With topics from Algebra and Pre-Calculus to

With topics from Algebra and Pre-Calculus to get you ready to the AP! (Key contains solved problems) Note: The purpose of this packet is to give you a review of basic skills. You are asked not to use the calculator, ecept on p. (8) and p.6-7(7-0all). Prepared by: Mrs. Trebat Fairfield Ludlowe High School Fairfield, Connecticut mtrebat@fairfield.k.ct.us

ALGEBRA TOPICS LAWS OF EXPONENTS What you need to know: Assume a and b are real numbers and m and n positive integers. (i) a m a n = a m+ (ii) ( ) (iv) a a ab a b n m m m m m m n n n m = (v) a = m ( a ) = a m = (iii) ( a ) n = a mn PRACTICE PROBLEMS (NO, please!) Simplify: ( ) ) c d ) ( a b)( ab )( ab) 4) Solve for n: = ( ) h h+ 5n 5 n ) r ( r ) 5) 4 6 = 8 n+ n n 6) Write in eponential form (no negative eponents): 6 4 8 y 7) Write in simplest radical form: 4 7 8 9 8) Simplify. Give your answer in eponential form: a a a 9) Using the property of eponents, show that 4 = (without the 7 9 help of a, of course!)

Solve (algebraic solution!): 0) ( + ) 4 = 8 ) 9 = 4 FACTORING POLYNOMIALS What you need to know: Perfect Square Trinomials: Difference of Squares: Sum of Cubes: a + ab + b = ( a + b) a ab + b = ( a b) a b = ( a + b)( a b) a + b = ( a + b)( a ab + b ) Difference of Cubes: a b = ( a b)( a + ab + b ) PRACTICE PROBLEMS Factor completely: ) 0 ) 4 + 7 ) a 0a + 5 4) 4a 4ab +b 5 5) 5 6y 6) 48 6 6 8) 7) y 6 64 z + + 9) 8p 0) ( y ) ( y) + + + ) 6 9 4y ) ( y) ( y) ) 6 7y y 4) 4 + 8 y 5y

SOLVING POLYNOMIAL INEQUALITIES Eample : 4 5 < 0 Solution: ( )( ) + 5 < 0 factor + - - 5 + mark the zeros; pick a test point to determine the sign of the polynomial in each interval The solution set of this conjunction is { : < < 5 }. Eample : + 8 + 0 < 0 ( + )( 5) < 0 Factor - + - 5 The solution set of this disjunction is: - Make a sign chart; mark the zeros; pick a test point in each interval and plug it into the factored form to determine the sign of each interval { : < or > 5} PRACTICE PROBLEMS ) + 7 + 0 > 0 ) 6 > 0 ) ( 4 )( 4) 0 4) < 0 4

PRE-CALCULUS TOPICS FUNCTIONS You are given a polynomial equation and one or more of its roots. Find the remaining roots. What you need to know: (i) The Remainder Theorem: When a polynomial P() is divided by (-a), the remainder is P(a); (ii) Factor Theorem: For a polynomial P(), (-a) is a factor if and only if P(a)=0; (iii) Use synthetic division when dividing a polynomial by a linear factor; long division will always work; (iv) Odd Function: A function f() is odd f(-) = -f(); (v) Even Function: A function f() is even f(-) = f(); Remarks: item (iv) means the graph of the function has rotational symmetry about the origin, and (v) means the graph has reflectional symmetry about the y-ais; Puzzled Let s try some Find the quotient and the remainder when the first polynomial is divided by the second. No calculators allowed! ) + 5 + ; ) 4 + 5 + + ; + ) Determine whether - or + are factors of 4 +. 00 99 5

4) When a polynomial P() is divided by 4, the quotient is + + and the remainder is. Find P(). (recall: P()=(divisor)(quotient)+remainder!) 5) You are given a polynomial equation and one or more of its roots. Find the 4 remaining roots. 9 + + 9 4 = 0; roots: = -, =. Verify algebraically whether the functions are odd or even. 5 6) f( ) = 5 7) h ( ) = + FINDING THE ASYMPTOTES OF A RATIONAL FUNCTION What you need to know: Vertical Asymptote: Set denominator = 0 and solve it. For values of near the asymptote, the y-values of the function approach infinity (i.e., the y-values get increasingly large ) Horizontal or Oblique Asymptote: when the values of get very large (approach infinity), the y-values tend to the asymptote. Below is an eample of a curve displaying asymptotic behavior: A curve can intersect its asymptote, even infinitely many times. This does not apply to vertical asymptotes of functions. 6

