coach stones’ expanded standard pre-calculus algorithm ... · coach stones’ expanded standard...

Coach Stones’ Expanded Standard Pre-Calculus Algorithm Packet

Section: P.1 Algebraic Expressions, Mathematical Models and Real Numbers

CLASSIFICATIONS OF NUMBERS NATURAL NUMBERS = N = {1,2,3,4,....} WHOLE NUMBERS = W = {0,1,2,3,4,....} INTEGERS = Z = {...., - 4, - 3, - 2, - 1,0,1,2,3,4,....} RATIONAL NUMBERS = Q = {NUMBERS THAT IN DECIMAL FORM EITHER REPEAT OR TERMINATE} IRRATIONAL NUMBERS = I = { NUMBERS THAT IN DECIMAL FORM DO NOT REPEAT OR TERMINATE} REAL NUMBERS = R = ANY NUMBER THAT IS RATIONAL OR IRRATIONAL

PROPERTIES OF REAL NUMBERS 1) COMMUTATIVE PROPERTY OF ADDITION a + b = b + a 2) COMMUTATIVE PROPERTY OF MULTIPLICATION a * b = b * a 3) ASSOCIATIVE PROPERTY OF ADDITION ( a + b ) + c = a + ( b + c ) 4) ASSOCIATIVE PROPERTY OF MULTIPLICATION ( a * b ) * c = a * ( b * c ) 5) ADDITIVE IDENTITY a + 0 = a 6) MULTIPLICATIVE IDENTITY a * 1 = a 7) ADDITIVE INVERSE a + ( - a ) = 0 8) MULTIPLICATIVE INVERSE a * ( 1 / a ) = 1 9) DISTRIBUTIVE PROPERTY a ( b + c ) = a * b + a * c and ( b + c ) a = b * a + c * a 10) REFLEXIVE PROPERTY OF EQUALITY a = a 11) SYMMETRIC PROPERTY OF EQUALITY if a = b then b = a 12) TRANSITIVE PROPERTY OF EQUALITY if a = b and b = c then a = c 13) MULTIPLICATIVE PROPERTY OF ZERO a * 0 = 0 14) ADDITION OR SUBTRACTION PROPERTY OF EQUALITY - YOU CAN ADD OR SUBTRACT THE SAME NUMBER TO BOTH SIDES OF AN EQUATION AND NOT CHANGE ITS’ EQUALITY 15) MULTIPLICATION OR DIVISION PROPERTY OF EQUALITY - YOU CAN MULTIPLY OR DIVIDE THE SAME NUMBER ON BOTH SIDES OF AN EQUATION AND NOT CHANGE ITS’ EQUALITY 16) SUBSTITUTION PROPERTY OF EQUALITY - WHEN YOU REPLACE A VALUE (OR AN EXPRESSION) IN AN EQUATION WITH A VALUE FOR WHICH IT IS EQUIVALENT. (i. e. - for any numbers a and b, if a = b then a may be replaced by b any where in that equation.)

ORDER OF OPERATIONS (PEMDAS) PLEASE (PARENTHESIS, BRACKETS, BRACES, DIVISION LINE, ABSOLUTE VALUE SYMBOLS, RADICALS, GREATEST INTEGER SYMBOLS) EXCUSE- (EXPONENTS) MY DEAR - (MULTIPLICATION AND DIVISION ARE DONE AT THE SAME TIME FROM LEFT TO RIGHT) AUNT SALLY (ADDITION AND SUBTRACTION AT THE SAME TIME FROM LEFT TO RIGHT)


Section: P.2 Exponents and Scientific Notation

MONOMIAL - A NUMBER, A VARIABLE, OR A PRODUCT OF NUMBERS AND VARIABLES

THE DEGREE OF A MONOMIAL IS THE SUM OF THE EXPONENTS THE VARIABLES IN THAT MONOMIAL. FOR EXAMPLE THE DEGREE OF THE MONOMIAL 6x2y3z4 I S 2 + 3 + 4 IS DEGREE 9

RULES FOR MONOMIALS NEGATIVE EXPONENTS a - n = 1 / a n PRODUCT OF POWERS a m * a n = a m + n POWER OF POWERS ( a m ) n = a m n POWER OF PRODUCT ( a b ) m = a m * b m POWER OF A MONOMIAL ( a m b n ) p = a m p * b n p QUOTIENT OF POWERS a m / a n = a m - n

ZERO EXPONENTS - ANY NUMBER OR VARIABLE (EXCEPT THE NUMBER ZERO) RAISED TO THE ZERO POWER EQUALS ONE

DEGREE OF A POLYNOMIAL IS EQUAL TO THE HIGHEST DEGREE OF IS’ MONOMIAL TERMS. FOR EXAMPLE THE DEGREE OF THE POLYNOMIAL 6x2y3 + 2xy2 - 7xy IS DEGREE 5 BECAUSE THE HIGHEST DEGREE OF ITS MONOMIAL TERMS IS FROM THE TERM 6x2y3 WHICH IS DEGREE 5


Section: P.3 Radicals and Rational Exponents

PRODUCT PROPERTY OF SQUARE ROOTS √ab = √a * √b

QUOTIENT PROPERTY OF SQUARE ROOTS √ a / b = √a / √b

PRIME NUMBERS ARE NUMBERS WHOSE ONLY FACTORS ARE ONE AND THEMSELVES. EXAMPLES : { 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 43, . . . } FOUR THINGS THAT MUST BE TRUE FOR A RADICAL TO BE IN SIMPLEST RADICAL FORM 1) THE RADICAND IS NOT A FRACTION. 2) NO RADICAL APPEARS IN THE DENOMINATOR. . 3) THE RADICAND HAS NO FACTORS RAISED TO A POWER GREATER THAN OR EQUAL TO INDEX 4) THE INDEX (OR ROOT) IS AS SMALL AS POSSIBLE

STEPS TO ADD OR SUBTRACT RADICALS EXPRESSIONS 1) SIMPLIFY EACH RADICAL TERM FIRST 2) THEN COMBINE (ADD OR SUBTRACT) TERMS WITH THE SAME ROOTS (INDEXES) AND THE SAME RADICANDS.


