* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Final Study Guide - da Vinci Institute
Survey
Document related concepts
Polynomial greatest common divisor wikipedia , lookup
Root of unity wikipedia , lookup
Horner's method wikipedia , lookup
Signal-flow graph wikipedia , lookup
System of linear equations wikipedia , lookup
Polynomial ring wikipedia , lookup
Elementary algebra wikipedia , lookup
Factorization of polynomials over finite fields wikipedia , lookup
Cubic function wikipedia , lookup
Eisenstein's criterion wikipedia , lookup
Quadratic equation wikipedia , lookup
Quartic function wikipedia , lookup
History of algebra wikipedia , lookup
System of polynomial equations wikipedia , lookup
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 β To find the zeros of a quadratic function π¦ = ππ₯ 2 + ππ₯ + π, solve a related quadratic equation 0 = ππ₯ 2 + ππ₯ + π. 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₯ β 28 = 0 2π₯ 2 + 8π₯ = 5π₯ + 20 π₯ 2 β 10π₯ + 25 = 0 π₯2 β 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 ππ₯ 2 = π o To solve equations by completing the square Practice: Solve by completing the square 1. π₯ 2 β 12π₯ + 7 = 0 2. π₯ 2 + 12 = 10π₯ 3. π₯ 2 + 2 = 6π₯ + 4 4. 25π₯ 2 + 30π₯ = 12 5. 9π₯ 2 β 12π₯ β 2 = 0 Algebra II B Final Study Guide ο· 4.7 β You can solve a quadratic equation ππ₯ 2 + ππ₯ + π = 0 in more than one way. In general, you can find a formula that gives values of π₯ in terms of π, π, and π. βπ±βπ2 β4ππ 2π 2 discriminant βπ β 4ππ 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 β 3π₯ + 1 = 0 2π₯ 2 β 5 = β3π₯ 6π₯ β 5 = βπ₯ 2 12π₯ + 9π₯ 2 = 5 7π₯ 2 β π₯ β 12 = 0 4.8 β The complex numbers are based on a number whose square is -1 o The imaginary unit π is the complex number whose square is -1. So, π 2 = β1, and π = ββ1 Example: Complex number operations Practice: perform the indicated operation using complex numbers 1. 2. 3. 4. 5. 6. Simplify ββ4 , ββ32 (2 + 4π) + (4 β π) (12 + 5π) β (2 β π) (8 + π)(2 + 7π) (β6π)2 (9 + 4π)2 Solve using the quadratic equation, include complex solutions 7. π₯ 2 + 2π₯ + 3 = 0 8. π₯ 2 β 2π₯ + 2 = 0 Algebra II B Final Study Guide Chapter 5: Polynomials and Polynomial Functions ο· 5.1 β A polynomial function has distinguishing βbehaviorsβ. 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· π¦ = 7π₯ + 3π₯ + 5 π¦ = 4π₯ + 5π₯ 2 + 8 π¦ = 5π2 + 3π3 + 1 π¦ = β3 β 6π₯ 5 β 9π₯ 8 π¦ = βπ₯ 3 β π₯ 2 + 3 π¦ = π₯ 3 β 14π₯ β 4 5.2 β 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₯ = β1, β2, β3, β4 Find the zeros of each function and state the multiplicity of each zero, if any. 4. 