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Transcript
Algebra II B Final Study Guide
Students will be responsible for understanding and using the following concepts and procedures from the
second semester of Algebra II. They should be able to demonstrate mastery of the following Essential
Understandings on the final.
The final exam will be half multiple choice and half free-response. Free-response questions will include both
traditional solving of equations, and explaining mathematical reasoning. Students may prepare their own
formula/note-sheet for use during the final exam. This must be checked and approved by the teacher prior to
the beginning of the exam.
Chapter 4: Quadratic Functions and Equations
ο‚·
4.5 – To find the zeros of a quadratic function 𝑦 = π‘Žπ‘₯ 2 + 𝑏π‘₯ + 𝑐, solve a related quadratic
equation 0 = π‘Žπ‘₯ 2 + 𝑏π‘₯ + 𝑐.
o You can solve some quadratic equations in standard form by factoring the quadratic expression
and using the Zero Product Property.
o To solve quadratic equations by factoring
Practice: Solve each equation by factoring
1.
2.
3.
4.
5.
6.
7.
8.
ο‚·
π‘₯ 2 + 6π‘₯ + 8 = 0
2π‘₯ 2 + 6π‘₯ + 4 = 0
2π‘₯ 2 = 8π‘₯
2π‘₯ 2 + π‘₯ βˆ’ 28 = 0
2π‘₯ 2 + 8π‘₯ = 5π‘₯ + 20
π‘₯ 2 βˆ’ 10π‘₯ + 25 = 0
π‘₯2 βˆ’ 4 = 0
π‘₯ 2 + 18 = 9π‘₯
4.6 - Completing a perfect square trinomial allows you to factor the completed trinomial as the square
of a binomial.
o You can solve an equation that contains a perfect square by finding square roots. The simplest
of this type of equation has the form π‘Žπ‘₯ 2 = 𝑐
o To solve equations by completing the square
Practice: Solve by completing the square
1.
π‘₯ 2 βˆ’ 12π‘₯ + 7 = 0
2.
π‘₯ 2 + 12 = 10π‘₯
3.
π‘₯ 2 + 2 = 6π‘₯ + 4
4.
25π‘₯ 2 + 30π‘₯ = 12
5. 9π‘₯ 2 βˆ’ 12π‘₯ βˆ’ 2 = 0
Algebra II B Final Study Guide
ο‚·
4.7 – You can solve a quadratic equation π‘Žπ‘₯ 2 + 𝑏π‘₯ + 𝑐 = 0 in more than one way. In general, you can
find a formula that gives values of π‘₯ in terms of π‘Ž, 𝑏, and 𝑐.
βˆ’π‘±βˆšπ‘2 βˆ’4π‘Žπ‘
2π‘Ž
2
discriminant βˆšπ‘ βˆ’ 4π‘Žπ‘
o
To solve quadratic equations using the Quadratic Formula π‘₯ =
o
To determine the number of real solutions using the
Practice: Solve each equations using the
Quadratic Formula
1.
2.
3.
4.
5.
ο‚·
π‘₯ 2 βˆ’ 3π‘₯ + 1 = 0
2π‘₯ 2 βˆ’ 5 = βˆ’3π‘₯
6π‘₯ βˆ’ 5 = βˆ’π‘₯ 2
12π‘₯ + 9π‘₯ 2 = 5
7π‘₯ 2 βˆ’ π‘₯ βˆ’ 12 = 0
4.8 – The complex numbers are based on a number whose square is -1
o
The imaginary unit 𝑖 is the complex number whose square is -1. So, 𝑖 2 = βˆ’1, and 𝑖 = βˆšβˆ’1
Example: Complex number operations
Practice: perform the indicated operation using
complex numbers
1.
2.
3.
4.
5.
6.