Section: P.4 Polynomials

MONOMIAL TERMS. FOR EXAMPLE THE DEGREE OF THE POLYNOMIAL 6x2y3 + 2xy2 - 7xy IS DEGREE 5 BECAUSE THE HIGHEST DEGREE OF ITS MONOMIAL TERMS IS FROM THE TERM 6x2y3 WHICH IS 5 BINOMIAL - TWO MONOMIALS COMBINED BY ADDITION OR SUBTRACTION

TRINOMIAL - THREE MONOMIALS COMBINED BY ADDITION OR SUBTRACTION POLYNOMIAL - ONE OR MORE MONOMIALS COMBINED BY ADDITION OR SUBTRACTION

LIKE TERMS- YOU CAN ONLY COMBINE ( ADD OR SUBTRACT) LIKE TERMS (EXAMPLES; TERMS WITH THE SAME VARIABLES RAISED TO THE SAME POWER OR RADICALS WITH THE SAME ROOTS AND THE SAME RADICANDS)

USE THE DISTRIBUTIVE PROPERTY TO MULTIPLY A MONOMIAL TIMES A POLYNOMIAL USE THE FOIL METHOD TO MULTIPLY TWO BINOMIALS TOGETHER F the first terms, O the outer terms, I the inner terms, and L the last terms.


Section: P.5 Factoring Polynomials

GREATEST COMMON FACTOR - THE LARGEST NUMBER THAT WILL EVENLY DIVIDE INTO TWO OR MORE NUMBERS

STEPS IN FACTORING BINOMIALS: I) TRY TO FACTOR OUT THE GREATEST COMMON FACTOR. 2) CHECK TO SEE IF THE BINOMIAL IS A “DIFFERENCE OF PERFECT SQUARES”`

1. IS THE FIRST TERM A PERFECT SQUARE? 2. IS THE ABSOLUTE VALUE OF THE SECOND TERM A PERFECT SQUARE? 3. IS IT A DIFFERENCE? (IS IT A SUBTRACTION EXPRESSION?)

Coach Stones’ Expanded Standard Pre-Calculus Algorithm Packet 4. THEN THE FACTORIZATION IS IN THE FORM OF ( “the square root of the first term” + “the square root of the second term” ) times ( “the square root of the first term” - “the square root of the second term” )

3) CHECK TO SEE IF THE BINOMIAL IS A “SUM OF PERFECT CUBES” 1. IS THE FIRST TERM A PERFECT CUBE? 2. IS THE SECOND TERM A PERFECT CUBE? 3. IS IT A SUM? (IS IT AN ADDITION EXPRESSION?) 4. THEN THE FACTORIZATION IS IN THE FORM OF ( a +b ) ( a 2 - ab + b 2 ) WHERE “ a ” REPRESENTS THE CUBE ROOT OF THE FIRST TERMS OF THE ORIGINAL BINOMIAL AND “ b ” REPRESENTS THE CUBE ROOT OF THE SECOND TERM OF THE ORIGINAL BINOMIAL

4) CHECK TO SEE IF THE BINOMIAL IS A “DIFFERENCE OF PERFECT CUBES” 1. IS THE FIRST TERM A PERFECT CUBE? 2. IS THE SECOND TERM A PERFECT CUBE? 3. IS IT A DIFFERENCE? (IS IT A SUBTRACTION EXPRESSION?) 4. THEN THE FACTORIZATION IS IN THE FORM OF ( a - b ) ( a 2 +ab + b 2 ) WHERE “ a ”REPRESENTS THE CUBE ROOT OF THE FIRST TERMS OF THE ORIGINAL BINOMIAL AND “ b ” REPRESENTS THE CUBE ROOT OF THE SECOND TERM OF THE ORIGINAL BINOMIAL

5) IF THE BINOMIAL IS NOT FACTORABLE USING ANY OF THESE METHODS THEN IT IS “ PRIME ”. STEPS IN FACTORING TRINOMIALS: 1. TRY TO FACTOR OUT THE GREATEST COMMON FACTOR. 2. CHECK TO SEE IF THE TRINOMIAL IS A “PERFECT SQUARE TRINOMIAL”

1. IS THE FIRST TERM A PERFECT SQUARE? 2. IS THE THIRD TERM A PERFECT SQUARE? 3. IS THE ABSOLUTE VALUE OF THE SECOND TERM EQUAL TO TWICE THE SQUARE ROOT OF THE FIRST TERM TIMES THE SQUARE ROOT OF THE THIRD TERM? 4. THEN THE FACTORIZATION IS IN THE FORM OF (a + b)2 OR (a - b)2. WHERE “a” REPRESENTS THE SQUARE ROOT OF THE FIRST TERM OF THE ORIGINAL TRINOMIAL AND “b” REPRESENTS THE SQUARE ROOT OF THE THIRD TERM OF THE ORIGINAL TRINOMIAL. (THE SIGN OF THE FACTORIZATION IS DETERMINED BY THE SIGN OF THE SECOND TERM OF THE ORIGINAL TRINOMIAL)