5. 6. 7. π¦ = 3π₯ 3 β 3π₯ π¦ = (π₯ β 2)2 (π₯ β 1) π¦ = (π₯ β 4)3 π¦ = π₯ 2 (π₯ + 1)4 Algebra II B Final Study Guide ο· 5.3 β If (π₯ β π) is a factor of a polynomial, then the polynomial has a value 0 when π₯ = π. If π is a real number, then the graph of the polynomial has (π, 0) as an x-intercept o To solve a polynomial equation by factoring: ο§ Write the equation in the form π(π₯) = 0 ο§ Factor π(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 β 27π₯ = 0 64π₯ 3 = β8 π₯ 4 β 12π₯ 2 = 64 π₯ 5 β 5π₯ 3 + 4π₯ = 0 5.4 β 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 β π₯ + 2) ÷ (π₯ β 1) 2. (3π₯ 2 + 7π₯ β 20) ÷ (π₯ + 4) 3. (π₯ 4 + 4π₯ 3 β π₯ β 4) ÷ (π₯ 3 β 1) Divide using synthetic division 4. 5. 6. 7. (π₯ 3 + 3π₯ 2 β π₯ β 3) ÷ (π₯ β 1) (π₯ 3 + 27) ÷ (π₯ + 3) (6π₯ 2 β 8π₯ β 2) ÷ (π₯ β 1) (3π₯ 3 + 17π₯ 2 + 21π₯ β 9) ÷ (π₯ + 3) Use synthetic division and the Remainder Theorem to find P(a) 8. π(π₯) = π₯ 3 + 7π₯ 2 + 4π₯; π = β2 9. π(π₯) = 2π₯ 3 + 4π₯ 2 β 10π₯ β 9; π = 3 Algebra II B Final Study Guide ο· 5.5 β The factors of the numbers ππ and π0 in π(π₯) = ππ π₯ π + ππβ1 π₯ πβ1 + β― + π1 π₯ + π0 can help you factor π(π₯) and solve the equation π(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 β 4π₯ + 1 = 0 2. 4π₯ 3 + 2π₯ β 12 = 0 3. 8π₯ 3 + 2π₯ 2 β 5π₯ + 1 = 0 Write a polynomial function with rational coefficients so that π(π₯) = 0 has the given roots. 4. β10π 5. β2π and β10 6. 4 + β5 and 8π Find all rational roots for π(π₯) = 0 7. π(π₯) = 2π₯ 3 β 5π₯ 2 + π₯ β 1 8. π(π₯) = 6π₯ 4 β 7π₯ 2 β 3 9. π(π₯) = 6π₯ 4 β 13π₯ 3 + 13π₯ 2 β 39π₯ β 15 10. π(π₯) = 2π₯ 3 β 3π₯ 2 β 8π₯ + 12 Algebra II B Final Study Guide ο· 5.6 β 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 β 3π₯ 2 + π₯ β 3 = 0 2. π₯ 5 + 3π₯ 3 β 4π₯ = 0 3. π₯ 4 β π₯ 3 β 5π₯ 2 β π₯ β 6 = 0 Find the zeros of each function. 4. π¦ = 2π₯ 3 β 3π₯ 2 β 18π₯ β 8 5. π(π₯) = π₯ 3 β 4π₯ 2 β π₯ + 22 6. π¦ = π₯ 3 β π₯ 2 β 3π₯ β 9 Algebra II B Final Study Guide Chapter 6: Radical Functions and Rational Exponents ο· 6.1 β 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· βπ. 25 β16π₯ 2 3 ββ27 5 β32π¦10 4 βπ₯ 20 π¦ 28 6.2 β You can simplify a radical expression when the exponent of one factor of the radicand is a multiple of the radicalβs 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. β50 β β75 3 3 ββ12 β ββ18 3. 4β2π₯ β 5β6π₯π¦ 2 4. 2β2π₯π¦ 2 β β4π₯ 2 π¦ 5 3 Divide and simplify 5. 6. β500 β5 β48π₯ 3 β3π₯π¦ 2 3 7. β250π₯ 7 π¦ 3 3 β2π₯ 2 π¦ 3 Algebra II B Final Study Guide ο· 6.3 β You can combine like radicals using properties of real numbers o To add and subtract radical expressions Practice: Add or subtract if possible 1. 