Simplify βˆšβˆ’4 , βˆšβˆ’32
(2 + 4𝑖) + (4 βˆ’ 𝑖)
(12 + 5𝑖) βˆ’ (2 βˆ’ 𝑖)
(8 + 𝑖)(2 + 7𝑖)
(βˆ’6𝑖)2
(9 + 4𝑖)2
Solve using the quadratic equation, include
complex solutions
7. π‘₯ 2 + 2π‘₯ + 3 = 0
8. π‘₯ 2 βˆ’ 2π‘₯ + 2 = 0
Algebra II B Final Study Guide
Chapter 5: Polynomials and Polynomial Functions
ο‚·
5.1 – A polynomial function has distinguishing β€œbehaviors”. You can look at its algebraic form and know
something about its graph. You can look at its graph and know something about its algebraic form.
o Terms: monomial, degree of a monomial, polynomial, degree of a polynomial
o To classify polynomials
o To graph polynomial functions and describe end behavior
Examples: Standard form of a polynomial function arranges the
Practice: write each polynomial in
terms by degree in descending numerical order
standard form, classify it by degree
and by number of terms, and
describe its end behavior.
1.
2.
3.
4.
5.
6.
ο‚·
𝑦 = 7π‘₯ + 3π‘₯ + 5
𝑦 = 4π‘₯ + 5π‘₯ 2 + 8
𝑦 = 5π‘Ž2 + 3π‘Ž3 + 1
𝑦 = βˆ’3 βˆ’ 6π‘₯ 5 βˆ’ 9π‘₯ 8
𝑦 = βˆ’π‘₯ 3 βˆ’ π‘₯ 2 + 3
𝑦 = π‘₯ 3 βˆ’ 14π‘₯ βˆ’ 4
5.2 – Finding the zeros of a polynomial function will help you factor the polynomial, graph the function,
and solve the related polynomial equation.
o To analyze the factored form of a polynomial
o To write a polynomial function from its zeros
Practice: write a polynomial with the following
zeros
1.
2.
3.
π‘₯ = 5, 6, 7
π‘₯ = 0, 0, 2, 3
π‘₯ = βˆ’1, βˆ’2, βˆ’3, βˆ’4
Find the zeros of each function and state the
multiplicity of each zero, if any.
4.
5.
6.
7.
𝑦 = 3π‘₯ 3 βˆ’ 3π‘₯
𝑦 = (π‘₯ βˆ’ 2)2 (π‘₯ βˆ’ 1)
𝑦 = (π‘₯ βˆ’ 4)3
𝑦 = π‘₯ 2 (π‘₯ + 1)4
Algebra II B Final Study Guide
ο‚·
5.3 – If (π‘₯ βˆ’ π‘Ž) is a factor of a polynomial, then the polynomial has a value 0 when π‘₯ = π‘Ž. If π‘Ž is a real
number, then the graph of the polynomial has (π‘Ž, 0) as an x-intercept
o To solve a polynomial equation by factoring:
 Write the equation in the form 𝑃(π‘₯) = 0
 Factor 𝑃(π‘₯). Use the Zero Product Property to find the roots
Practice: Find real or imaginary solutions of
each equation by factoring
1.
2.
3.
4.
5.
ο‚·
π‘₯ 3 + 64 = 0
π‘₯ 3 + 6π‘₯ 2 βˆ’ 27π‘₯ = 0
64π‘₯ 3 = βˆ’8
π‘₯ 4 βˆ’ 12π‘₯ 2 = 64
π‘₯ 5 βˆ’ 5π‘₯ 3 + 4π‘₯ = 0
5.4 – You can divide polynomials using steps that are similar to the long-division steps that you use to
divide whole numbers
o To divide polynomials using long division
o To divide polynomials using synthetic division
Practice: Divide using long division
1. (π‘₯ 3 + 3π‘₯ 2 βˆ’ π‘₯ + 2) ÷ (π‘₯ βˆ’ 1)
2. (3π‘₯ 2 + 7π‘₯ βˆ’ 20) ÷ (π‘₯ + 4)
3. (π‘₯ 4 + 4π‘₯ 3 βˆ’ π‘₯ βˆ’ 4) ÷ (π‘₯ 3 βˆ’ 1)
Divide using synthetic division
4.
5.
6.
7.
