Station 1
A weight 2 feet above the ground on a carnival strength test is shot in the air with an initial velocity
of 32 ft/sec where t represents the time in seconds the weight is in the air.
Approximately how long does the weight take to reach a height of 10 feet on the way up?
Solve both graphically and another algebraic way.
Station 1 key
Graphically:
Alternate method: (quadratic formula)
10 = −16𝑡 2 + 32𝑡 + 2
−16𝑡 2 + 32𝑡 − 8 = 0 (now divide by -8)
2𝑡² − 4𝑡 + 1 = 0
𝑥=
4 ± √16 − 4(2)(1)
4
𝑥=
4 ± 2√2 2 ± √2
=
= 0.2928 𝑎𝑛𝑑 1.707
4
2
0.2928 represents the way up
Station 2
A rocket is launched from atop a 63-foot cliff with an initial velocity of 102 ft/s.
a. What is the vertical motion equation that represents this situation?
b. How long will the rocket take to hit the ground (not the cliff) after it is launched? Round to the nearest
hundredth. Use your calculator.
c. At what time does the rocket reach its maximum height?
d. There is a bird in the sky 75 feet above the cliff. Is it possible for the rocket to injure the bird? If yes, then
at what time after the rocket is launched should the bird be worried? Use the table to tell between what time
intervals the bird might need to put on a safety helmet.
Station 2 key
a. ℎ = −16𝑡 2 + 102𝑡 + 63
b. use 0 for h, t = 6.94 seconds
c. 3.19 seconds
d. 63 + 75 = 138, so we are looking for the times when the parabola is at 138 feet. If you look at
the table, the y value is 138 between 0 and 1 second and also between 5 and 6 seconds.
Station 3
1. The discriminant of a function is 32. Which could be a graph of the function?
A
C
B
D
2. When does the equation 𝑥 2 = 𝑑 have no solutions?
A always
B when d 0 C Never
D when d 0
3. Determine how many solutions each equation has:
a) 4𝑥 2 − 5𝑥 + 7 = 10
b) −3𝑥 2 − 𝑥 + 4 = 5
c) −𝑥 2 − 2𝑥 − 1 = 0
Station 3 key
1. C
2. B
3. A) 𝑏² − 4𝑎𝑐 > 0 𝑠𝑜 𝑡ℎ𝑒𝑟𝑒 𝑎𝑟𝑒 2 𝑟𝑒𝑎𝑙 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠
B) 𝑏² − 4𝑎𝑐 < 0 𝑠𝑜 𝑡ℎ𝑒𝑟𝑒 𝑎𝑟𝑒 𝑧𝑒𝑟𝑜 𝑟𝑒𝑎𝑙 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠
C) 𝑏² − 4𝑎𝑐 = 0 𝑠𝑜 𝑡ℎ𝑒𝑟𝑒 𝑖𝑠 1 𝑟𝑒𝑎𝑙 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
Station 4
Solve using any method except graphing:
You are helping to install a square patio for a service project. Around the patio will be a
sidewalk that is 3 feet wide. The total area of the patio and sidewalk will be 400 square
feet. What are the dimensions of the patio?
Station 4 key
Solve using square roots:
(𝑥 + 6)2 = 400
√(𝑥 + 6)2 = √400
𝑥 + 6 = ±20
𝑥 = ±20 − 6 𝑠𝑜 𝑥 = 14 𝑎𝑛𝑑 𝑥 = −26
𝑤𝑒 𝑑𝑜𝑛′ 𝑡 𝑢𝑠𝑒 − 26, 𝑠𝑜 𝑥 = 14 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑎𝑡𝑖𝑜 𝑎𝑟𝑒 14 𝑓𝑒𝑒𝑡 𝑏𝑦 14 𝑓𝑒𝑒𝑡
Station 5
Solve using the quadratic formula. Write answers in simplest radical form.
1. 6𝑎2 − 2𝑛 − 8 = −5
2. 4𝑏 2 − 3𝑏 = 10
3. 5𝑓 2 − 𝑓 − 1 = 0
Station 5 key
1. 6𝑎2 − 2𝑛 − 8 = 5
𝑥=
2 ± √4 − 4(6)(−3) 2 ± √76 2 ± 2√19 1 ± √19
=
=
=
12
12
12
6
2. 4𝑏 2 − 3𝑏 = 10
𝑥=
3±√9−4(4)(−10)
3. 5𝑓 2 − 𝑓 − 1 = 0
𝑥=
1±√1−4(5)(−1)
8
10
=
=
3±√169
8
1±√21
10
=
3±13
8
𝑠𝑜 𝑥 = 2 𝑎𝑛𝑑 𝑥 =
−5
4
Station 6
1
The function 𝑔(𝑥 ) is graphed below. If 𝑔(𝑥 ) = (𝑥 + 2)2 − 2, what are the values of x when 𝑔(𝑥 ) = 6?
2
Describe 2 different methods of answering this question.
Station 6 key
Graphically:
When g(x) = 6, what are the x values? What are the x values when y = 6? Look at the graph to find the
appropriate x values at y = 6. The x values are 2 and -6.
Algebraically:
6=
1
(𝑥 + 2)2 − 2
2
1
6 = (𝑥 2 + 4𝑥 + 4) − 2
2
1
6 = ( 𝑥 2 + 2𝑥 + 2) − 2
2
1 2
𝑥 + 2𝑥 − 6 = 0
2
𝑥 2 + 4𝑥 − 12 = 0
(𝑥 + 6)(𝑥 − 2) = 0
𝑥 = −6 𝑎𝑛𝑑 2
Station 7
Solve by completing the square. Round to the nearest hundredth if necessary.
1. 4𝑓 2 + 16𝑓 − 100 = 0
2. 𝑥² + 8𝑥 + 5 = −4
3. 3𝑥² + 6𝑥 + 5 = 12
Station 7 key
1. 4𝑓 2 + 16𝑓 − 100 = 0
𝑓 2 + 4𝑓 − 25 = 0
𝑓 2 + 4𝑓 + 4 = 25 + 4
(𝑓 + 2)2 = 29
√(𝑓 + 2)2 = √29
𝑓 + 2 = ±√29
𝑓 = −2 ± √29 = −7.39 𝑎𝑛𝑑 3.39
2. 𝑥² + 8𝑥 + 5 = −4
𝑥² + 8𝑥 + 16 = −9 + 16
(𝑥 + 4)² = 7
√(𝑥 + 4)2 = √7
𝑥 + 4 = ±√7
𝑥 = −4 ± √7 = −6.65 𝑎𝑛𝑑 − 1.35
3. 3𝑥² + 6𝑥 + 5 = 12
𝑥 2 + 2𝑥 + 1 =
(𝑥 + 1)2 =
7
+1
3
10
3
10
10
𝑥 + 1 = ±√ = −1 ± √ = −2.83 𝑎𝑛𝑑 0.83
3
3
Station 8
Solve using square roots. Write answers in simplest radical form.
Don’t forget to rationalize the denominator!
1. 5𝑚² = 12
2. 9𝑥 2 = 200
3. 3𝑏² = 25
Station 8 key
1. 5𝑚² = 12
√𝑚² = √
12
5
±2√3 √5 ±2√15
𝑚=
∙
=
5
√5 √5
2. 9𝑥 2 = 200
√𝑥² = √
200
9
±10√2
𝑥=
3
3. 3𝑏² = 25
√𝑏² = √
𝑏 = ±√
25
3
25 ±5 √3 ±5√3
=
∙
=
3
3
√3 √ 3