To find the horizontal or oblique asymptote, divide the numerator by the denominator. The asymptote is given by the quotient function. PRACTICE PROBLEMS (No, please!) ) For the functions below determine: a) the domain; b) the -and y- intercepts; c) the vertical, horizontal or oblique asymptotes; and d) whether R() is even or odd (show algebraic analysis). ) R( ) = + ) r( ) = 5 +. (eplain what happens at =- ) 4 8 ) Consider f( ) =. a) is f even or odd? b) eplain why there are no + vertical asymptotes; c) What is the domain of f? Range? d) Given the equation of its horizontal asymptote. 7

4) Find a quadratic function f() with y-intercept - and -intercepts (-,0) and,0. RATIONAL ALGEBRAIC EXPRESSIONS AND FUNCTIONS What you need to know: (i) A rational function is a function of the form are polynomial functions and q( ) 0 for every. p( ), where p and q q( ) (ii) To simplify a rational epression, factor both numerator and denominator. Cancel out common factors. (iii) To find the domain of a rational function: solve q() = 0. The zeroes of the polynomial q() must be ecluded from the domain. (iv) To find the zeroes of a rational function: set p() = 0 and solve it. The zeros of p() are the zeros of the polynomial function (provided they are not also zeros of the denominator q()!) In #-: Find (a) the domain of each function; and (b) the zeros, if any. No calculators, please! ) g ( ) = + 9 4 ( + )( )( ) ) g ( ) = 6 + 8 8

) h ( ) = + + Simplify: (attention: just perform the indicated operation ) 4) + 5 + 6 4 5) + + ( )( +) 6) y + y 4 4 y = Find the constants A and B that make the equation true. 9 A B 7) = + 6 + 9

Solve the fractional equations. If it has no solution, say so 8) = + 5 a 9) = a + a 6 a EQUATIONS CONTAINING RADICALS Eample : Solve 5 = = 5 9 + 4 = 5 isolate the radical term square both sides 9 7 + 4 = 0 Solve for ( 4)(9 ) = 0 = 4 or = these are candidate solutions! 9 Now check each candidate above by plugging them into the original equation. The only possible solution is = 4. PRACTICE PROBLEMS: Solve. If an equation has no solution, say so. (hint: sometimes you may have to square it twice ) ) + 5 = ) + + 4 = 5 0

FUNCTION OPERATIONS AND COMPOSITION; INVERSE FUNCTIONS What you need to know: The composition of a function g with a function f is defined as: h ( ) = gf ( ( )) The domain of h is the set of all -values such that is in the domain of f and f() is in the domain of g. Functions f and g are inverses of each other provided: f( g( )) = and g( f( )) = The function g is denoted as f, read as f inverse. When a function f has an inverse, for every point (a,b) on the graph of f, there is a point (b,a) on the graph of f. This means that the graph of f can be obtained from the graph of f by changing every point (,y) on f to the point (y,). This also means that the graphs of f and are reflections about the line y=. Verify that the functions f f( ) = and g( ) = are inverses! Horizontal Line Test: If any horizontal line which is drawn through the graph of a function f intersects the graph no more than once, then f is said to be a one-to-one function and has an inverse. Let s try some problems! Let f and g be functions whose values are given by the table below. Assume g is one-to-one with inverse g. X f() g() 6 9 0 4 4-6 ) f( g()) ) g (4) ) f( g (6)) 4) f ( f ( g ())) 5) g( g ())

6) If f( ) = + 5 and g( ) = + a, find a so that the graph of f o g crosses the y-ais at. TRIGONOMETRY What you need to know: Trig values for selected angles: Radians Degrees Sin Cos Tan 0 0 0 0 π 6 0 π 4 45 π 60 π 90 0 Und. Key Trig Identities formulas you should know: () i sinθ + cosθ = ( ii ) tan θ + = sec θ ( iv ) + cot θ = csc θ ( v) sin( + y) = sin cosy + cos sin y ( vi ) sin( y ) = sin cosy cos sin y ( vii ) cos( + y ) = cos cosy sin siny ( viii ) cos( y ) = cos cosy + sin sin y ( i ) sinθ = sinθcosθ ( ) cos cos sin θ = θ θ = cos θ = sin θ or