(π‘₯ 3 + 3π‘₯ 2 βˆ’ π‘₯ βˆ’ 3) ÷ (π‘₯ βˆ’ 1)
(π‘₯ 3 + 27) ÷ (π‘₯ + 3)
(6π‘₯ 2 βˆ’ 8π‘₯ βˆ’ 2) ÷ (π‘₯ βˆ’ 1)
(3π‘₯ 3 + 17π‘₯ 2 + 21π‘₯ βˆ’ 9) ÷ (π‘₯ + 3)
Use synthetic division and the Remainder Theorem
to find P(a)
8. 𝑃(π‘₯) = π‘₯ 3 + 7π‘₯ 2 + 4π‘₯; π‘Ž = βˆ’2
9. 𝑃(π‘₯) = 2π‘₯ 3 + 4π‘₯ 2 βˆ’ 10π‘₯ βˆ’ 9; π‘Ž = 3
Algebra II B Final Study Guide
ο‚·
5.5 – The factors of the numbers π‘Žπ‘› and π‘Ž0 in 𝑃(π‘₯) = π‘Žπ‘› π‘₯ 𝑛 + π‘Žπ‘›βˆ’1 π‘₯ π‘›βˆ’1 + β‹― + π‘Ž1 π‘₯ + π‘Ž0 can help you
factor 𝑃(π‘₯) and solve the equation 𝑃(π‘₯) = 0.
o To solve equations using the Rational Root Theorem
o To use the Conjugate Root Theorem
Practice: Use the Rational Root Theorem to
list all possible rational roots for each
equation. The find any actual rational roots.
1. π‘₯ 3 βˆ’ 4π‘₯ + 1 = 0
2. 4π‘₯ 3 + 2π‘₯ βˆ’ 12 = 0
3. 8π‘₯ 3 + 2π‘₯ 2 βˆ’ 5π‘₯ + 1 = 0
Write a polynomial function with rational
coefficients so that 𝑃(π‘₯) = 0 has the given
roots.
4. βˆ’10𝑖
5. βˆ’2𝑖 and √10
6. 4 + √5 and 8𝑖
Find all rational roots for 𝑃(π‘₯) = 0
7. 𝑃(π‘₯) = 2π‘₯ 3 βˆ’ 5π‘₯ 2 + π‘₯ βˆ’ 1
8. 𝑃(π‘₯) = 6π‘₯ 4 βˆ’ 7π‘₯ 2 βˆ’ 3
9. 𝑃(π‘₯) = 6π‘₯ 4 βˆ’ 13π‘₯ 3 + 13π‘₯ 2 βˆ’
39π‘₯ βˆ’ 15
10. 𝑃(π‘₯) = 2π‘₯ 3 βˆ’ 3π‘₯ 2 βˆ’ 8π‘₯ + 12
Algebra II B Final Study Guide
ο‚·
5.6 – The degree of a polynomial equation tells you how many roots the equation has.
o
To use the Fundamental Theorem of Algebra to solve polynomial equations with complex
solutions
Practice: Find all the roots of each
equation.
1. π‘₯ 3 βˆ’ 3π‘₯ 2 + π‘₯ βˆ’ 3 = 0
2. π‘₯ 5 + 3π‘₯ 3 βˆ’ 4π‘₯ = 0
3. π‘₯ 4 βˆ’ π‘₯ 3 βˆ’ 5π‘₯ 2 βˆ’ π‘₯ βˆ’ 6 = 0
Find the zeros of each function.
4. 𝑦 = 2π‘₯ 3 βˆ’ 3π‘₯ 2 βˆ’ 18π‘₯ βˆ’ 8
5. 𝑓(π‘₯) = π‘₯ 3 βˆ’ 4π‘₯ 2 βˆ’ π‘₯ + 22
6. 𝑦 = π‘₯ 3 βˆ’ π‘₯ 2 βˆ’ 3π‘₯ βˆ’ 9
Algebra II B Final Study Guide
Chapter 6: Radical Functions and Rational Exponents
ο‚·
6.1 – Corresponding to every power, there is a root. For example, just as there are squares (second
powers), there are square roots. Just as there are cubes (third powers), there are cube roots, and so on.
o Finding roots of various degrees
Practice: Find each real root
1.
