due date: third day of class estimated time: 10 hours (for planning purposes only; work until you finish)

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1 Honors PreCalculus Summer Work 016 due date: third day of class estimated time: 10 hours (for planning purposes only; work until you finish) Dear Honors PreCalculus Students, This assignment is designed for you to review all of your algebra skills and make sure you are well prepared for the start of Honors PreCalculus in the fall. The assignment is in two parts: written problems in this packet, and problems on ixl.com. See the next page for detailed instructions. We anticipate that the work will take about 10 hours to complete. Both the written work and the online work are due on the third day of class. At that time you will have a short quiz with problems taken directly from both this packet and the ixl sections listed on the next page. A portion of the quiz grade will be based on the completeness of your summer work. We are really looking forward to the upcoming school year and hope you are, too! Please any of us if you have questions. Barbara.Filler@stewardschool.org Annie.MacKimmie@stewardschool.org Todd.Serr@stewardschool.org Jennifer.Maitland@stewardschool.org Karen.Hudson@stewardschool.org Page 1 of 19

2 Assessment of Summer Work 1. On the third day of class, you will submit your written work, and your teacher will check your online work.. You will have a quiz made up of problems taken directly from the on-line and written work. 3. Your written and online work will be evaluated for completeness; that grade will form part of the quiz grade. On-Line Work 1. Log in to ixl.com with the username and password you were given in your math class. Students new to Steward: your parents will receive an with your username and password.. Work the problems listed below, until you earn a score of at least 60% for each section. If you find that the name of the section does not match the numbers below, go by the section name. 3. If you are having difficulty of any sort with ixl, reach out to a teacher at one of the addresses above. 1. click this tab. find these sections 3. work these problem sets Algebra II Complex numbers Quadratic functions Logarithms Exponential and logarithmic functions H.1 - Introduction to complex numbers H.3 - Complex conjugates I. 5 - Match quadratic functions and graphs I.9 - Complete the square R.1 - Convert between exponential and logarithmic form: rational bases S.1 - Domain and range of exponential and logarithmic functions Written Work 1. Complete each problem on the following pages.. Do the problems on separate paper. Do the problems in order and number them clearly. Exception: print the graphing problems (p.8-9)rather than drawing your own graphs. 3. Please show all of your work as was required of you throughout the entire school year. Remember: no work, no credit. Page of 19

3 Section 1: Radicals Simplifying radicals involves two concerns. 1. No radicand (expression left under a radical) has a perfect square/cubic/etc factor.. There is no radical in a denominator (you must rationalize). Examples: Simplify = 4 3 = 3. 9 = 3 = 3 5 = = = + 6 = + 6 = This word should match the root value of the radical. See problem 3 below. Some things to think about in the above examples: 1. You can use factor trees to help you if you want.. Examples and 3 use the trick of creatively multiplying by Example 3 uses the concept of a conjugate. A conjugate occurs with binomials when the sign between them changes. Simplify. 1) 3 ) (3ww) 6 5 3) ) 7aa 4 bb 9 5) ) ) 5 3 ( 5 + 3) 8) 4 3 9) 10) 1 7 Page 3 of 19

4 Section : Imaginary and Complex Numbers Imaginary numbers allow us to imagine what an even root of a negative number might be, if it were allowed. Recall the imaginary number i : i = 1 and ii = 1 1 = 1 Complex numbers are expressions that contain both a real and imaginary parts: a + bi Examples: = 1 10 = ii 10. 5ii(4 ii) = 0ii 10ii = 0ii 10( 1) = 0ii ii 3. = 5+ii = 5ii 1 = 1 5ii 3ii 3ii 3 3 ii ii = 5ii+ii 3ii Some things to think about in the above examples: 1. Treat i like any other variable when you +,,,. BUT remember ii = 1 3. Since i = 1 it cannot be in a denominator Simplify. 11) 64 1) 3 13) 3ii( + 4ii) 6(7ii + 1) 14) (3 6ii) 15) (x + i)(x i) 16) 3 4ii ii Page 4 of 19

5 Section 3: Geometry Recall these relationships for right triangles: Pythagorean Theorem: aa + bb = cc right triangles right triangles Find the value of x. 17) 18) Page 5 of 19

6 19) Find the value of x and y. The triangles are ) 1) Find the value of x and y. The triangles are ) 3) Page 6 of 19

7 Section 4: Linear Equations Slope Intercept Form: y = mx + b Point-Slope Form: yy yy 1 = mm(xx xx 1) Vertical line: xx = aa (slope is undefined) Horizontal line: yy = bb (slope = 0) Standard Form: Ax + By = C Slope: yy yy 1 xx xx 1 and AA BB 4) State the slope and y-intercept of 3xx yy = 10. 5) State the x- and y-intercepts of xx 4yy = 5. 6) Rewrite in standard form: yy = xx ) Write in slope-intercept form the equation of the line passing through (-, 5) with a slope of -6. 8) Write in point slope form the equation of the line passing through (8, -13) and (4, 0). 9) Write in standard for the equation of the line with an x-intercept of 3 and a y-intercept of ) Write the equation of the line passing through (-1, ) with an undefined slope. Page 7 of 19

