1

2

3

4

5

m0.5

m

2,3

2.5

?

The graphs to the right illustrate a fountain shooting a jet of water with a laser beam illuminating the jet in two places, one of them being (2 m ,3 m).

Determine the height of the other point illuminated by the laser.

Equation of the jet of water

Solution of the system

The height of the other point illuminated by water is 1.75 meters.

x1 = 0.5; x1 = 2.5; (x,y) = (2,3)

y = a(x – x1)(x – x2)

3 = a(2 – 0.5)( 2 – 2.5)

3 = a(1.5)(-0.5)

3 = -0.75a

a = -4

y = -4(x – 0.5)(x – 2.5)

y = -4(x2 – 2.5x – 0.5x + 1.25)

y = -4x2 + 12x – 5

Equation of the laser beam

122

0213

3210

12

12

2211

xxyy

m

),()y,x();,()y,x(

1

11

xy

bmxy

b;m

x + 1 = -4x2 + 12x – 5

4x2 - 12x + x + 5 + 1 = 0

4x2 - 11x + 6 = 0

4x2 - 3x - 8x + 6 = 0

x(4x – 3) – 2(4x – 3) = 0

(4x – 3)(x – 2) = 0

4x – 3 = 0 OR x – 2 = 0

243

xORx

m.y

y

xy

751

143

1

The proportion of men to women employed by a company has varied over the years of operation. The number of men was 428 in the year 2000 and reached a maximum of 500 three years later following a second degree curve. The number of women in 2000 was 100 and increased at a rate of 25 per years after that. men

wom

en

year

# of employe

es

x: number of years after 2000

y: number of employees

If the trends continue, in what year will the number of men and women be the same? How many will be employed by the company at that time?

m = 25; b = 100

y = 25x + 100

Equation for women

Equation for men

Vertex: (3,500); Other point: (0,428)

y = a(x – h)2 + k

428 = a(0 – 3)2 + 500

428 – 500 = 9a

9a = -72

a = -8

y = -8(x – 3)2 + 500

y = -8(x2 – 6x + 9) + 500

y = -8x2 + 48x - 72 + 500

y = -8x2 + 48x + 42825x + 100 = -8x2 + 48x + 428

8x2 - 48x + 25x – 428 + 100 = 0

8x2 - 23x – 328 = 0

8x2 - 64x + 41x – 328 = 0

8x(x – 8) + 41(x – 8) = 0

(x – 8)(8x + 41) = 0

x – 8 = 0 OR 8x + 41 = 0

x = 8 OR x = -5.125

Solution of the System

y = 25x + 100

y = 25(8) + 100

= 200 + 100 = 300 employees

After 8 years, there were 300 men and 300 women employed for a total of 600 employees

A cold front sweeps into the town of Amos causing the temperatures to drop steadily from 1°C at 2h to -4°C at 21h and continue. As yet unaffected by this cold front, on the same day Gaspe experiences temperatures indicated in the table to the right.

4

8

12

16

20

24

-2.4°C

0.4°C

2.4°C

3.6°C

4°C

3.6°C

GASPE

Time Temp

For how long is the temperature of Gaspe warmer than that of Amos during that time?x = time

y = temperature

Equation for Amos

195

22114

42112

12

12

2211

xxyy

m

),()y,x();,()y,x(

19295

29519

1910519

1051919

2511921

195

1

1

xy

xy

xy

xy

)x()y(xyxxyy

m

Solution of the System

Vertex: (20,4); Other point: (8, 0.4)

y = a(x – h)2 + k

0.4 = a(8 – 20)2 + 4

0.4 – 4 = 144a

144a = -3.6

a = -0.025

y = -0.025(x – 20)2 + 4

y = -0.025(x2 – 40x + 400) + 4

y = -0.025x2 + x - 10 + 4

y = -0.025x2 + x - 6

Equation for Gaspe

114194750295

6025019

295

2

2

xx.x

xx.x

0.475x2 - 19x – 5x + 114 + 29 = 0

0.475x2 - 24x + 143 = 0

a = 0.475; b = -24; c = 143

Δ = b2 – 4ac

= (-24)2 – 4(0.475)(143)

= 476 – 271.7

=304.3

916643950566

9504441

950441724

95044172447502

3304242

21

21

21

.xOR.x..

xOR..

x

..

xOR.

.x

).(.)(

ab

x

Gaspe was warmer than Amos between 6.91 hours and 43.6 hours: 43.6 – 6.9 = 36.1 hours.

A display is constructed to take pictures of kids with Santa Claus.

Two symmetrical braces extend up from the floor to support the seat and attach to a piece of board above the seat. The display is 2.25 meters high and 1.5 meters wide at its base. The braces are 42 cm apart at the floor and 58 cm apart at the seat which is 48 cm above the floor.