2.
3.
4.
5.
ο‚·
βˆšπ‘œ. 25
√16π‘₯ 2
3
βˆšβˆ’27
5
√32𝑦10
4
√π‘₯ 20 𝑦 28
6.2 – You can simplify a radical expression when the exponent of one factor of the radicand is a multiple
of the radical’s index. You can simplify powers that have the same exponent. Similarly, you can simplify
the product of radicals that have the same index
o To multiply and divide radical expressions
Practice: Multiply, if possible. Then
simplify.
1.
2.
√50 βˆ™ √75
3
3
βˆšβˆ’12 βˆ™ βˆšβˆ’18
3.
4√2π‘₯ βˆ™ 5√6π‘₯𝑦 2
4.
2√2π‘₯𝑦 2 βˆ™ √4π‘₯ 2 𝑦 5
3
Divide and simplify
5.
6.
√500
√5
√48π‘₯ 3
√3π‘₯𝑦 2
3
7.
√250π‘₯ 7 𝑦 3
3
√2π‘₯ 2 𝑦
3
Algebra II B Final Study Guide
ο‚·
6.3 – You can combine like radicals using properties of real numbers
o
To add and subtract radical expressions
Practice: Add or subtract if possible
1. 6√18 + 3√50
3
3
2. √54 + √16
3
3
3. 3√81 βˆ’ 2√54
Mutliply and simplify
4. (3 + √5)(1 + √5)
5. (3 βˆ’ 4√2)(5 βˆ’ 6√2)
6. (4 βˆ’ 2√3)(4 + 2√3)
Rationalize the denominator
4
7.
1+√3
5+√3
8.
ο‚·
2βˆ’βˆš3
6.4 – You can write a radical expression in an equivalent form using a fractional (rational) exponent
instead of a radical sign.
o To simplify expressions with rational exponents
Practice: simplify each expression
1
1.
362
2.
102 βˆ™ 102
3.
34 βˆ™ 274
4.
643 βˆ™ 643
5.
3
1
1
1
1
2
2
√6
√36
Write in simplest form
2
6. (π‘₯ 3 )βˆ’3
7. 161.5
1
8.
(βˆ’27π‘₯ βˆ’9 )3
9.
(π‘₯ 2 𝑦 βˆ’3 )βˆ’6
1
2
Algebra II B Final Study Guide
ο‚·
6.5 – Solve a square root equation may require that you square each side of the equation. This can
introduce extraneous solutions
o To solve square root and other radical equations
Practice: solve the following radical
equations. Check for extraneous solutions.
2.
3√π‘₯ + 3 = 15
√3π‘₯ + 4 = 4
3.
(π‘₯ + 5)3 = 4
4.
5.
(π‘₯ + 3)2 βˆ’ 1 = π‘₯
√π‘₯ + 7 βˆ’ π‘₯ = 1
6.
(7π‘₯ + 6)2 βˆ’ (9 + 4π‘₯)2 = 0
1.
2
1
1
ο‚·
1
6.7 – The inverse of a function may not be a function. An inverse performs the opposite operation in the
opposite order of the original relation or function.
o To find the inverse (𝑓 βˆ’1) of a relation or function
Practice: Find the inverse of each function.
State if the inverse is a function.
1.
2.
3.
4.
5.
𝑦 = 3π‘₯ + 1
𝑦 = π‘₯2 + 4
𝑓(π‘₯) = π‘₯ 3
𝑓(π‘₯) = √2π‘₯ βˆ’ 1 + 3
𝑓(π‘₯) = √π‘₯ + 3
Algebra II B Final Study Guide
Chapter 7: Exponential and Logarithmic Functions
ο‚·
7.1 – You can represent repeated multiplication with a function of the form 𝑦 = π‘Žπ‘ π‘₯ where b is a
positive number other than 1. This is called an exponential function.
o To model exponential growth and decay with graphs and equations
Practice: Determine whether the function
represents exponential growth or decay, and
then find the y-intercept.