3. THE SEVEN-STEP METHOD. 1. PUT THE TRINOMIAL IN DESCENDING ORDER. 2. FIND THE PRODUCT OF THE COEFFICIENT OF THE FIRST TERM AND THE COEFFICIENT OF THE THIRD TERM. 3. LIST ALL THE PAIRS OF INTEGERS THAT ARE FACTORS OF THE PRODUCT FOUND IN STEP 2. 4. FIND THE SUM OF EACH PAIR OF INTEGER FACTORS FOUND IN STEP 3 (UNTIL YOU FIND THE PAIR OF FACTORS WHOSE SUM IS EQUAL TO THE MIDDLE TERM OF THE ORIGINAL TRINOMIAL.) 5. USE THIS PAIR OF FACTORS TO SUBSTITUTE INTO THE ORIGINAL TRINOMIAL USING THE FORM a x2 + ( ? + ? ) x + c 6. DISTRIBUTE THE VARIABLE THROUGH THE PARENTHESIS. YOU WILL THEN HAVE A FOUR-NOMIAL. 7. THE FOLLOW THE STEPS FOR FACTORING A FOUR-NOMIAL

4. IF THE TRINOMIAL IS NOT FACTORABLE USING ANY OF THESE METHODS THEN IT IS “ PRIME ”.


STEPS FOR FACTORING A FOUR-NOMIAL BY GROUPING: 1. TRY TO FACTOR OUT THE GREATEST COMMON FACTOR. 2. TRY TO FACTOR THE FOUR NOMIAL BY GROUPING.

a. GROUP THE FOUR-NOMIAL INTO PAIRS b. FACTOR OUT THE GREATEST COMMON FACTOR OF EACH PAIR, TO FORM TWO BINOMIALS.

3. TRY TO FACTOR THE FOUR NOMIAL BY GROUPING THE FOUR - NOMIAL INTO BINOMIAL AND MONOMIAL PAIRS.

a. GROUP THE FOUR-NOMIAL INTO THE FORM ( x + a ) 2 - b 2 SO THAT THE FINAL b. FACTORIZATION IS IN THE FORM OF [ ( x + a ) + b2 ] * [ ( x + a ) - b 2

Section: P.6 Rational Expressions

RATIONAL EXPRESSION - AN ALGEBRAIC FRACTION WHOSE NUMERATOR AND DENOMINATOR ARE POLYNOMIALS

STEPS FOR MULTIPLYING RATIONAL EXPRESSIONS OR FRACTIONS 1) COMPLETELY FACTOR EACH NUMERATOR AND DENOMINATOR 2) CANCEL ANY LIKE TERMS THAT APPEAR IN BOTH THE NUMERATOR AND DENOMINATOR 3) ALL REMAINING TERMS IN THE NUMERATOR STAY IN THE NUMERATOR OF YOUR PRODUCT AND ALL REMAINING TERMS IN THE DENOMINATOR STAY IN THE DENOMINATOR OF YOUR PRODUCT

STEPS FOR DIVIDING RATIONAL EXPRESSIONS OR FRACTIONS 1) INVERT THE SECOND RATIONAL EXPRESSION OR FRACTION (THE RATIONAL EXPRESSION OR FRACTION YOU ARE DIVIDING BY) 2) CHANGE THE DIVISION SIGN TO A MULTIPLICATION SIGN 3) THEN USE THE RULES FOR MULTIPLYING RATIONAL EXPRESSION OR FRACTIONS.

STEPS FOR ADDING (OR SUBTRACTING) FRACTIONS 1) CHECK TO SEE IF THEY ALREADY HAVE A COMMON DENOMINATOR. (IF NOT, FIND THE LEAST COMMON MULTIPLE OF THE DENOMINATORS AND THEN MAKE EQUIVALENT FRACTIONS WITH THIS LEAST COMMON MULTIPLE AS THE NEW DENOMINATOR.) 2) THEN ADD (OR SUBTRACT) THE NUMERATORS AND WRITE THE SUM (OR DIFFERENCE) OVER THAT COMMON DENOMINATOR. 3) SIMPLIFY YOUR ANSWER. (REDUCE THE FRACTION BY EITHER FACTORING OUT THE GREATEST COMMON FACTOR FROM BOTH THE NUMERATOR AND DENOMINATOR OR IF IT IS AN IMPROPER FRACTION THEN WRITE YOUR ANSWER AS A MIXED NUMBER)

EXCLUDED VALUE - VALUES THAT HAVE TO BE EXCLUDED FROM THE POSSIBLE ANSWERS FOR A RATIONAL EXPRESSION BECAUSE THEY WOULD ALLOW THE DENOMINATOR OF THE RATIONAL EXPRESSION TO BE EQUAL TO ZERO. THIS WOULD MAKE THE RATIONAL EXPRESSION UNDEFINED IN MATHEMATICS.


Section: 1.1 Graphs and Graphing Utilities

Steps for graphing functions:

1. Solve the function for ‘y”. 2. Make a table and choose some values for “x” and solve for the corresponding values for “y”. 3. Plot the points on a coordinate plane and connect the points with either straight line or a smooth

curve depending on how the points line up.

