Name: ____________________________ Date: __________________

Unit #4 Test – Vertex Form VF1 I can identify the vertex and axis of symmetry and explain the

roles of a, h, and k as transformations applied to the base curve y = x2 to create y = a(x – h)2 + k.

13 1. What are two things that “a” in the equation of a parabola tell us? (2 marks)

2. For the following equations of quadratics in vertex form, state the vertex and the axis of symmetry. (4 marks)

a. y = (x +3)2 + 5 b. y = -6x2 – 3

3. For the following equations of quadratics in vertex form, describe the sequence of transformations that you would apply to the graph of y = x2. (7 marks)

a. 32

� � 21y (x ) +2

b. b. y = 3(x + 8)2 + 9

a If there Is a Stretchor Compression

• The direction of opening

Vertex :f3 , 5) Vertex (9-3)A. 0.5 .

×= -3 A. 0 's

.×=o

Vertical compression , factor is }reflection in x - axis

�4� translated. }auinptsunfhtVertical stretch factor of 3

�3� translated 8 units lefttranslated 9 unite Up

Name: ____________________________ Date: __________________

VF2 I can sketch the graph of y = a(x – h)2 + k by applying transformations to the graph y = x2.

5 4. Graph 1 � �2y 2(x-3) by applying the transformations to the base curve y = x2 in

the correct order. (5 marks)

* in

* *§

¥±¥as a nooare*•o

.fm#*M.y..uxs5t'

=D Dos

.

*onose oa *

. .

Name: ____________________________ Date: __________________

VF3 I can determine the equation, in vertex form y = a(x – h)2 + k, of a given parabola.

21 5. Determine the equation of a quadratic relation in vertex form given that the vertex

is at (5, 6) and it passes through the point (1, 18). Show your work. (4 marks)

6. Express y = -5(x + 1)2 + 9 in standard form. Show your work. (3 marks)

7. A campground charges $20.00 to camp for one night. They average 56 people each night. A recent survey indicated that for every $1.00 decrease in the nightly price, the number of camping sites rented increases by seven. What price will maximize nightly revenue? What is the greatest revenue? Show your work. (4 marks)

h, K X Y

y=a(xh ) 'tk

18 =a( 1- 5) 2+6 12=16 a

18=4-45+6 l,÷=a18.61%8169+6

a.%==aa

Y=oK×→5t6

Expand I simplify

y.SI#D2t9y=-SHtDH+Dt9

=- 51×41×+1×+1 )t9

=- sWt2xtDt9

=- 5×2 - lox - stg Y= -5×2-10×+4

Let × represent # 4$11 price

R=#of people)( price per person)decreases

13=156+7/20 . $K)56+7,1=0 20-4=0

7X = -56 20=1×1×20

11=-8a o ,s

.

x= 2021

X =

20¥× =

12g

X = 6

R "#FalseYDCYIIYD

= (56+ 42) ( 20 - 6)= of8) ( l 4)

=$1372

Name: ____________________________ Date: __________________

8. The underside of a bridge forms a parabolic arch. The arch has a maximum height of 36 m and a width of 48 m. Can a sailboat pass under the bridge, 6 m from the axis of symmetry, if the top if its mast is 33 m above the water? Justify your solution by showing all of your calculations. (5 marks)

9. Express y = 5x2 – 10x + 8 in vertex form using partial factoring. Show your work. (5 marks)

so

( 24,36 )

co

#f)Vertex ( h ,k)

T.net#(l8i9( 24,36)

-248 - Point ( x ,y)( 0,9

×=ota48_ y=a( xhttkX= 24 0 =a( 0-245+36

i. y =- 0.062541-24336 0=576 a +36

- 36 = 576 a

-36/5,6 = a a=- 0.0625

%g=-0.062430-245+36y

= 5×2-10×+8y

= -0,062465+36

Y= -0.062436436

Y= -2,25+36 LetY=8y=33.75

8=5×2-10×+833,75>33

g. q = 5×2 . IOX÷ sailboat fits

0=5×2 .

