perweek at an average price of $6400 each. Honest John wouldlike to increase the average price by $300. However, he knowsthat his sales would fall by one car if he does. If the dealer’s(Honest John’s) cost per car is $4000, for what price should hesell the cars to maximize profits?

h t t= − + +3 9814

2 .


Module 2

6. A rectangular field is to be enclosed by a fence anddivided into three smaller plots by two fences parallel toone of the sides. Find the dimensions of the largest suchfield if 800 m of fencing is available.

7. Tom Katz’s company sold tickets to a U2 concert. If 1000or less people purchased tickets, the price was $60.00 perticket. For every 100 tickets sold over 1000, the ticket wasdiscounted by $3.00. What ticket price will give amaximum income?

8. A dealer finds that he can sell 800 radios at $60.00 perset. For every $2.00 drop in price, he can sell 50 moreradios. At what price per radio should he sell to receive amaximum cash return?


Review Module 2 before attempting the review questionsbeginning on the next page. These questions should help youconsolidate your knowledge as you prepare for the Module 2Section 1 Assignment.


Module 2

Review

1. A graph of a quadratic function is shown below.

a) What is the domain of the graph?b) What is the range of the graph?c) What are the coordinates of the vertex?d) What is the equation of the axis of symmetry?e) What are the zeros of the function?f) Where are the x-intercepts of the function?g) What is the y-intercept?h) Does the curve have a maximum and what is its value?i) Does the curve have a minimum and what is that value?j) Name two points that are reflections.

1 2 3 4 5

1

2

3

4

5

−1

−2

−3

−1−2−3 x

y

Principles of Mathematics 11 Section 1, Review 87

Module 2

2. Given the following functions,

state:a) values of a, h, kb) axis of symmetryc) domaind) direction of openinge) vertexf) minimum/maximum and valuesg) the description of the transformation with respect to y = x2

h) zerosi) x-interceptsj) rangek) y-interceptl) sketch the graphm) the new equation if the parabola is moved 2 units upward

and 3 units to the left

3. Express in standard form.

4. Find the values of a, h, and k that will make the followingstatement true:

4x2 – 20x + 5 = a(x – h)2 + k

a)

b)

c)

d)

e)

f x x x

f x x x

y x x

y x x

f x x x

b gb g

b g

= − −

= − − +

= + −

= − −

= − −

2 8 5

3 6 2

12

2 1

2 3 7

01 2 1

2

2

2

2

2.

i)

ii)

iii)

y x

y x

f x x

= − + −

= − +

= + −

2 1 3

12

1 3

2 1 3

2

2

2

b gb g

b g b g

88 Section 1, Review Principles of Mathematics 11

Module 2

5. If the value of a for a quadratic function is a = 1, and if the x-intercepts of the function are at (5, 0) and (1, 0), find thecoordinates of the vertex of the function.

6. Find the coordinates of the vertex and the axis of symmetryfor f(x) = x

2– 8x + 2 without completing the square.

7. Find the quadratic function having a vertex at (2, –8) andpassing through the point (4, 2).

8. If P(2, 4) is on the quadratic function y = (x – 1)2

+ p, find theminimum value of the function.

9. A riverboat with a capacity of 200 passengers is to bechartered for an excursion. The price of one ticket is $30 if100 people or less buy tickets, but the riverboat companyagrees to reduce the price of every ticket by 20 cents for eachticket sold in excess of 100. What number of passengers willproduce the largest income? Show algebraically.

10. A rancher has 200 m to fence a rectangular field. Whatdimensions will give the maximum area? Show algebraically.


Now do the Section Assignment which begins on the next page.

May I Use My Graphing Calculator on this Assignment?If you have a hand-held graphing calculator, you may! In fact,you may use a hand-held graphing calculator on all theassignments and tests in this course.

Principles of Mathematics 11 Section 1, Review 89

Module 2

Notes

90 Section 1, Review Principles of Mathematics 11

Module 2

PRINCIPLES OF MATHEMATICS 11

Section Assignment 2.1

Principles of Mathematics 11 Section Assignment 2.1 91

Version 04 Module 2

General Instructions for Assignments

These instructions apply to all the Assignments, but will not be reprinted eachtime. Remember them for future sections.

(1) Treat this assignment as a test, so do not refer to your Module or notes orother materials. A scientific calculator is permitted.

(2) Where questions require computations or have several steps, show your work.

(3) Always read the question carefully to ensure you answer what is asked. Oftenunnecessary work is done because a question has not been read correctly.

92 Section Assignment 2.1 Principles of Mathematics 11

Module 2

Section Assignment 2.1

Quadratic Functions

Part A: Multiple Choice (1 mark each)

Place the letter of the correct answer in the space to the right.

1. Determine the vertex for the equation f(x) = –3x2 – 4.

a) (–3, –4) b) (0, –3)

c) (0, –4) d) (–3, 0)

For questions 2 to 4, use the function f(x) = –1(x + 2)2 + 3.