Section: 1.2 Linear and Rational Equations

STEPS TO SOLVE AN EQUATION 1) SIMPLIFY BOTH SIDES OF THE EQUAL SIGN (USING THE ORDER OF OPERATIONS) 2) ISOLATE THE CHOSEN VARIABLE ON ONE SIDE OF THE EQUAL SIGN (USING ADDITION OR SUBTRACTION. ) 3) GET THE CHOSEN VARIABLE BY ITSELF. (WITH NOTHING ADDED TO OR SUBTRACTED FROM IT) 4) AND GET THE COEFFICIENT OF THE CHOSEN VARIABLE TO EQUAL POSITIVE ONE. (BY USING MULTIPLICATION OR DIVISION) 5) CHECK YOUR ANSWER BACK INTO THE ORIGINAL EQUATION TO MAKE SURE THAT IT IS A TRUE STATEMENT

Steps for solving Rational Equations 1. Make sure all of the terms are written as rational expressions. (If not make them into rational expressions by putting them over one.) 2. Find the least common denominator of all of the terms in the rational equation. 3. Multiply the entire rational equation by the least common denominator. (This should cause all of the denominators to be canceled out and leave only an equation to solve.) 4. Solve the resulting equation for the desired variable. 5. Check your answer back into the original rational equation to make sure it is the correct answer. (Make sure that none of your solutions are excluded values.)


Section: 1.3 Models and Application


Section: 1.4 Complex Numbers

IMAGINARY NUMBERS BY DEFINITION √ -1 = i then i 2 = - 1 i 3 = - i

i 4 = 1 and i 0 = 1

COMPLEX NUMBERS ARE ALWAYS WRITTEN IN THE a + bi FORM WHERE “ a “ IS THE REAL NUMBER PART AND “ bi “ IS THE IMAGINARY PART

TWO COMPLEX NUMBERS ARE EQUAL IF AND ONLY IF THE REAL NUMBER PARTS ARE EQUAL AND THE IMAGINARY NUMBER PARTS ARE EQUAL

THE CONJUGATE OF “ a + bi “ is “ a - bi “

Section: 1.5 Quadratic Equations

GOOD STANDARD QUADRATIC EQUATION FORM IS y = a x2 + b x + c

GREATEST COMMON FACTOR - THE LARGEST NUMBER THAT WILL EVENLY DIVIDE INTO TWO OR MORE NUMBERS STEPS IN FACTORING BINOMIALS: I) TRY TO FACTOR OUT THE GREATEST COMMON FACTOR. 2) CHECK TO SEE IF THE BINOMIAL IS A “DIFFERENCE OF PERFECT SQUARES”`

1. IS THE FIRST TERM A PERFECT SQUARE? 2. IS THE ABSOLUTE VALUE OF THE SECOND TERM A PERFECT SQUARE? 3. IS IT A DIFFERENCE? (IS IT A SUBTRACTION EXPRESSION?) 4. THEN THE FACTORIZATION IS IN THE FORM OF ( “the square root of the first term” + “the square root of the second term” ) times ( “the square root of the first term” - “the square root of the second term” )

3) CHECK TO SEE IF THE BINOMIAL IS A “SUM OF PERFECT CUBES” 1. IS THE FIRST TERM A PERFECT CUBE? 2. IS THE SECOND TERM A PERFECT CUBE? 3. IS IT A SUM? (IS IT AN ADDITION EXPRESSION?) 4. THEN THE FACTORIZATION IS IN THE FORM OF ( a +b ) ( a 2 - ab + b 2 ) WHERE “ a ” REPRESENTS THE CUBE ROOT OF THE FIRST TERMS OF THE ORIGINAL BINOMIAL AND “ b ” REPRESENTS THE CUBE ROOT OF THE SECOND TERM OF THE ORIGINAL BINOMIAL

4) CHECK TO SEE IF THE BINOMIAL IS A “DIFFERENCE OF PERFECT CUBES” 1. IS THE FIRST TERM A PERFECT CUBE? 2. IS THE SECOND TERM A PERFECT CUBE? 3. IS IT A DIFFERENCE? (IS IT A SUBTRACTION EXPRESSION?) 4. THEN THE FACTORIZATION IS IN THE FORM OF ( a - b ) ( a 2 + ab + b 2 ) WHERE “ a ”REPRESENTS THE CUBE ROOT OF THE FIRST TERMS OF THE ORIGINAL BINOMIAL AND “ b ” REPRESENTS THE CUBE ROOT OF THE SECOND TERM OF THE ORIGINAL BINOMIAL 5) IF THE BINOMIAL IS NOT FACTORABLE USING ANY OF THESE METHODS THEN IT IS “ PRIME ”.


STEPS IN FACTORING TRINOMIALS: 1. TRY TO FACTOR OUT THE GREATEST COMMON FACTOR. 2. CHECK TO SEE IF THE TRINOMIAL IS A “PERFECT SQUARE TRINOMIAL”

1. IS THE FIRST TERM A PERFECT SQUARE? 2. IS THE THIRD TERM A PERFECT SQUARE? 3. IS THE ABSOLUTE VALUE OF THE SECOND TERM EQUAL TO TWICE THE SQUARE ROOT OF THE FIRST TERM TIMES THE SQUARE ROOT OF THE THIRD TERM? 4. THEN THE FACTORIZATION IS IN THE FORM OF (a + b)2 OR (a - b)2. WHERE “a” REPRESENTS THE SQUARE ROOT OF THE FIRST TERM OF THE ORIGINAL TRINOMIAL AND “b” REPRESENTS THE SQUARE ROOT OF THE THIRD TERM OF THE ORIGINAL TRINOMIAL. (THE SIGN OF THE FACTORIZATION IS DETERMINED BY THE SIGN OF THE SECOND TERM OF THE ORIGINAL TRINOMIAL)

