49Cumulative Review Chapters 1-5—Solutions©P DO NOT COPY.

CUMULATIVE REVIEW Chapters 1-5 ,pages 419—424

1. Identify this series as arithmetic or geometric, then determine itssum.

2. Rationalize the denominator.

a) b)

3. Factor each polynomial expression.

a) b)

c) 9(2x - 3)2- 64(y + 2)2

2x2+ 3.5x - 16x2

- 14x - 12

3

2√

7 - 4

2√

5 + 3

5√

2 - 2√

5

4√

3

2

-5 -54 -

516 -

564 - . . .

1

The series is infinite and geometric because each term is times thepreceding term.

Use: Substitute: t1 � �5,

Sˆ

� �6.6

�5

1�14

Sˆ

�

r �14S

ˆ�

t1

1� r

14

�5√

6 � 2√

1512

�(5

√2 � 2

√5)

4√

3#

√3√3

Guess and test factors of3 with factors of �6.� 2(3x � 2)(x � 3)

� 2(3x2 � 7x � 6)Guess and test factors of 4 withfactors of �2.� 0.5(4x � 1)(x � 2)

� 0.5(4x2 � 7x � 2)

�4√

35 � 6√

7 � 8√

5 � 1211

�4√

35 � 6√

7 � 8√

5 � 1220 � 9

�(2

√7 � 4)

(2√

5 � 3)#

(2√

5 � 3)

(2√

5 � 3)

� (6x � 8y � 7)(6x � 8y � 25)

� [3(2x � 3) � 8(y � 2)] [3(2x � 3) � 8(y � 2)] � [3(2x � 3)]2 � [8(y � 2)]2

05_ch05_pre-calculas11_wncp_solution.qxd 5/28/11 11:47 AM Page 49

4. Solve by factoring, then verify each solution.

a) b)

5. Solve each equation by completing the square.

a) b)

6. Solve each quadratic equation.

a) b)

7. For the equation , determine the values of k if theequation has no real roots.

4x2- 3x + k = 0

1.2x2= 2.3x + 4.7x2

+ 12x + 8 = 0

-5x2+ 7x + 4 = 0

14 x2

+12x = 1

√x + 2 + 4 = x5x2

+ 15 = x(x + 17)

50 Cumulative Review Chapters 1-5—Solutions DO NOT COPY. ©P

x � �1_√

5

x � 1 � —√

5

(x � 1)2 � 5 x2 � 2x � 1 � 4 � 1

x2 � 2x � 4

x � 0.7_√

1.29

x � 0.7 � —√

1.29

(x � 0.7)2 � 1.29 x2 � 1.4x � 0.49 � 0.8 � 0.49 x2 � 1.4x � 0.8 � 0

Use:

Substitute: a � 1, b � 12, c � 8

x � �6_2√

7

x ��12_4

√7

2

x ��12_

√112

2

x ��12_

√122 �4(1)(8)

2(1)

x ��b_

√b2 �4ac

2aUse:

Substitute: a � 1.2, b � �2.3, c � �4.7

x �2.3_

√27.85

2.4

x �2.3_

√(�2.3)2 �4(1.2)(�4.7)

2(1.2)

x ��b_

√b2 �4ac

2a

1.2x2 � 2.3x � 4.7 � 0

x � 3 or x � 1.25 (x � 3)(4x � 5) � 0 4x2 � 17x � 15 � 0

5x2 � 15 � x2 � 17x

When x � 2, L.S. R.S.When x � 7, L.S. � R.S.The solution is x � 7.

�

x � 7 or x � 2 (x � 7)(x � 2) � 0 x2 � 9x � 14 � 0

x � 2 � x2 � 8x � 16

√

x � 2 � x � 4

In b2 � 4ac, substitute: a � 4, b � �3, c � kb2 � 4ac � (�3)2 � 4(4)(k)

� 9 � 16k

k> 916

16k>9 For no real roots, 9 � 16k<0

05_ch05_pre-calculas11_wncp_solution.qxd 5/28/11 11:47 AM Page 50

8. A person dives from a 10-m high platform. His approximate heightabove the water, h metres, after t seconds is modelled by thisfunction: h = -4.9t2

+ 2.2t + 10. Use graphing technology todetermine the time, to the nearest tenth of second, when the diverhits the water. How could you check that your answer is correct?

9. For the quadratic function y = -3(x - 2)2+ 5

a) Identify:

i) the coordinates of the vertex ii) the domain

iii) the range

iv) the equation of the axis of symmetry

v) the intercepts, to the nearest tenth where necessary

b) Sketch a graph of the function.

4

51Cumulative Review Chapters 1-5—Solutions©P DO NOT COPY.

x0 2 4 6

2

�2

4

6y

y � �3(x � 2)2 � 5

In a graphing calculator, input: y � �4.9x2 � 2.2x � 10The diver hits the water when h � 0; that is, the positive horizontalintercept. This is x � 1.6705. . .The diver hits the water after approximately 1.7 s.I check the answer by substituting t � 1.6705. . . in the equation of thefunction to verify that the solution satisfies the equation.