6β18 + 3β50 3 3 2. β54 + β16 3 3 3. 3β81 β 2β54 Mutliply and simplify 4. (3 + β5)(1 + β5) 5. (3 β 4β2)(5 β 6β2) 6. (4 β 2β3)(4 + 2β3) Rationalize the denominator 4 7. 1+β3 5+β3 8. ο· 2ββ3 6.4 β 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 β 102 3. 34 β 274 4. 643 β 643 5. 3 1 1 1 1 2 2 β6 β36 Write in simplest form 2 6. (π₯ 3 )β3 7. 161.5 1 8. (β27π₯ β9 )3 9. (π₯ 2 π¦ β3 )β6 1 2 Algebra II B Final Study Guide ο· 6.5 β 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. 3βπ₯ + 3 = 15 β3π₯ + 4 = 4 3. (π₯ + 5)3 = 4 4. 5. (π₯ + 3)2 β 1 = π₯ βπ₯ + 7 β π₯ = 1 6. (7π₯ + 6)2 β (9 + 4π₯)2 = 0 1. 2 1 1 ο· 1 6.7 β 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 (π β1) of a relation or function Practice: Find the inverse of each function. State if the inverse is a function. 1. 2. 3. 4. 5. π¦ = 3π₯ + 1 π¦ = π₯2 + 4 π(π₯) = π₯ 3 π(π₯) = β2π₯ β 1 + 3 π(π₯) = βπ₯ + 3 Algebra II B Final Study Guide Chapter 7: Exponential and Logarithmic Functions ο· 7.1 β You can represent repeated multiplication with a function of the form π¦ = ππ 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. π¦ = 9(3)π₯ π¦ = 1.5π₯ π(π₯) = 2(0.65)π₯ π(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 β exponential functions can have an irrational base. One important irrational base is the number π. These are called natural base exponential functions and describe continuous growth or decay. o To graph exponential fnctions that have base π 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 β The exponential function π¦ = π π₯ in one-to-one, so its inverse π₯ = π π¦ is a function. To express βy as a function of xβ for the inverse, write π¦ = log π 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 = 10β1 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. π¦ = log 4 π₯ 8. π¦ = log10 π₯ 9. π¦ = log(π₯ β 6) ο· 7.4 β 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 β log π 3. log 3 4 + log 3 π¦ + 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 β 2 log 3 3 7.5 β 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¦β2 = 20 2 log π₯ = β1 2 log(π₯ + 1) = 5 2 log π₯ + log 4 = 2 7.6 β The functions π¦ = π π₯ and π¦ = ln π₯ are inverse functions. Just as before, this means that if π = π π , then π = ln π 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 β 2 ln π + ln π 3 Solve each equation. Check your answers. 3. ln 3π₯ = 6 4. 2 ln 2π₯ 2 = 1 5. ln(4π₯ β 1) = 36 Use natural logarithms to solve each equation. 