3. THE SEVEN-STEP METHOD. 1. PUT THE TRINOMIAL IN DESCENDING ORDER. 2. FIND THE PRODUCT OF THE COEFFICIENT OF THE FIRST TERM AND THE COEFFICIENT OF THE THIRD TERM. 3. LIST ALL THE PAIRS OF INTEGERS THAT ARE FACTORS OF THE PRODUCT FOUND IN STEP 2. 4. FIND THE SUM OF EACH PAIR OF INTEGER FACTORS FOUND IN STEP 3 (UNTIL YOU FIND THE PAIR OF FACTORS WHOSE SUM IS EQUAL TO THE MIDDLE TERM OF THE ORIGINAL TRINOMIAL.) 5. USE THIS PAIR OF FACTORS TO SUBSTITUTE INTO THE ORIGINAL TRINOMIAL USING THE FORM a x2 + ( ? + ? ) x + c 6. DISTRIBUTE THE VARIABLE THROUGH THE PARENTHESIS. YOU WILL THEN HAVE A FOUR-NOMIAL. 7. THE FOLLOW THE STEPS FOR FACTORING A FOUR-NOMIAL

4. IF THE TRINOMIAL IS NOT FACTORABLE USING ANY OF THESE METHODS THEN IT IS “ PRIME ”. STEPS FOR FACTORING A FOUR-NOMIAL BY GROUPING: 1. TRY TO FACTOR OUT THE GREATEST COMMON FACTOR. 2. TRY TO FACTOR THE FOUR NOMIAL BY GROUPING.

a. GROUP THE FOUR-NOMIAL INTO PAIRS b. FACTOR OUT THE GREATEST COMMON FACTOR OF EACH PAIR, TO FORM TWO BINOMIALS.

3. TRY TO FACTOR THE FOUR NOMIAL BY GROUPING THE FOUR - NOMIAL INTO BINOMIAL AND MONOMIAL PAIRS.

a. GROUP THE FOUR-NOMIAL INTO THE FORM ( x + a ) 2 - b 2 SO THAT THE FINAL b. FACTORIZATION IS IN THE FORM OF [ ( x + a ) + b2 ] * [ ( x + a ) - b 2


STEPS FOR SOLVING QUADRATIC EQUATIONS USING THE FACTORING METHOD 1) MAKE SURE THAT THE EQUATION IS IN “ GOOD STANDARD QUADRATIC EQUATION FORM SET EQUAL TO ZERO “ (THAT IS ax2 + bx + c = 0 FORM) 2) FACTOR THE POLYNOMIAL COMPLETELY 3) SET EACH FACTOR THAT CONTAINS A VARIABLE EQUAL TO ZERO AND SOLVE FOR THE VARIABLE 4) CHECK EACH OF THE POSSIBLE ANSWERS BACK INTO THE ORIGINAL EQUATION TO MAKE SURE THAT IT IS A SOLUTION. (IF NONE OF THE POSSIBLE ANSWERS WORK WHEN CHECKED BACK INTO THE ORIGINAL EQUATION THEN THERE IS “ NO SOLUTION “ TO THE EQUATION.)

STEPS FOR SOLVING QUADRATIC EQUATIONS USING THE “ COMPLETING THE SQUARE “ METHOD 1) MAKE SURE THAT THE EQUATION IS IN “ GOOD STANDARD QUADRATIC EQUATION FORM “ (THAT IS ax2 + bx + c = 0 FORM) 2) CHECK TO MAKE SURE THAT THE COEFFICIENT OF THE QUADRATIC TERM (OF THE SQUARED TERM IS EQUAL TO ONE. (IF NOT MAKE IT ONE USING MULTIPLICATION OR DIVISION.) 3) ISOLATE THE CONSTANT 4) FIND 1/2 OF THE COEFFICIENT OF THE LINEAR TERM (OF THE DEGREE ONE TERM). SQUARE THIS PRODUCT AND ADD THIS NUMBER TO BOTH SIDES OF THE EQUAL SIGN. 5) FACTOR THE RESULTING “ PERFECT SQUARE TRINOMIAL “ AND SIMPLIFY THE OTHER SIDE OF THE EQUAL SIGN. 6) TAKE THE SQUARE ROOT OF BOTH SIDES OF THE EQUATION (REMEMBERING TO USE A “ +/- “ SIGN TO ADJUST FOR THE ABSOLUTE VALUE OF THE BINOMIAL.) 7) SOLVE FOR THE VARIABLE. 8) CHECK YOUR ANSWERS TO MAKE SURE THEY MAKE THE ORIGINAL STATEMENT TRUE.

STEPS FOR SOLVING QUADRATIC EQUATIONS USING THE “ QUADRATIC FORMULA “ 1) MAKE SURE THAT THE EQUATION IS IN “ GOOD STANDARD QUADRATIC EQUATION FORM “ (THAT IS ax2 + bx + c = 0 FORM) 2) FIND THE VALUES FOR “ a “, “ b “, and “ c “ 3) TO FIND THE VALUE OF THE DISCRIMINANT AND THEREFORE KNOW THE NUMBER AND NATURE OF THE ROOTS OF THE QUADRATIC EQUATION. SUBSTITUTE THE VALUES FOR “ a “, “ b “, and “ c “ INTO THE FORMULA b2 - 4 a c

1. IF THE VALUE IS A POSITIVE NUMBER AND A PERFECT SQUARE THEN THERE ARE 2 REAL RATIONAL ROOTS OF THE QUADRATIC EQUATION 2. IF THE VALUE IS A POSITIVE NUMBER AND NOT A PERFECT SQUARE THEN THERE ARE 2 REAL IRRATIONAL ROOTS OF THE QUADRATIC EQUATION 3. IF THE VALUE IS THE NUMBER “ ZERO “ THEN THERE IS 1 REAL ROOT OF THE QUADRATIC EQUATION 4. IF THE VALUE IS A NEGATIVE NUMBER THEN THERE ARE 2 IMAGINARY ROOTS OF THE QUADRATIC EQUATION

4) TO FIND THE ROOTS SUBSTITUTE FOR “ a “, “ b “, and “ c “ INTO THE QUADRATIC

FORMULA ( x = - b + / - √ b2 - 4 a c ) / 2 a

5) CHECK YOUR ANSWERS TO MAKE SURE THEY MAKE THE ORIGINAL STATEMENT TRUE.


