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Honors Algebra 2
MP2 Progress Check Review
1. In a bacterial culture, the number π΅ of bacteria is modeled by the equation π΅ = 15,000π0.27π‘,
where π‘ represents the number of hours since noon.
(a) How many bacteria (nearest whole number) will be present at 5:00 pm?
(b) How many hours (correct to three decimal places) will it take for there to be 120,000
bacteria?
(c) What is the average rate of change of the population over the first 5 hours?
π΅(5) β 57861
120,000 = 15,000π0.27π‘
8 = π0.27π‘
0.27π‘ = ln 8
π‘ β 7.702 hrs
π΅(5) β π΅(0)
5 β 0=
57861.38296 β 15000
5β 8572 bacteria/hour
KEY
2
2. Solve each equation. Your answer must be exact or expressed as a decimal to at least three
places beyond the decimal point.
(a) 5 β 10π₯ = 2750 (b) π2π‘ = 142 (c) 5 = 7 + 3π‘
3. Write a function rule given the functionβs value in the table.
(a)
(b) (c) (d)
π₯ π(π₯) π₯ π(π₯) π₯ β(π₯) π₯ π(π₯)
0 1 5 1 β3 1
8 1 0
1 1
3 25 2 β2
1
4
1
10 1
2 1
9 125 3 β1
1
2
1
100 2
3 1
27 625 4 0 1
1
1000 3
4 1
81 3125 5 1 2
1
10000
4
5 1
243 15625 6 2 4
1
100000 5
4. Which functions represent exponential decay? Check all that apply.
πΏ(π‘) = 2.6πβ0.4π‘
π¦(π‘) = 2000(1.03)π‘
β(π₯) = 0.1βπ₯
π(π) = 50(0.71)π
π(π₯) = 752(4)π₯
10π₯ = 550
π₯ = log 550 β 2.740
2π‘ = ln 142
π₯ =1
2ln 142 β 2.478
β2 = 3π‘
no real solutions
π(π₯) = (1
3)
π₯
π(π₯) = log5 π₯ β(π₯) = 2π₯ π(π₯) = β log π₯
3
5. Consider the functions π(π₯) = βπ₯2 + 4 and π(π₯) = 2π₯+1 β 2 and complete the characteristics.
(a) The asymptote of π(π₯) is ________________________.
(b) The domain and range of π(π₯) Domain ___________________ Range ___________________
(c) Domain and range of π(π₯) Domain ___________________ Range ___________________
(d) End behavior of π(π₯) ________________________________
________________________________
(e) End behavior of π(π₯) ________________________________
________________________________
(f) π¦ β intercept of π(π₯) ___________________
(g) π¦β intercept of π(π₯) ___________________
(h) π₯β intercepts of π(π₯) ___________________
(i) Graph π(π₯). (j) Graph π(π₯).
6. The solutions of the quadratic equation π(π₯) = 0 are π₯ = 3 and π₯ = 7. Complete the following.
(a) _________________________ and _________________________ are factors of π(π₯).
(b) The points _____________ and _____________ are the locations of π₯-intercepts of the graph of π(π₯).
(c) _____________ and _____________ are the zeros of π(π₯).
(d) The equation of the axis of symmetry for π(π₯) is _________________.
π¦ = β2
(ββ, β)
(ββ, β)
(ββ, 4]
(β2, β)
as π₯ β ββ, π(π₯) β ββ
as π₯ β β, π(π₯) β ββ
as π₯ β ββ, π(π₯) β β2
as π₯ β β, π(π₯) β β
(0,4)
(0,0)
π₯ = Β±2
(π₯ β 3) (π₯ β 7)
(3,0) (7, 0)
π₯ = 3 π₯ = 7
π₯ = 5
4
7. Consider the functions π, π, β, and π and their graphs below.
π(π₯) = 2π₯ π(π₯) = log3 π₯ β(π₯) = 10βπ₯ π(π₯) = β ln π₯
Several properties are listed below. For each property, write the function(s) that have this property. You may use π, π, β, or π as your answers.
(a) ______________________ The graph of the function has a horizontal asymptote of π¦ = 0.
(b) ______________________ The function has a range of all real numbers.
(c) ______________________ The function is increasing on its entire domain.
(d) ______________________ The graph of the function has a π¦-intercept at the point (0, 1).
(e) ______________________ The domain of the function is the positive real numbers.
(f) ______________________ The graph of the function has a vertical asymptote of π₯ = 0.
(g) ______________________ The function has a domain of all real numbers.
(h) ______________________ The function is decreasing on its entire domain.
(i) ______________________ The graph of the function has an π₯-intercept at the point (1, 0).
(j) ______________________ The range of the function is the positive real numbers.
8. One solution to the quadratic equation π₯2 β 6π₯ + 13 = 0 is π₯ = 3 + 2π. Which statement is true?
Select all that apply.
(π₯ β (3 + 2π)) is a factor of π₯2 β 6π₯ + 13.
3 + 2π is a zero of the function π(π₯) = π₯2 β 6π₯ + 13.
π₯ = 3 β 2π Is also a solution to π₯2 β 6π₯ + 13 = 0.
The other solution to π₯2 β 6π₯ + 13 = 0 is a real number.