The coordinates are: (2, 5) The domain is: x ç �

The graph opens down.The range is: , y ç �y ◊ 5

The equation is: x � 2

In For the x-intercept, substitute y � 0

The x-intercepts are: and For the y-intercept, substitute x � 0y � �3(0 � 2)2 � 5y � �7

x � 0.7x � 3.3

x � 2—√

53

x � 2 � —√

53

(x � 2)2 �53

0 � �3(x � 2)2 � 5

y � �3(x � 2)2 � 5

Plot a point at the vertex, then sketch a graph congruent to .y � �3x2

05_ch05_pre-calculas11_wncp_solution.qxd 5/28/11 11:47 AM Page 51

10. The graph of a quadratic function passes through J(-3, 2) and K(0, 11), and the equation of its axis of symmetry is x = -2.Determine an equation of the function.

11. The arch of a bridge over a river has the shape of a parabola. Thebase of the arch is 30 m wide and its maximum height is 6 m.

a) Determine an equation that models the arch.

b) Determine the height of the arch at a horizontal distance of 6 mfrom its centre.

52 Cumulative Review Chapters 1-5—Solutions DO NOT COPY. ©P

30 m

6 m

Sample response: Visualize a coordinate grid superimposed on the arch, with the origin at one end of the base of the arch.An equation of the arch has the form:The vertex of the parabola has coordinates (15, 6).Substitute:

An equation that models the arch is: y � �275 x (x � 30)

a � �275

6 � �225a6 � a(15)(15 � 30)

x � 15, y � 6

y � ax(x � 30)

One point that is 6 m from the centre of the arch has x-coordinate 9.Substitute in the equation of the arch.

The arch is 5.04 m high.y � 5.04y � �

275 (9)(9 � 30)

x � 9

An equation has the form: y � a(x � p)2 � qFrom the equation of the axis of symmetry: p � 2So, an equation has the form: y � a(x � 2)2 � qSubstitute: x � �3, y � 2

2 � a(�3 � 2)2 � qa � q � 2 ➀➀

Substitute: x � 0, y � 1111 � a(0 � 2)2 � q

4a � q � 11 ➁➁

Solve the linear system formed by equations ➀➀ and ➁➁.Subtract equation ➀➀ from equation ➁➁.3a � 9

a � 3Substitute a � 3 in equation ➀➀.q � �1 So, an equation is: y � 3(x � 2)2 � 1

05_ch05_pre-calculas11_wncp_solution.qxd 5/28/11 11:47 AM Page 52

12. Write the equation y = 0.5x2- 4x + 7 in standard form; then

identify the equation of the axis of symmetry and the coordinates ofthe vertex of the graph of the function.

13. Graph each inequality.

a)

b) c) y ≥ -(x + 1)2- 3x - 2y > 4

3x2- 2x ≤ 21

5

53Cumulative Review Chapters 1-5—Solutions©P DO NOT COPY.

x0�2 4

�4

�6

yGraph of x � 2y � 4

x � 2y � 4x

20�2�4 4

�2

�4

�6

yGraph of y � �(x � 1)2 � 3

The equation of the axis of symmetry is , and the coordinatesof the vertex are (4, �1).

x � 4y � 0.5(x � 4)2 � 1

y � 0.5(x2 � 8x � 16) � 8 � 7

y � 0.5(x2 � 8x � 16 � 16) � 7

Rearrange the inequality.

Graph the related function.When When Draw a broken line. Shadethe region below the line.

y � 0, x � 4x � 0, y � �2

y<0.5x � 22y<x � 4

The graph of the related quadraticfunction is congruent to y � �x2 andhas vertex (�1, �3).The curve is solid and the regionabove is shaded.

�1 1�2�3�4�5 0 2 3 4 5

Solve:

When , such as x � �3, L.S. � 33; R.S. � 21;

so does not satisfy the inequality.

When , such as x � 0, L.S. � 0; R.S. � 21;

so x � 0 does satisfy the inequality.

The solution is: , x ç ��73 ◊ x ◊ 3

�73<x<3

x � �3

x<�73

x � �73 or x � 3

(3x � 7)(x � 3) � 03x2 � 2x � 21 � 0

05_ch05_pre-calculas11_wncp_solution.qxd 5/28/11 11:47 AM Page 53

14. Use technology to graph each system of equations. Write thecoordinates of the points of intersection to the nearest tenth.

a) y = 2x - 6 b) y = - 0.5x 2+ 4

y = 0.5x 2- 3x y = 2x 2

+ x - 6

15. Solve this system of equations algebraically.

y = 2x2- 4x - 2.5

y = -0.5(x + 1)2+ 2

54 Cumulative Review Chapters 1-5—Solutions DO NOT COPY. ©P

Input each pair of equations in a graphing calculator.The coordinates of the The coordinates of the points of intersection are points of intersection areapproximately: approximately:(1.4, �3.2), (8.6, 11.2) (�2.2, 1.6), (1.8, 2.4)

From equation ➁➁, substitute in equation ➀➀.

So, or Substitute each value of x in equation ➁➁.When : When

The solutions are: (�0.8, 1.98) and (2, �2.5)y � �2.5y � 1.98y � 2(2)2 � 4(2) � 2.5y � 2(�0.8)2 � 4(�0.8) � 2.5

x � 2x � �0.8

x � 2x � �0.8

(5x � 4)(x � 2) � 0 5x2 � 6x � 8 � 0

2.5x2 � 3x � 4 � 0 2x2 � 4x � 2.5 � �0.5x2 � x � 0.5 � 2

2x2 � 4x � 2.5 � �0.5(x � 1)2 � 2

y � 2x2 � 4x � 2.5

➀➀

➁➁

05_ch05_pre-calculas11_wncp_solution.qxd 5/28/11 11:47 AM Page 54