Section: 1.6 Other Equations

STEPS FOR SOLVING A RADICAL EQUATION WHERE THE VARIABLE IS IN THE RADICAND 1. ISOLATE THE RADICAL THAT CONTAINS THE VARIABLE (IF THERE ARE MORE THAN ONE RADICALS THAT CONTAIN VARIABLES ISOLATE ONE OF THEM) 2. GET RID OF THE RADICAL BY RAISING BOTH SIDES OF THE EQUATION TO THE APPROPRIATE POWER (REMEMBER TO INCLUDE ABSOLUTE VALUES WHERE NECESSARY) 3. IF THERE IS STILL A RADICAL THAT CONTAINS A VARIABLE REPEAT STEPS ONE AND TWO 4. SOLVE THE REMAINING EQUATIONS 5. CHECK YOUR ANSWERS BACK INTO THE ORIGINAL EQUATION TO SEE IF THEY MAKE THE EQUATION TRUE (ONLY INCLUDE IN YOUR FINAL ANSWER, SOLUTIONS THAT MAKE THE ORIGINAL EQUATION TRUE. ALL OTHER ANSWERS ARE “EXTRANEOUS SOLUTIONS” AND SHOULD NOT BE INCLUDED IN THE FINAL ANSWER.)

STEPS IN SOLVING FOR A VARIABLE IN A NON-QUADRATIC EQUATION USING A QUADRATIC TECHNIQUE: 1. CHECK TO SEE IF IT IS IN THE FORM OF AN EXPRESSION IN DESCENDING ORDER SET EQUAL TO ZERO. (IF NOT MAKE IT ONE.) 2. CHECK TO SEE IF YOU WRITE THE EQUATION SO IT LOOKS LIKE A QUADRATIC EQUATION AND THEN USE THE QUADRATIC FORMULA TO SOLVE IT.

1. CHECK TO SEE IF YOU SQUARE THE VARIABLE IN THE MIDDLE TERM WILL YOU GET THE VARIABLE IN THE FIRST TERM. IF SO, THEN REWRITE THE EQUATION SO IT IS IN “QUADRATIC EQUATION FORM” AND USE THE QUADRATIC FORMULA TO SOLVE FOR THE VARIABLE. MAKE SURE TO CHECK YOUR ANSWERS BACK INTO THE ORIGINAL EQUATION TO MAKE SURE THAT THE MAKE THE ORIGINAL EQUATION TRUE. 2. IF YOU SQUARE THE VARIABLE IN THE MIDDLE TERM AND IT DOES NOT EQUAL THE VARIABLE IN THE FIRST TERM THEN, TRY USING THE FACTORING METHOD.

1. FACTOR THE POLYNOMIAL COMPLETELY. 2. SET ALL FACTORS THAT CONTAIN A VARIABLE EQUAL TO ZERO AND SOLVE EACH. 3. MAKE SURE TO CHECK YOUR ANSWERS BACK INTO THE ORIGINAL EQUATION TO MAKE SURE THAT THE MAKE THE ORIGINAL EQUATION TRUE.

ABSOLUTE VALUE | a | = a and | - a | = a (THE DEFINITION OF ABSOLUTE VALUE IS THE DISTANCE SOME NUMBER IS FROM ZERO) (THAT IS WHY THE ABSOLUTE VALUE OF A NUMBER CANNEVER BE A NEGATIVE NUMBER BECAUSE IT REPRESENTS A DISTANCE AND DISTANCES ARE ALWAYS POSITIVE)


STEPS FOR SOLVING ABSOLUTE VALUE EQUATIONS 1) ISOLATE THE ABSOLUTE VALUE TERM 2) TAKE WHATEVER IS INSIDE THE ABSOLUTE VALUE SYMBOLS (CALLED THE “ARGUMENT”) AND SET IT EQUAL TO WHATEVER IS ON THE OTHER SIDE OF THE EQUAL SIGN AND SOLVE FOR THE VARIABLE 3) THEN TAKE WHATEVER IS INSIDE THE ABSOLUTE VALUE SYMBOLS (CALLED THE “ARGUMENT”) AND SET IT EQUAL TO THE OPPOSITE OF WHATEVER IS ON THE OTHER SIDE OF THE EQUAL SIGN AND SOLVE FOR THE VARIABLE 4) CHECK THESE TWO ANSWERS BACK INTO THE ORIGINAL ABSOLUTE VALUE EQUATION TO SEE WHICH ONE OR ONES WORK (IF NEITHER VALUE WORKS BACK IN THE ORIGINAL EQUATION THEN THERE IS “NO SOLUTION” TO THE PROBLEM)

Section: 1.7 Linear and Absolute Value Inequalities

IMPORTANT FACTS ABOUT INEQUALITIES - IF YOU MULTIPLY OR DIVIDE “BOTH SIDES” OF AN INEQUALITY BY A NEGATIVE NUMBER YOU HAVE TO REVERSE THE INEQUALITY SYMBOL