8 Section 5: Graphing Print this section. Graph the equations and systems. xx + yy = 4 31) 3xx 4yy = 1 3) xx yy = 33) yy < 4xx 34) yy = xx ) yy > xx 1 36) yy = (xx) + 5 Page 8 of 19

9 37) yy = ee xx 38) xx = 39) yy = (xx + 3) Page 9 of 19

10 Section 6: Systems of Equations Systems of equations occur when two or more equations need to work together. They can be solved in four ways: graphically, by substitution, by elimination, or with matrices. Consider the following system: xx + yy = 5 xx 4yy = 7 Three of the above four ways of solving this system are illustrated below. Graphically Substitution Elimination Graph both lines. The point where the lines intersect is the solution. 4 (3, -1) If the lines are parallel, there will be no solution. If the lines coincide, there will be an infinite number of solutions. 5 Express one variable in terms of the other and substitute. xx + yy = 5 yy = 5 xx xx 4(5 xx) = 7 xx 0 + 8xx = 7 9xx = 7 xx = 3 Then use xx = 3 to find y: (3) + yy = 5 yy = 1 Solution: (3, 1) Eliminate one variable by adding equations. Use multiplication if needed. 4(xx + yy = 5) 8xx + 4yy = 0 8xx + 4yy = 0 + xx 4yy = 7 9xx = 7 xx = 3 Then use xx = 3 to find y. Solution: (3, 1) Solve each system of equations. Use each method at least once. xx + yy = 4 40) 3xx + yy = 1 3xx + yy = 6 41) xx yy = 4 4) xx 5yy = 13 6xx + 3yy = 10 43) 5xx 3yy = 15 10xx 6yy = 5 Page 10 of 19

11 Section 7: Exponents Product Rules To multiply when two bases are the same, add the exponents. To multiply when two bases are different and the exponents are the same, multiply the bases first. aa nn aa mm = aa nn+mm = 3 +3 = 3 5 = 43 aa nn bb nn = (aaaa) nn 3 4 = (3 4) = 1 = 144 Quotient Rules To divide when two bases are the same, subtract the denominator s exponent from the numerator s exponent. To divide when two bases are different and the exponents are the same, divide the bases first. aa nn aamm = aann mm = 34 = 3 = 9 aa nn bb nn = aa bb nn 3 3 = 6 3 = 3 = 8 Power Rule To raise a power to a power, multiply the exponents. (aa nn ) mm = aa nnnn ( ) 3 = 3 = 6 = 64 Zero Rule A base (other than zero) raised to the zero power is equal to 1. aa 0 = = 1 One Rules A base raised to the power of one is equal to itself. 1 raised to any integer power is equal to 1. aa 1 = aa 4 1 = 4 1 nn = = 1 Negative Exponents Rules A base raised to a negative exponent is equal to 1 divided by the base raised to the corresponding positive exponent. aa nn = 1 aa nn 4 = 1 4 = 1 16 Visit rapidtables.com for more exponent rules. Page 11 of 19

12 Simplify. Use only positive exponents when needed. 44) 8ee 0 45) 46) 5aa aa 1 eeff 1 ee 1 47) (nn 3 pp ) 4 (nnnn) 48) 3xx xx 49) (h 3 ) 5 50) ( bb 3 cc 4 ) 5 51) 4mm(3aa mm) Page 1 of 19

13 Section 8: Polynomials So far with polynomials you have learned to do the following: Combine like terms Distribute Multiply (if two binomials, then FOIL) Factor (including ONGF, swing method, and grouping) Solve (set = 0 and use Multiplicative Property of Zero; may have to use quadratic equation) Graph (parabola, including calculating and interpreting discriminant) Combine like terms. 5) (6x 5x+ 7) + (4x 3x+ ) 53) 3 3 (14a a + 5 a) (11a 5a + 10) Find each product. Simplify the answer. 54) 55) 3 x(x 7x+ 1) x( x 5) + xx ( + ) 56) (4m 1)(5m+ 8) 57) (4x + 3) 58) ( x+ 6 y)( x 6 y) 59) ( x+ 3)( x 5x+ 7) Write a polynomial expression to represent the area of the rectangle below. 60) 3aa + 4 5aa 6 Factor completely. 4 61) 5a + 15a 6) 3 7 p 3 p 63) x x 63 Page 13 of 19

14 64) 3a + 8a+ 4 65) 6 5x x 66) 9cc + 30cc ) 16nn ) mmmm mmmm + ssss ssss Solve each equation. 69) x + 3x 8 = 0 70) x + 9x+ 4= 0 71) 36x = 49 Page 14 of 19