10×0=5×(-1-2)Ya In T

X=1 5-1=0 x -2=0

Y. 5×2-10×+8 X=0 x=2

Y=5( 1)2-100+8 ( 98 ) ( 2,9

y=s -10+8

A .gs . -1=0+23y=3 verteh,3)

y=s( x-D 't3× = '

Name: ____________________________ Date: __________________

SQ2 I can express y = ax2 + bx + c in the form y = a(x – h)2 + k by completing the square in situations involving no fractions.

7

10. Complete the square to write the quadratic relation y = 7x2 – 56x + 16 in vertex form. (4 marks)

11. Suppose the quadratic relation y = -4x2 + 3x + 9 was written in the form y = a(x – h)2 + k. What is the value of a? (1 mark)

12. What values would you add and subtract to make the expression x2 + 28x a perfect square? (2 marks)

SQ3 I can develop the quadratic formula and use it to interpret real or non-real roots of quadratic equations.

6

13. Use the quadratic formula to determine the solutions to the equation 8x2 – 2x – 16 = 0. (6 marks)

2 42

b b acxa

� r �

y -7×2-56×+16

⇐7ft.fr/j+l6EEIyY=7(x2-8×+16-16 ) +16 76

Y=7(x2 -8×+16 ) -112+16

yes. ' # 4)

2 -96 yefyyy

a=4

add and subtract 196↳

EH= 142= 196

000agex. t⇒±fgaiY

" "

tox=2 t.fm

=

×=2± -4+5,2

ix.2t.rs#2tydx=syyT6x=2yx=.xhx=2tb2I

x. 1.54 x= -1.3

Name: _______________________ Date: _____________

3.5 cm

4.5 cmx

x

12.2 cm

23q

Unit 7:Trigonometry–Quiz #14

TR2 I can define the sine, cosine, and tangent ratios and use them and the Pythagorean theorem to solve for side lengths and angles in right triangles.

10 1. Solve for x to one decimal place. (3 marks)

a) 7xsin39 q x ≐ ____________________

b) 10xcos65 q x ≐ ____________________

c) x31tan49 q x ≐ ____________________

2. Solve for �A to the nearest degree. (3 marks)

a) 85sinA �A ≐ __________________

b) 2213cosA �A ≐ __________________

c) 2519tanA �A ≐ __________________

3. Calculate x to the nearest tenth of a centimetre. Show your work. (2 marks)

4. Calculate x to the nearest degree. Show your work. (2 marks)

p x=7(sin39D4.4

a ×= 101<0565)4.2

×= 3¥49 26.9

A. sii '( %)zq°

< A. cos"('%2)

g4:(< A # an"µbs)

HYP ldecvid place* OPP SOH caittoacos 25=491

hyp= 12.2 cos23°=±adj 12.2adj =×

x= 12,2425)°PP So # CAH TOA X= 11.2cm

→|↳Popp -3,5

adj adj=4 's< x=tan"(3§s)

tan×=o#adj

tax =3=< X= 38

'

4. s

Name: _______________________ Date: _____________

Doorway

Parking Lot

Camera

TR3 I can explore the development of the sine law and cosine law within acute triangles and use them to solve for side lengths and angles.

10

1. Find the measure of angle T to the nearest degree. Show your work. (3 marks)

2. Find the length of x to the nearest tenth of a metre. Show your work. (3 marks)

3. A security camera needs to be set so that its angle of view includes the area from a doorway to the edge of a parking lot. The doorway is 18 m from the camera. The edge of the parking lot is 25 m from the camera. The doorway is 30 from the edge of the parking lot. What angle of view is needed from the camera? (4 marks)

Sino = sinsi35.4-3=22 side lengths sina.es#=zi

+ I opp angle i. sin / qwshe = 0.88176

e=sii' ( 0.88176 )0=620

.

a2sb2ti.2bc@sDx2.82ty2-2tokYkos4i2s.des & X

's 64+121-120

angle between 1/3=185-120 ×=Fscosinelaw X2= 65 X÷8 .|m

30 a a2=b3+E2bc@sA)18

c osA=5ti . a'

-

*zsb -

2 be

A cosa = 257182-302LA=as" (49/900) Tie

< A= g)°

( OSA=44qoo

Name: _______________________ Date: _____________

Name: _______________________ Date: _____________