Quadratics 1 - Test

Solutions

1. Factor using diagram

a) x2 + 11x + 30

= (x + 6)(x + 5)

b) 6x2 – 17x + 5

= (3x – 1)(2x – 5)

* x 6

x x2 6x

5 5x 30

* 3x -1

2x 6x2 -2x

-5 -15x 5

1. Factor using diagram [2 – KU each]

c) 8x2y2 + 16xy – 36x3y

= 2·4·xxyy + 4·4xy - 4·9xxxy

= 4xy(2xy + 4 – 9xx)

= 4xy(2xy + 4 – 9x2)

* 2x -7

2x 4x2 -14x

+7 14x -49

d) 4x2 – 49

= (2x – 7)(2x + 7)

2. Complete the table for the parabola.[4 – KU]

Axis of Symmetry:

x = -2

Optimal Value:

Maximum y = 4

Vertex: (-2, 4)

Zeros: (-4, 0) and (0, 0)

3. Find the vertex for the quadratic relation.[3 – KU]

y = 2(x + 1)(2x – 6)

0 = 2(x + 1)(2x – 6)

x + 1 = 0

x = -1

2x – 6 = 0

2x = 6

x = 3

Axis of Symmetry: x = (-1 + 3) / 2 = 1

Optimal Value y = 2(1 + 1)(2(1) – 6)

= 2(2)(-4)

= -16

Therefore the vertex is (1, -16).

4. Is the point (-2, -5) on the parabola?[2 – App]

y = 2x2 + 2x - 12

LS = y

= -5

RS = 2x2 + 2x – 12

= 2(-2)2 + 2(-2) – 12

= 2(4) -4 – 12

= 8 – 4 – 12

= -8

Therefore, since LS ≠ RS … the point (-2, -5) is NOT on the parabola.

5. Write the equation of a parabola with one zero. [1 – TI]

• y = (x – 3)2

• y = (x + 2)2

• y = x2

• y = -x2

• y = (x + 10000)2

• Etc…

6. Explain the difference between roots and zeros. [2 – KU]

• Roots – equation is form 0 =

• Roots – 2 points on a parabola with the same y value (equal distance from the axis of symmetry)

• Zeros – equation is form y =

• Zeros are x-intercepts (where parabola crosses the x-axis.

7. Find the roots of the equation.[3 – KU]

10 = 3x2 + x

0 = 3x2 + x – 10

0 = (x + 2)(3x – 5)

* 3x -5

x 3x2 -5x

2 6x -10

x + 2 = 0

x = -2

3x – 5 = 0

3x = 5

x = 5/3

Therefore, the roots are -2 and 5/3.

8. Create a cuba-link diagram for a parabola with a second difference of 1. [2 – TI]

X Y

1 -22

2 -16

3 -11

4 -7

5 -4

6 -2

+6

+5

+4

+3

+2

-1

-1

-1

-1

9. Application Question #1 [6 – App]Miss. Nieuwenhuis’ 2D Math class created a fabulous cuba-link video that demonstrates their use in

the 2D classroom. They sold 115 videos for $4 each. They were curious if the price of $4 maximized their revenue. They researched and discovered for every 50 cent increase in price, they would sell 10 less videos.

Did they maximize their revenue? Discuss your findings?

# of videos (x)

Price (y) Revenue

x·y

115 4 460

105 4.50 472.50

95 5 475

85 5.5 467.50

Slope = 05.0

10

5.0

run

rise

Linear Equation: y = -0.05x + b

5 = -0.05(95) +b

5 = -4.75 + b

9.75 = b

y = -0.05x + 9.75

Equation for Revenue:

R = x(-0.05x + 9.75)

9. Application Question (continued)

Find Zeros: 0 = x(-0.05x + 9.75)

x = 0

0 = -0.05x + 9.75

-9.75 = -0.05x

195 = x

Find Axis of Symmetry: x = (0 + 195) / 2 = 97.5

Optimal Value: R = 97.5(-0.05(97.5) + 9.75)

R = 97.5(4.875)

R = 475.3125

Therefore, they did not maximize their profit. They would have needed to sell 97.5 (or approximate 98) videos at a price of $4.88 in order to make a maximum profit of $475.31. However, their profit of $460 was not that far off, so they did quite well with their price of $4.

10. Application Question [6 – App]

Miss. Nieuwenhuis’ 2D Math class wanted to determine the quadratic equation that would represent something thrown out of our classroom window. So, we grabbed Chris B’s cell phone and pitched it out the window. Our observations determined the following:

• The window’s height is 22m• The cell phone landed on the ground after

5s• The maximum height of the cell phone was

26m after 2s

There are two ways to approach this problem:

1. Use the two points (0,22) and (4, 22)

2. Use the zeros (-1, 0) and (5, 0)

You can determine the other point using the axis of symmetry.

Method #1 – (0, 22), (4, 22) Method #2 – (-1, 0), (5, 0)

224

22)4(

1

44

22426

22)2)(2(26

22)42)(2(26

22)4)((

2

xxy

xxy

a

a

a

a

a

xxay

9

130

9

104

9

26

)54(9

26

)5)(1(9

269

26

926

)3)(3(26

)52)(12(26

)5)(1(

2

2

xxy

xxy

xxy

a

a

a

a

xxay

These equations are not equivalent only because Miss. N chose bad numbers!

Question 11 – Alternate [4 – App]The city of Ottawa is talking about building a bridge between Riverside South and Barrhaven to try to

solve some traffic issues. They know that the maximum height the bridge will be is 20m, and that is needs to span a distance of 110m.

a) Write a quadratic equation that represents the bridge.b) They want to add in two additional supports that are 20m away from the center of the bridge. How

high will these supports need to be (to 2 decimal places).

a) Method #1

Zeros: (-55, 0) and (55, 0)

a) Method #2

Zeros: (0, 0) and (110, 0)

)55)(55(605

4605

43025

20

302520

)55)(55(20

)550)(550(20

)55)(55(

xxy

a

a

a

a

a

xxay

)110(605

4605

4

)55)(55(20

)11055)(55(20

)110)((

xxy

a

a

a

xxay

b) Substitute in a value for x.

If you used method #1 you can use x=20 or x=-20.

If you used method #2 you can use x = 35 or x = 75.

Either way … you get a support height of 17.36m my

y

y

36.17

)35)(75(605

4

)5520)(5520(605

4