STEPS FOR SOLVING AN INEQUALITY 1) SIMPLIFY BOTH SIDES OF THE INEQUALITY SIGN (USING THE ORDER OF OPERATIONS) 2) ISOLATE THE CHOSEN VARIABLE ON ONE SIDE OF THE INEQUALITY SIGN (USING ADDITION OR SUBTRACTION.) 3) GET THE CHOSEN VARIABLE BY ITSELF. (WITH NOTHING ADDED TO OR SUBTRACTED FROM IT) 4) AND GET THE CHOSEN VARIABLES’ COEFFICIENT EQUAL TO POSITIVE ONE. ( BY USING MULTIPLICATION OR DIVISION) (REMEMBER THAT IF YOU MULTIPLY OR DIVIDE BOTH SIDES OF A INEQUALITY BY A NEGATIVE NUMBER YOU MUST FLIP THE INEQUALITY SYMBOL) 5) CHECK YOUR ANSWER BACK INTO THE ORIGINAL INEQUALITY TO MAKE SURE THAT IT IS A TRUE STATEMENT

STEPS FOR SOLVING ABSOLUTE VALUE INEQUALITIES ON A NUMBER LINE 1) ISOLATE THE ABSOLUTE VALUE TERM (MEANING TO GET THIS TERM ON ONE SIDE OF THE INEQUALITY SYMBOL BY ITSELF WITH ITS’ COEFFICIENT EQUAL TO ONE) 2) NOW WE MUST LOOK AT THE POSITIVE POSSIBILITY

1. WRITE WHATEVER IS INSIDE THE ABSOLUTE VALUE SYMBOLS USING THE SAME INEQUALITY SYMBOL AND WHATEVER IS ON THE OTHER SIDE OF THE INEQUALITY SYMBOL 2. SOLVE FOR THE VARIABLE

3) NOW WE MUST LOOK AT THE NEGATIVE POSSIBILITY 1. WRITE WHATEVER IS INSIDE THE ABSOLUTE VALUE SYMBOLS USING THE OPPOSITE INEQUALITY SYMBOL AND THE OPPOSITE OF WHATEVER IS ON THE OTHER SIDE OF THE INEQUALITY SYMBOL 2. SOLVE FOR THE VARIABLE

4) TO SEE THE SOLUTION IN GRAPH FORM GRAPH BOTH OF YOUR SOLUTIONS ON THE NUMBER LINE


2-1 Section: 2.1 Basics of Functions and Their Graphs

THE “DOMAIN” OF A RELATION IS THE SET OF ALL “X” VALUES THAT SATISFY THAT RELATION

THE “RANGE” OF A RELATION IS THE SET OF ALL “Y” VALUES THAT SATISFY THAT RELATION

A RELATION IS A FUNCTION IF AND ONLY IF EACH ELEMENT OF THE DOMAIN OF THAT RELATION GOES TO ONE AND ONLY ONE ELEMENT OF THE RANGE OF THAT RELATION FUNCTIONAL NOTATION - THE EQUATION y = 3x + 2 CAN BE WRITTEN I N FUNCTIONAL NOTATION AS f (x) = 3x + 2


TO FINDING THE SOLUTION FOR A FUNCTION AT A GIVEN DOMAIN VALUE SUBSTITUTE THAT VALUE INTO THE FUNCTION WHERE EVER THERE IS AN “ x ”

EXAMPLE IF f (x) = 4 x + 5 THEN f (6) = 4 (6) + 5 OR f (6) = 24 + 5

f (6) = 29

A DISCRETE FUNCTION IS A FUNCTION WHOSE POINTS ARE COUNTABLE (DISCRETE) AND THE GRAPH MAY CONSISTS OF POINTS IN A PLANE THAT ARE NOT CONNECTED BY A LINE.

A CONTINUOUS FUNCTION IS A FUNCTION WHICH CONTAINS INFINITELY MANY POINTS AND THE GRAPH CONSISTS A LINE OR A SMOOTH CURVE.

Section: 2.2 More Functions


Section: 2.3 Linear Functions and Slope

THE SLOPE OF A LINE IF GIVEN TWO POINTS ON THE LINE IS THE RELATION OF THE LINES’ RISE OVER ITS’ RUN AND CAN BE FOUND GIVEN TWO POINTS USING THE FORMULA

SLOPE FORMULA m = ( y2 - y1 ) / ( x2 - x1 )

LINEAR EQUATIONS IN SLOPE-INTERCEPT FORM ARE ALWAYS IN THE FORM OF y = m x + b WHERE “ m ” IS THE SLOPE AND “ b ” IS THE “ y “ - INTERCEPT

LINEAR EQUATIONS IN “POINT-SLOPE” FORM ARE ALWAYS IN THE FORM OF ( y - y1 ) = m ( x - x1 ) WHERE “ m ” IS THE SLOPE AND “ x1 ” IS THE “ X “ VALUE IN THE GIVEN ORDERED PAIR AND “ y1” IS THE “ Y “ VALUE IN THE GIVEN ORDERED


LINEAR EQUATIONS IN FUNCTIONAL NOTATION FORM ARE ALWAYS IN THE FORM OF f ( x ) = m x + b WHERE “ m ” IS THE SLOPE AND “ b ” IS THE “ y “ – INTERCEPT


Section: 2.4 More on Slope

LINEAR EQUATIONS IN SLOPE-INTERCEPT FORM ARE ALWAYS IN THE FORM OF y = m x + b WHERE “ m ” IS THE SLOPE AND “ b ” IS THE “ y “ - INTERCEPT