1.
2.
3.
4.
𝑦 = 9(3)π‘₯
𝑦 = 1.5π‘₯
𝑓(π‘₯) = 2(0.65)π‘₯
𝑓(π‘₯) = 0.45(3)π‘₯
Find each amount after the specified time
5. A population of 120,000 grows 1.2%
per year for 15 years.
6. A population of 115 bears decreases
1.25% yearly for 5 years.
ο‚·
7.2 – exponential functions can have an irrational base. One important irrational base is the number 𝑒.
These are called natural base exponential functions and describe continuous growth or decay.
o To graph exponential fnctions that have base 𝑒
Practice: Find the amount in a
continuously compounded account for
the given conditions:
1. Principal: $2000, annual interest
rate: 5.1%, time: 3 years
2. Principal $400, annual interest
rate: 7.6% time: 1.5 years
3. Principal: $950, annual interest
rate: 6.5% time: 10 years
ο‚·
7.3 – The exponential function 𝑦 = 𝑏 π‘₯ in one-to-one, so its inverse π‘₯ = 𝑏 𝑦 is a function. To express β€œy
as a function of x” for the inverse, write 𝑦 = log 𝑏 π‘₯
o To write and evaluate logarithmic expressions
o To graph logarithmic functions
Algebra II B Final Study Guide
Practice: Write the equation in
logarithmic form
1. 103 = 1000
1
2. 10 = 10βˆ’1
Evaluate each logarithm
3.
4.
5.
6.
log 2 16
log 4 8
log 49 7
log 10,000
Find the inverse of each function.
7. 𝑦 = log 4 π‘₯
8. 𝑦 = log10 π‘₯
9. 𝑦 = log(π‘₯ βˆ’ 6)
ο‚·
7.4 – logarithms and exponents have corresponing properties
o
To use the properites of logarithms to simplify or expand expressions
Practice: Write each expression as a single
logarithm
1. log 7 + log 2
2. 4 log π‘š βˆ’ log 𝑛
3. log 3 4 + log 3 𝑦 + log 3 8π‘₯
Use the Change of Base formula to evaluate
each expression
4. log 2 9
5. log 3 54
6. log 7 30
Use the Properties of Logarithms to evaluate
each expression
Change of Base Formula:
ο‚·
7. log 6 12 + log 6 3
8. log 3 27 βˆ’ 2 log 3 3
7.5 – you can use logarithms to solve exponential equations. You can use exponents to solve logarithmic
equations.
o To solve exponential and logarithmic equations
Algebra II B Final Study Guide
Practice: Solve each equation. Round to the
nearest ten-thousandths if necessary. Check your
answers
1.
2.
3.
4.
5.
6.
7.
ο‚·
2π‘₯ = 8
32π‘₯ = 27
3π‘₯+2 = 272π‘₯
12π‘¦βˆ’2 = 20
2 log π‘₯ = βˆ’1
2 log(π‘₯ + 1) = 5
2 log π‘₯ + log 4 = 2
7.6 – The functions 𝑦 = 𝑒 π‘₯ and 𝑦 = ln π‘₯ are inverse functions. Just as before, this means that if π‘Ž = 𝑒 𝑏 ,
then 𝑏 = ln π‘Ž and vice versa.
o To evaluate and simplify natural logarithmic expressions
o To solve equations using natural logarithms.
Practice: Write each expression as a single natural
logarithm.
1. ln 9 + ln 2
1
2. ln π‘Ž βˆ’ 2 ln 𝑏 + ln 𝑐
3
Solve each equation. Check your answers.
3. ln 3π‘₯ = 6
4. 2 ln 2π‘₯ 2 = 1
5. ln(4π‘₯ βˆ’ 1) = 36
Use natural logarithms to solve each equation.
6. 𝑒 π‘₯ = 18
7. 𝑒 π‘₯+1 = 30
8. 𝑒 2π‘₯ = 10
Algebra II B Final Study Guide
8. 1 ± 𝑖
Answers:
4.5
5.1
1.