6. π π₯ = 18 7. π π₯+1 = 30 8. π 2π₯ = 10 Algebra II B Final Study Guide 8. 1 ± π Answers: 4.5 5.1 1. 2. 3. 4. 5. 6. 7. 8. π₯ π₯ π₯ π₯ π₯ π₯ π₯ π₯ = β2, β4 = β2, β1 = 0, 4 = β4, 3.5 = β4, 2.5 =5 = β2, 2 = 3, 6 1. 10π₯ + 5 linear binomial; down,up 2. 5π₯ 2 + 4π₯ + 8 quadratic trinomial; up,up 3. 3π3 + 5π2 + 1 cubic trinomial; down, up 4. β9π₯ 8 β 6π₯ 5 β 3 5. βπ₯ 3 β π₯ 2 + 3 6. π₯ 3 β 14π₯ β 4 4.6 5.2 1. π₯ = 6 ± β29 2. π₯ = 5 ± β13 3. π₯ = 3 ± β11 3 21 5 5 4. π₯ = β ± β 5. π₯ = 2 3 ± 1. 2. 3. 4. 5. 6. 7. β6 3 4.7 1. 2. 3±β5 5.3 2 5 1. β4, 2 ± 2πβ3 2. β9, 0, 3 β ,1 2 3. β3 ± β14 4. 5 1 β , 1 1±πβ3 3. β , 1. 2. 3. 4. 5. 6. 7. π₯ 2 4π₯ + 3 π : 5 3π₯ β 5 π₯+4 π₯ 2 + 4π₯ + 3 π₯ 2 + 4π₯ + 3 6π₯ β 2 π : 4 3π₯ 2 + 8π₯ β 3 2 4 4. ±4, ±2π 5. 0, ±1, ±2 3 3 5. β1.24, 1.38 4.8 π¦ = π₯ 3 β 18π₯ 2 + 107π₯ β 210 π¦ = π₯ 4 β 5π₯ 3 + 6π₯ 2 π¦ = π₯ 4 + 10π₯ 3 + 35π₯ 2 + 50π₯ + 24 -1, 0, 1 π₯ = 2 ππ’ππ‘πππππππ‘π¦ 2, π₯ = 1 π₯ = 4 ππ’ππ‘πππππππ‘π¦ 3 π₯ = 0 ππ’ππ‘ππππππ‘π¦ 2, π₯ = β1 ππ’ππ‘πππππππ‘π¦ 4 5.4 1. 2. 3. 4. 5. 6. 7. ±2π, ±4πβ2 6 ± 3π 10 ± 6π 9 ± 58π -36 65 ± 72π β1 ± πβ2 Algebra II B Final Study Guide 8. π(β2) = 12 9. π(3) = 51 3 7. 5π₯ βπ₯ 2 π¦ 2 6.3 5.5 1. 33β2 1. ±1 no rational roots 2. 3. 4. 5. 6. 7. 8. 9. 10. 3 2. 5β2 3. Cannot subtract, can simplify expression to No rational roots -1, ¼, ½ π(π₯) = π₯ 2 + 100 π(π₯) = (π₯ 2 + 4)(π₯ 2 β 10) π(π₯) = π₯ 4 β 8π₯ 3 + 75π₯ 2 β 512π₯ + 704 no rational roots no rational roots 5 2 3 , β 3 3 9 β3 β 6β2 1 3 4. 5. 6. 7. 8. 8 + 4β5 63 β 38β2 4 1. 2. 3. 4. 6 10 3 256 β2 + 2β3 13 + 7β3 , -2, 2 2 6.4 5.6 1. 3, ±π 2. 0, ±1, ±2π 3. β2, 3, ±π 1 2 4. β2, β , 4 6 5. β2, 3 ± πβ2 6. 3, β1 ± πβ2 5. 6.1 6. 1. 2. 3. 4. 5. π₯2 7. 64 0.5 4x -3 2π¦ 2 π₯5π¦7 6.2 β7776 (You have to rationalize this 6 denominator!) 1 8. β 9. 3 π₯3 4 π¦ π₯3 6.5 1. 25β6 2. 6 3. 40π₯π¦β3 4. 4π₯π¦ 2 βπ¦ 5. 10 6. 16π₯ π¦ 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. π β1 (π₯) = π₯ β ; yes 2. π β1 (π₯) = βπ₯ β 4 ; no 3. 4. 5. 6. 7. 8. 3 3. π β1 (π₯) = βπ₯ ; yes 1 4. π β1 (π₯) = (π₯ 2 β 6π₯ + 10); yes 2 5. π β1 (π₯) = (π₯ β 3)2 ; yes π4 π log 3 32π₯π¦ β 3.17 β 3.631 β 1.748 2 1 7.5 7.1 1. 3 1. 2. 3. 4. 5. 6. 2. Growth, π¦ β πππ‘ππππππ‘: (0, 9) Growth, π¦ β πππ‘ππππππ‘: (0, 1) Decay, π¦ β πππ‘ππππππ‘: (0, 2) Growth, π¦ β πππ‘ππππππ‘: (0, 0.45) 3. 4. 5. 6. 7. 143,512 β 108 πππππ 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 = β1 3. 4 4. 3 2 5. ½ 6. 7. 8. 9. 4 π¦ = 4π₯ π¦ = 10π₯ π¦ = 10π₯ + 6 7.4 1. log 14 1. ln 18 3 2. ln 3. 4. 5. 6. 7. 8. π βπ π2 134.476 ±0.908 1.078 × 1015 β 2.890 β 2.401 β 1.151