15 Section 9: Polynomials, continued: Using Discriminants The quadratic formula allows you to solve any quadratic for all its real and imaginary roots. xx = bb± bb 4aaaa aa The number under the radical in the quadratic formula (b 4ac) is called the discriminant (D). The value of the discriminant can tell you what kinds of roots you will have. If b 4ac > 0, you will have two real roots. 4 If b 4ac = 0, you will have one real root. If b 4ac < 0, you will have two imaginary roots The graph touches the x-axis twice. The graph touches the x-axis once. The graph does not touch the x-axis. For each quadratic equation on the next page, (a) find the value of the discrimant (b) describe the nature of the roots (c) solve the equation. Use the following as an example. ex. x + x + 3 = 0 (a) First, determine the values for a, b, and c: a = 1, b =, c = 3 Then, find the value of the discriminant (D). DD = DD = 4 8 DD = 88 (b) Since D < 0, there are two imaginary roots. (c) Use the quadratic formula to solve the equation. xx = ± = ± ii = 11 ± ii Page 15 of 19

16 7) xx 9xx + 14 = 0 (a) value of the discriminant (b) nature of the roots (c) roots 73) 5xx xx + 4 = 0 (a) value of the discriminant (b) nature of the roots (c) roots Page 16 of 19

17 Section 10: Using Function Notation Evaluating a function at a particular value is done by plugging that value in. example: For ff(xx) = xx 5, f(-3) = (-3) 5 = (9)-5 = 13 Finding the inverse of a function involves switching x for y and solving again for y. The symbol for the inverse of a function is ff 1 (xx). example: The inverse of ff(xx) = xx 5 is found by solving xx = yy 5 for y. xx + 5 = yy xx+5 = yy ± xx+5 = yy or ff 1 (xx) = ± xx+5 Composite functions are written as (ffffff)(xx) or ff(gg(xx)) and are read f of g of x. example: Given ff(xx) = xx and gg(xx) = xx 1, ff gg(xx) = (xx 1) = 3 xx The composite of a function and its inverse results in x. ff ff 1 (xx) = xx Use the following function definitions for problems ff(xx) = xx gg(xx) = hh(xx) = xx 74) f(-4) 75) g(x + h) 76) f(x + ) 77) f[g(x)] 78) h[f(3)] 79) h[g(x)] 80) g - 1 (x) 81) h - 1 (x) Page 17 of 19

18 Section 11: Algebraic Rational Expressions Rational expressions involve fractions. To multiply or divide, factor the numerator and denominator completely. Cancel common factors if they appear in both the numerator and donominator. Remember to change division to multiplication by using the reciprocal of the second fraction. xx + 10xx xx xx xx3 + 4xx 1xx xx + xx 15 = xx + 10xx xx xx xx + xx 15 (xx + 7)(xx + 3) (xx + 5)(xx 3) xx 3 = + 4xx 1xx 1(xx + 5)(xx 1) xx(xx 3)(xx + 7) = xx + 3 xx(xx 1) To add or subtract you have to find a least common denominator. 3xx+1 + 5xx 4 = 3xx+1 + 5xx 4 xx +xx xx+4 xx(xx+) (xx+) The common denominator is x(x+). Multiply the first fraction by and the second fraction by xx xx. (3xx+1) + xx(5xx 4) = 6xx++5xx 4xx = 5xx +xx+ xx(xx+) xx(xx+) xx(xx+) xx(xx+) Complex fractions can be simplified by rewriting the numerator as one fraction and the denominator as one fraction, then treating it as a division problem xx xx 1 = xx+1 xx xx xx = xx+1 xx xx xx = xx+1 xx Simplify. 8) 5mm 3 + mm mm 3mm 83) xx 5 xx + 5xx 84) 10aa 5 1aa 3aa 5aa 3 85) xx 5xx + 6 3xx + 1 xx + 4 xx 86) 6h h + 6h 5h h 7h 87) xx 5 xx 3 88) bb aa aa + bb aa + bb aabb 89) xx 1 xx + xx + 4xx + 4 3xx 3 90) 1 xx xx ) xx xx 1 xx xx Page 18 of 19

19 Section 1: Logarithmic and Exponential Functions Log Rules 1. log bb 1 = 0. log bb bb = 1 3. log bb bb xx = xx 4. log bb xx = log bb yy, if and only if xx = yy 5. log bb (uuuu) = log bb uu + log bb vv To solve an exponential or logarithmic equation, isolate the function and then change forms. Example Solve log 3 xx = 6. First, isolate the log: o log 3 xx = 3 Then change to exponential form: o 3 3 = xx Then evaluate: o 7 = xx 6. log bb uu vv = log bb uu log bb vv 7. log bb uu nn = nn log bb uu Expand and simplify, if possible. 9) log ) log 5 5xx 94) ln 3xx yy Condense and simplify, if possible. 95) log 7 xx + 3 log 7 yy 5 log ) ln xx 1 ln 4 Solve, rounding decimals to three places as needed. 97) 3 xx = 10 98) 5 + log 8 xx = 1 99) ee xx = 4 100) 00ee.03tt = 100 Page 19 of 19