2.
3.
4.
5.
6.
7.
8.
π‘₯
π‘₯
π‘₯
π‘₯
π‘₯
π‘₯
π‘₯
π‘₯
= βˆ’2, βˆ’4
= βˆ’2, βˆ’1
= 0, 4
= βˆ’4, 3.5
= βˆ’4, 2.5
=5
= βˆ’2, 2
= 3, 6
1. 10π‘₯ + 5 linear binomial; down,up
2. 5π‘₯ 2 + 4π‘₯ + 8 quadratic trinomial;
up,up
3. 3π‘Ž3 + 5π‘Ž2 + 1 cubic trinomial; down,
up
4. βˆ’9π‘₯ 8 βˆ’ 6π‘₯ 5 βˆ’ 3
5. βˆ’π‘₯ 3 βˆ’ π‘₯ 2 + 3
6. π‘₯ 3 βˆ’ 14π‘₯ βˆ’ 4
4.6
5.2
1. π‘₯ = 6 ± √29
2. π‘₯ = 5 ± √13
3. π‘₯ = 3 ± √11
3
21
5
5
4. π‘₯ = βˆ’ ± √
5. π‘₯ =
2
3
±
1.
2.
3.
4.
5.
6.
7.
√6
3
4.7
1.
2.
3±βˆš5
5.3
2
5
1. βˆ’4, 2 ± 2π‘–βˆš3
2. βˆ’9, 0, 3
βˆ’ ,1
2
3. βˆ’3 ± √14
4.
5 1
βˆ’ ,
1
1±π‘–βˆš3
3.
βˆ’ ,
1.
2.
3.
4.
5.
6.
7.
π‘₯ 2 4π‘₯ + 3 𝑅: 5
3π‘₯ βˆ’ 5
π‘₯+4
π‘₯ 2 + 4π‘₯ + 3
π‘₯ 2 + 4π‘₯ + 3
6π‘₯ βˆ’ 2 𝑅: 4
3π‘₯ 2 + 8π‘₯ βˆ’ 3
2
4
4. ±4, ±2𝑖
5. 0, ±1, ±2
3 3
5. βˆ’1.24, 1.38
4.8
𝑦 = π‘₯ 3 βˆ’ 18π‘₯ 2 + 107π‘₯ βˆ’ 210
𝑦 = π‘₯ 4 βˆ’ 5π‘₯ 3 + 6π‘₯ 2
𝑦 = π‘₯ 4 + 10π‘₯ 3 + 35π‘₯ 2 + 50π‘₯ + 24
-1, 0, 1
π‘₯ = 2 π‘šπ‘’π‘™π‘‘π‘–π‘π‘™π‘–π‘π‘–π‘‘π‘¦ 2, π‘₯ = 1
π‘₯ = 4 π‘šπ‘’π‘™π‘‘π‘–π‘π‘™π‘–π‘π‘–π‘‘π‘¦ 3
π‘₯ = 0 π‘šπ‘’π‘™π‘‘π‘–π‘π‘™π‘–π‘π‘‘π‘¦ 2, π‘₯ =
βˆ’1 π‘šπ‘’π‘™π‘‘π‘–π‘π‘™π‘–π‘π‘–π‘‘π‘¦ 4
5.4
1.
2.
3.
4.
5.
6.
7.
±2𝑖, ±4π‘–βˆš2
6 ± 3𝑖
10 ± 6𝑖
9 ± 58𝑖
-36
65 ± 72𝑖
βˆ’1 ± π‘–βˆš2
Algebra II B Final Study Guide
8. 𝑃(βˆ’2) = 12
9. 𝑃(3) = 51
3
7. 5π‘₯ √π‘₯ 2 𝑦 2
6.3
5.5
1. 33√2
1. ±1 no rational roots
2.
3.
4.
5.
6.
7.
8.
9.
10.