Without using a calculator or table, find each value: π 7π. cos( π ). sin. cos 6 π 5π π 4. sin 5. cos 6.sin 6 4 7. Find the slope and equation of the line with y-intercept = 4 and inclination 45. (the inclination of a line is the angle α, where 0 α 80, that is measured from the positive -ais to the line. If a line has slope m, then m = tanα ). 8. Find the inclination of the line given the equation: + 5y = 8 (you ll need a calculator here ) Solve the trigonometric equations algebraically by using identities and without the use of a calculator. You must find all solutions in the interval 0 θ π : 9. sin θ = 0 0. cos tan = cos. sin = sin. cos = cos. cos + cos = 0 4. sin = cos

EXPONENTIAL AND LOGARITHMIC FUNCTIONS LOGARITHMS What you need to know Definition of Logarithm with base b: For any positive numbers b and y with b, we define the logarithm of y with base b as follows: log b y = if and only if b = y LAWS OF LOGARITHMS (for M, N, b > 0, b ) ( i) logb MN = logb M + logbn M ( ii ) logb = logb M logbn N ( iii ) log M = log N if and only if M = N b k ( iv ) log M = k log M b logb M ( ) v b = M b b The logarithmic and eponential functions are inverse functions. Eample: Consider f( ) = and g( ) = log. Verify that (,8) is on the graph of f and (8,) on the graph of g. In addition, log ( ( )) f g = = and g( f( )) = log ( ) = which verifies that f and g are indeed inverse functions. Evaluate. The first is to be used as an eample. (No Calculators, please!) log 64 =. (because 4 = 64). 5 4 log = 5. 5. log 4 = 4. log 9 = 4 4 log 5 = 6. log 5 8 4 = Solve for. 7.log 64 = 8. = log ( 7) 4

log 5(log ) = 0 log 6(log (log )) = 0 9. 0. 4. log (log 64) = log 4 + log 8 = log. Show that by evaluating the logarithms. Make a generalization based on your findings. Write the given epression as a rational number or as a single logarithm. (Use the laws of logarithms above!). log 8 + log5 log 4 4. log48 log7 5. (log M logn log P ) 6. log8 80 log8 5 If and s. log = rand log 5 = s, 8 8 log 75 log8 0. 7. 8. 8 epress the given logarithm in terms of r 9. Solve for : log ( + 8) = log + log 6 Solve (no calculator please!) 0. = 8. 4 = 8. 4 = 6 5

. 6 = 0 (hint: 6 ( ) ( ) 6 = ) Simplify (recall: the notation ln stands for the natural logarithm base e) 4. ln ln ln a)ln e b) e c) e d) e Solve for (answers can be left in terms of e). 5. ln ln( ) = ln 6. (ln ) 6(ln ) + 9 = 0 We conclude with the following definition: EXPONENTIAL GROWTH MODEL: Consider an initial quantity Y o that changes eponentially with time t. At any time t the amount Y after t units of time may be given by: kt Y = Yoe Where k>0 represents the growth rate and k<0 represents the decay rate. 7. Growth of Cholera Suppose that the cholera bacteria in a colony grows according to the kt eponential model P = Pe ). The colony starts with bacterium and doubles o in number every half hour. How many bacteria will the colony contain at the end of 4 hours? 8. Bacteria Growth A colony of bacteria is grown under ideal conditions in a laboratory so that the population increases eponentially with time. At the end of hours 6

there are 0,000 bacteria. At the end of 5 h there are 40,000 bacteria. How many bacteria were present initially? The half-life of a radioactive element is the time required for half of the radioactive nuclei present in a sample to decay. The net problem shows that the half-life is a constant that depends only on the radioactive substance and not on the number of radioactive nuclei present in the sample. 9) Find the half-life of a radioactive substance with decay equation kt y = y e and show that the half-life depends only on k. (hint: set up o kt the equation ye o = y0. Solve it algebraically for t!) Conclusion: The half-life of a radioactive substance with rate constant k (k>0) is Half-life = 0) Using Carbon-4 Dating Scientists who do carbon-4 dating use 5700 years for its half-life. Find the age of a sample in which 0% of the radioactive nuclei originally present have decayed. 7