2. State the axis of symmetry.

a) x = 3 b) x = –2

c) y = –2 d) y = 3

3. State the domain.

a) {x|x ∈ ℜ} b) {x|y ∈ ℜ}

c) {x|x ∈ ℜ, y ≤ –1} d) {y|y ∈ ℜ, y ≤ –1}

4. Find the y-intercept.

a) 3 b) –3

c) 1 d) –1

5. If the graph of a function h(x) = 2(x – 1)2

– 1 ismoved 2 units to the right and 3 units up, state theequation of the resulting graph.

a) h(x) = 2(x – 3)2 + 2 b) h(x) = 2(x + 3)2 + 2

c) h(x) = 2(x – 2)2 + 3 d) h(x) = 2(x – 1)2 – 1


Module 2

Total Value: 70 marks(Mark values in brackets)

Module 2

6. For what value of k is the expression x2

– 5x + k a perfect square trinomial?

a) b)

c) d)

7. Which curve is the graph of y = 4x – x2?

a) b)

c) d)

8. Describe the line illustrated using interval notation.

a) (–∞, –2) ∪ [0, 3] b) (–∞, 3]

c) {x|x < –2 or 0 ≤ x ≤ 8} d) {x|x ≤ 3}

9. What is the y-intercept of y = ax2 + bx + c.

a) a b)

c) b d) c

−ba2

−2 0 3l l

y

xO

x

y

O

x

y

O

x

y

O

52

254

−254

−52

94 Section1 Assignment 2.1 Principles of Mathematics 11

10. Which of the following equations best describes theaccompanying graph?

a) y = (x + 2)2 + 1

b) y = (x + 2)2 – 1

c) y = (x – 2)2 + 1

d) y = (x – 2)2 – 1

Part B (marks in brackets)

Answer the following questions in the space provided. To receivefull marks, solutions should be legible, well presented, andmathematically correct.

1. Consider the relation defined by {(–1,4), (0,3), (2,4), (–1,3)}.

a) Give the domain of the relation.

b) Give the range of the relation.

c) Is the relation a function? Explain your answer.

2. Consider the relation graphed below.

a) Give the domain.

b) Give the range.

c) Is this a function?Explain youranswer.

3. Indicate if each of the following is a quadratic function or not.

a) f(x) = 7x – 2 + x2 b) y = (x + 2)2 – 1

c) d) y2 + x = 4= 2

32 –y

x

2 4

-2

x

y


Module 2

(3)

(3)

(4)

4. Give the zeros for each of the following functions.

a) f(x) = x2 – 4x + 3

b) g(x) = 2x2 – 8

5. Given the graph below, answer the following questions.

a) What are the coordinates of the vertex?

b) What is the domain?

c) What is the range?

d) What is the equation of the axis of symmetry?

x

y


Module 2

(4)

(1)

(1)

(1)

(1)

e) State whether it has a maximum or minimum value andwhat that value is.

f) What are the x-intercepts?

g) What is the y-intercept?

h) Name the coordinates of the reflection of the point (0, 3).

6. Consider the parabola y = 2x2 – 4x – 6.

a) Write the function in the form y = a(x – h)2 + k.


Module 2

(2)

(1)

(1)

(1)

(3)

b) Find the coordinates of the vertex.

c) Find the y-intercept.

d) Find the x-intercepts.

e) State the domain and range.

f) Write an equation for the axis of symmetry.


Module 2

(2)

(1)

(1)

(2)

(1)

g) Sketch the graph.

7. What is the maximum rectangular area that can be enclosedby 120 m of fencing if one of the sides of the rectangle is anexisting wall? (Show algebraically.)


Module 2

(2)

(4)

existing wall

x x

120 − 2x

8. Peter Peter has 40 pumpkins to sell. He will sell them all ifhe sets his price at $1.00 each, but will sell one less pumpkinfor each 5 cent increase in price. What price should hecharge in order to maximize his earnings? (Showalgebraically.)

9. Find the equation of the quadratic function with a vertex atV(1, 3) passing through the point P(2, 1). Write your answer

in the form y = ax2 + bx + c.


Module 2

(5)

(3)

10. Given f(x) = a(x – h)2

+ k, what is the effect of the followingon the graph of the function:

a) size of a

b) sign of a

c) value of h

11. Find the x-intercept and y-intercept of the line 2x – 3y – 12 = 0.


Module 2

(3)

(2)

12.Consider the parabola y = x2

– 2x – 15

a) Give the x-intercepts.

b) Give the vertex.

c) Give the y-intercept.

d) What is the equation of the axis of symmetry?

e) Sketch the graph showing three points.

Turn to the Module 2 Section Assignment Answer Key to mark your work.


Module 2

(2)

(2)

(1)

(1)

(2)

(Total: 70)

y

x

Documents

Module 2: Algebra - Open School BCmedia.openschool.bc.ca/ocrmedia/pmath11/pdf/pma11_… · · 2015-06-22Module 2: Algebra Section 1: Quadratic Functions ... Module 2, Section 1