How long are the braces? How far apart are the braces at the top where they attach to the piece of board?

x

2.2

5 m

1.5 m

42cm

?

y

seat

bracesEquation of the brace

6080480

2102900480

4802900210

12

12

2211

.

...

.xxyy

m

).,.()y,x();,.()y,x(

2616

2106012100

6

1

1

.xy

).x()y(.x

y

xxyy

m

Vertex: (0, 2.25); Other point: (0.75, 0)

y = a(x – h)2 + k

0 = a(0.75 – 0)2 + 2.25

0 – 2.25 = 0.5625a

0.5625a = -2.25

a = -4y = -4x2 + 2.25

Equation of the parabola

Solution of the System

6x – 1.26 = -4x2 + 2.25

4x2 + 6x - 3.51 = 0

a = 4; b = 6; c = -3.51

Δ = b2 – 4ac

= 62 – 4(4)(-3.51)

= 36 + 56.16

= 92.16

9514508

615863

8696

869642

169262

21

21

21

.xOR.x

.xOR

.x

.xOR

.x

)(.

ab

x

y = 6x – 1.26

= 6(0.45) – 1.26

= 2.7 – 1.26 = 1.44

End points of right brace: (0.21,0); (0.45,1.44)

x

2.2

5 m

1.5 m

42cm

?

y

seat

braces

(0.21,0)

(0.45,1.44)

Length of right brace:

(x1, y1) = (0.21,0);

(x2, y2) = (0.45,1.44)

m..

..

).().(

).()..(

)yy()xx(D

46113122

0736205760

441240

044121045022

22

212

212

x

2.2

5 m

1.5 m

42cm

?

y

seat

braces

(-0.45,1.44)

(0.45,1.44)Length of right

brace:

=|0.45-(-0.45)|

=0.9 m

End points of right brace: (-0.45,1.44); (0.45,1.44)

A new business is starting and the manager is analyzing his expenses and revenues. The monthly expenses start at $1950 and increase to a maximum of $2550 after 5 months at which point the expenses start to decline. The revenues increase at a constant rate of $30 per month and reach $2256 after 12 months.

At what point will there be a zero deficit? ( i.e. revenue = expenses)

expenses

revenue

month

$

m = 30; ( x1 , y1) = (12,2256)Equation for revenue

Equation for expensesVertex: (5,2550); Other point:

(0,1950)

y = a(x – h)2 + k

1950 = a(0 – 5)2 + 2550

1950 – 2550 = 25a

25a = -600

a = -24189630

225636030

12302256112

225630

1

1

xy

xy

)x()y(x

yxxyy

my = -24(x – 5)2 + 2550

y = -24(x2 - 10x + 25) + 2550

y = -24x2 + 240x - 600 + 2550

y = -24x2 + 240x + 1950Solution of the System

30x + 1896 = -24x2 + 240x + 1950

24x2 - 240x + 30x – 1950 + 1896 = 0

24x2 - 210x – 54 = 0

4x2 - 35x – 9 = 0

4x2 + 1x - 36x – 9 = 0

x(4x + 1) – 9(4x + 1) = 0

(4x + 1)(x – 9) = 0

4x + 1 = 0 OR x - 9 = 0

x = -0.25 OR x = 9

There is a zero deficit after 9 months.

Plans for a highway indicate that it will cross a river in 2 places.

Use the coordinates from the graph provided to determine the length of the section of the highway to be built between the bridges.

(9,4.2)

E

N

(5,1)

53

0,

1 unit = 1 km

River

HighwayBridges

Equation for Highway

Equation for River

Vertex: (5,1); Other point: (9,4.2)

y = a(x – h)2 + k

4.2 = a(9 – 5)2 + 1

4.2 - 1 = 16a

16a = 3.2

a = 0.2

40963

096024

249600

12

12

2211

.

.

..xxyy

m

).,()y,x();.,()y,x(

6040

6040

.x.y

bmxy

.b;.m

y = 0.2(x – 5)2 + 1

y =0.2(x2 – 10x + 25) + 1

y = 0.2x2 – 2x + 5 + 1

y = 0.2x2 – 2x + 6Solution of the System

0.2x2 – 2x + 6 = 0.4x + 0.6

0.2x2 – 2x – 0.4x + 11 – 0.6 = 0

0.2x2 – 2.4x + 5.4 = 0

x2 – 12x + 27 = 0

(x – 3)(x – 9) = 0

x – 3 = 0 OR x – 9 = 0

x = 3 OR x = 9

y = 0.4(3) + 0.6

= 1.2 + 0.6 = 1.8

(3, 1.8)y = 0.4(9) + 0.6

= 3.6 + 0.6 = 4.2

(9, 4.2)

km..

.

).()(

)..()(

)yy()xx(D

4667641

76536

426

81243922

22

212

212