The equation(π₯ β (3 + 2π)) (π₯ β ( 3 β 2π)) =0 is equivalent to π₯2 β 6π₯ + 13 = 0.
π, β
π, π
π, π
π, β
π, π
π, π
π, β
β, π
π, π
π, β
5
9. Each function below is a transformation of the function π(π₯) = log2 π₯. After each given
transformation, write the rule.
(a) The graph of function π is the graph of π(π₯) = log2 π₯ translated two units up.
π(π₯) = log2(π₯) + 2
(b) The asymptote of the graph of function β has the equation π₯ = 2.
β(π₯) = log2(π₯ β 2)
(c) The graph of function π is the graph of π(π₯) = log2 π₯ reflected across the π₯-axis.
π(π₯) = β log2 π₯
(d) From the graph of π‘(π₯) to the right.
π‘(π₯) = log2(π₯ + 3)
(e) The graph of function π(π₯) is the graph of
π(π₯) = log2 π₯ reflected across the π¦-axis
π(π₯) = log2(βπ₯)
10. Evaluate, expressing your answer in the form π + ππ.
(a) (3 + π) + (6 β 5π) (b) (4 + 7π) β (2 β 6π)
(c) (3 β 2π)(4 + 9π) (d) (4 β 3π)(4 + 3π)
(e) (7 + 5π)2 (f) π + π2 + π3 + π4
(g) π62 (h) π(π + 4)2 β π5(2π β 1)
= 9 β 4π = 4 + 7π β 2 + 6π = 2 + 13π
= 12 + 27π β 8π β 18π2
= 12 + 19π β 18(β1) = 30 + 19π
= 16 + 12π β 12π β 9π2
= 16 β 9(β1) = 25
= 49 + 35π + 35π + 25π2
= 49 + 70π + 25(β1) = 24 + 70π = π + (β1) + (βπ) + 1 = 0
= π60π2 = (π4)15π2 = 115π2 = π2 = β1 = π(π2 + 8π + 16) β π(2π β 1)
= π(8π + 15) β 2π2 + π
= 8π2 + 15π + 2 + π = β6 + 16π
6
11. Solve the following equations over the set of complex numbers, writing your answer in terms of
π if necessary. Show how you determined your solutions.
(a) π₯2 = β25 (b) (π₯ + 3)2 = β16
(c) π₯2 β 6π₯ = β15 (d) 5π₯2 + 2π₯ + 3 = 0
(e) 2π₯2 + π₯ + 20 = (π₯ + 1)2 (f) β2π₯ + 5 = π₯ + 1
12. Write an exponential function in terms of time π‘ (π‘ in years) for each situation.
(a) There are 300 bacteria at time 0. The bacteria has a growth rate of 70% per year.
(b) Jack puts $500 into a savings account. It earns interest at a nominal annual rate of 6% per
year, compounded monthly.
(c) The number of deer in a forest is decreasing at an annual rate of 8%. There are currently
700 deer in the forest.
(d) You have a mutual fund which yields 5% annually, compounded continuously, and your
balance is $2000 today.
π₯ = Β±ββ25
π₯ = Β±5π
π₯ + 3 = Β±ββ16
π₯ = β3 Β± 4π
π₯2 β 6π₯ + 9 = β6
(π₯ β 3)2 = β6
π₯ β 3 = Β±ββ6
π₯ = 3 Β± πβ6
π₯ =β2 Β± β22 β 4(5)(3)
2(5)=
β2 Β± β4 β 60
10
=β2 Β± ββ56
10=
β2 Β± πβ56
10
2π₯2 + π₯ + 20 = π₯2 + 2π₯ + 1
π₯2 β π₯ + 19 = 0
π₯ =1 Β± β(β1)2 β 4(1)(19)
2(1)
=1 Β± β1 β 76
2=
1 Β± πβ75
2
2π₯ + 5 = (π₯ + 1)2
2π₯ + 5 = π₯2 + 2π₯ + 1
4 = π₯2
π₯ = β2 or π₯ = 2
π΄(π‘) = 300(1.70)π‘
π΄(π‘) = 500 (1 +0.06
12)
12π‘
π΄(π‘) = 700(0.92)2
π΄(π‘) = 2000π0.05π‘
7
13. A pizza, heated to a temperature of 4000πΉ, is taken out of an oven and placed in a 750πΉ room at
time π‘ = 0 minutes. The temperature of the pizza is changing such that its decay constant, k, is
0.324.
(a) Write an equation for the temperature π‘ minutes after it was taken out of the oven.
(b) At what time is the temperature of the pizza 150Β° F and, therefore, safe to eat? Give you
answer in minutes.
(c) Over time, the pizza will approach what temperature value?
14. Choose the appropriate number from the Number Bank below that will make each equation true.
Each number should be used exactly once.
Number Bank:
e 1000 0 β1 10 3
log3 (1
3) = ln 1 = log = 3
log 10 = 1 log27 =1
3 ln = 1
β1 0 1000
10
3 π
π(π‘) = 75 + 325πβ0.324π‘
150 = 75 + 325πβ0.324π‘
75 = 325πβ0.324π‘
75
325= πβ0.324π‘
β0.324π‘ = ln (75
325)
π‘ =1
β0.324ln (
75
325) β 4.526 min
75Β° F
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