3
2. 5√2
3. Cannot subtract, can simplify expression to
No rational roots
-1, ¼, ½
𝑃(π‘₯) = π‘₯ 2 + 100
𝑃(π‘₯) = (π‘₯ 2 + 4)(π‘₯ 2 βˆ’ 10)
𝑃(π‘₯) = π‘₯ 4 βˆ’ 8π‘₯ 3 + 75π‘₯ 2 βˆ’ 512π‘₯ + 704
no rational roots
no rational roots
5
2
3
, βˆ’
3
3
9 √3 βˆ’ 6√2
1
3
4.
5.
6.
7.
8.
8 + 4√5
63 βˆ’ 38√2
4
1.
2.
3.
4.
6
10
3
256
βˆ’2 + 2√3
13 + 7√3
, -2, 2
2
6.4
5.6
1. 3, ±π‘–
2. 0, ±1, ±2𝑖
3. βˆ’2, 3, ±π‘–
1
2
4. βˆ’2, βˆ’ , 4
6
5. βˆ’2, 3 ± π‘–βˆš2
6. 3, βˆ’1 ± π‘–βˆš2
5.
6.1
6.
1.
2.
3.
4.
5.
π‘₯2
7. 64
0.5
4x
-3
2𝑦 2
π‘₯5𝑦7
6.2
√7776
(You have to rationalize this
6
denominator!)
1
8. βˆ’
9.
3
π‘₯3
4
𝑦
π‘₯3
6.5
1. 25√6
2. 6
3. 40π‘₯π‘¦βˆš3
4. 4π‘₯𝑦 2 βˆšπ‘¦
5. 10
6.
16π‘₯
𝑦
1.
2.
3.
4.
5.
6.
16
4
3, -13
1
2
1
Algebra II B Final Study Guide
6.7
2. log
1
1
3
3
1. 𝑓
βˆ’1 (π‘₯)
= π‘₯ βˆ’ ; yes
2. 𝑓
βˆ’1 (π‘₯)
= √π‘₯ βˆ’ 4 ; no
3.
4.
5.
6.
7.
8.
3
3. 𝑓 βˆ’1 (π‘₯) = √π‘₯ ; yes
1
4. 𝑓 βˆ’1 (π‘₯) = (π‘₯ 2 βˆ’ 6π‘₯ + 10); yes
2
5. 𝑓
βˆ’1 (π‘₯)
= (π‘₯ βˆ’ 3)2 ; yes
π‘š4
𝑛
log 3 32π‘₯𝑦
β‰ˆ 3.17
β‰ˆ 3.631
β‰ˆ 1.748
2
1
7.5
7.1
1. 3
1.
2.
3.
4.
5.
6.
2.
Growth, 𝑦 βˆ’ π‘–π‘›π‘‘π‘’π‘Ÿπ‘π‘’π‘π‘‘: (0, 9)
Growth, 𝑦 βˆ’ π‘–π‘›π‘‘π‘’π‘Ÿπ‘π‘’π‘π‘‘: (0, 1)
Decay, 𝑦 βˆ’ π‘–π‘›π‘‘π‘’π‘Ÿπ‘π‘’π‘π‘‘: (0, 2)
Growth, 𝑦 βˆ’ π‘–π‘›π‘‘π‘’π‘Ÿπ‘π‘’π‘π‘‘: (0, 0.45)
3.
4.
5.
6.
7.
143,512
β‰ˆ 108 π‘π‘’π‘Žπ‘Ÿπ‘ 
7.2
3
2
2
5
0.272
0.316
315.2
5
7.6
1. $2330.65
2. $448.30
3. $1819.76
7.3
1. log 1000 = 3
2. log
1
10
= βˆ’1
3. 4
4.
3
2
5. ½
6.
7.
8.
9.
4
𝑦 = 4π‘₯
𝑦 = 10π‘₯
𝑦 = 10π‘₯ + 6
7.4
1. log 14
1. ln 18
3
2. ln
3.
4.
5.
6.
7.
8.
π‘Ž βˆšπ‘
𝑏2
134.476
±0.908
1.078 × 1015
β‰ˆ 2.890
β‰ˆ 2.401
β‰ˆ 1.151