LINEAR EQUATIONS IN “POINT-SLOPE” FORM ARE ALWAYS IN THE FORM OF ( y - y1 ) = m ( x - x1 ) WHERE “ m ” IS THE SLOPE AND “ x1 ” IS THE “ X “ VALUE IN THE GIVEN ORDERED PAIR AND “ y1” IS THE “ Y “ VALUE IN THE GIVEN ORDERED STEPS FOR WRITING A QUADRATIC FUNCTION INTO “VERTEX FORM” (THE FORM OF y = a ( x - h ) 2 + k 1) CHECK TO MAKE SURE THE EQUATION IS WRITTEN IN “GOOD STANDARD QUADRATIC EQUATION FORM” ( y = ax2 + bx + c) INCLUDING ANY NEEDED SPACE HOLDERS 2) GROUP THE QUADRATIC AND LINEAR TERMS USING PARENTHESIS 3) FACTOR “ a “ (THE COEFFICIENT OF THE QUADRATIC TERM) OUT OF BOTH THE QUADRATIC AND LINEAR TERMS 4) COMPLETE THE SQUARE INSIDE THE PARENTHESIS 5) SUBTRACT THE PRODUCT OF “ a “ (THE COEFFICIENT OF THE QUADRATIC TERM) AND THE NUMBER ADDED INSIDE THE PARENTHESIS TO COMPLETE THE SQUARE FROM THE CONSTANT 6) FACTOR THE COMPLETED SQUARE AND SIMPLIFY THE CONSTANT 7) WRITE THE QUADRATIC IN THE FORM y = a ( x - h ) 2 + k . PERPENDICULAR LINES ARE LINES WHOSE SLOPES ARE NEGATIVE RECIPROCALS OF ONE ANOTHER

PARALLEL LINES ARE LINES WHOSE SLOPES ARE THE EQUAL TO ONE ANOTHER.


Section: 2.6 Composite Functions


Section: 2.7 Inverse Functions

INVERSE RELATIONS - TWO RELATIONS ARE INVERSE RELATIONS IF AND ONLY IF WHENEVER ONE

IF A RELATION CONTAINS THE ELEMENT ( a , b ), THE INVERSE RELATION CONTAINS THE ELEMENT ( b , a ).

Steps for finding the inverse of a function 1. Rewrite the function out of functional notation (replace the f(x) with “ y “ ) 2. Switch the “ x” ‘s and “ y “ ‘s (where ever there was an “ x “ put a “ y “ and where ever there was an “ y “ put an “ x “) 3. Solve this new equation for “ y “ 4. Write the new equation back into inverse functional notation (use f - 1 (x) in place of f (x) )


Section: 2.8 Distance/Midpoint Formulas

THE FORMULA FOR THE MIDPOINT OF A LINE SEGMENT ON A NUMBER LINE IS P = [ ( x1 + x2 ) / 2]

THE FORMULA FOR THE MIDPOINT OF A LINE SEGMENT IN A COORDINATE PLANE IS ( xm , ym ) = ( [ ( x1 + x2 ) / 2 ] , [ ( y1 + y2 ) / 2 ] )


STEPS FOR WRITING AN EQUATION FOR A CIRCLE FROM “GENERAL FORM” x2 + y2 + Dx + Ey + F = 0 INTO “STANDARD FORM” (x - h )2 + ( y - k )2 = r2 1) ISOLATE THE CONSTANT 2) USING PARENTHESES, GROUP THE x2 AND THE x TERMS 3) USING PARENTHESES, GROUP THE y2 AND THE y TERMS 4) FACTOR OUT THE COEFFICIENT OF THE x2 TERM FROM BOTH THE x2 AND THE x TERMS AND COMPLETE THE SQUARE INSIDE THE PARENTHESIS 5) FACTOR OUT THE COEFFICIENT OF THE y2 TERM FROM BOTH THE y2 AND THE y TERMS AND COMPLETE THE SQUARE INSIDE THE PARENTHESIS 6) ADD TO THE CONSTANT THE PRODUCT OF THE VALUE FACTORED OUT IN STEP 4 AND THE NUMBER YOU ADDED IN INSIDE THE “ x “ PARENTHESIS TO COMPLETE THE SQUARE. 7) ADD TO THE CONSTANT THE PRODUCT OF THE VALUE FACTORED OUT IN STEP 5 AND THE NUMBER YOU ADDED IN INSIDE THE “ y “ PARENTHESIS TO COMPLETE THE SQUARE. 8) WRITE INTO THE STANDARD FORM OF A CIRCLE ( x - h )2 + ( y - k )2 = r2 MAKING SURE THAT THE COEFFICIENTS OF THE ( x - h )2 AND THE ( y - k )2 ARE BOTH 1 IF NOT USE MULTIPLICATION OR DIVISION TO MAKE THEM 1 INFORMATION ABOUT CIRCLES EQUATION OF A CIRCLE IN “STANDARD FORM” ( x - h )2 + ( y - k )2 = r2 CENTER ( h , k ) AND THE RADIUS = r

Section: 3.1 Quadratic Functions

GOOD STANDARD QUADRATIC EQUATION FORM IS y = a x 2 + b x + c

STEPS FOR SOLVING QUADRATIC EQUATIONS BY GRAPHING 1) CHECK TO MAKE SURE THAT THE EQUATION IS WRITTEN IN GOOD STANDARD FORM y = a x 2 + b x + c 2) FIND THE AXIS OF SYMMETRY BY USING THE EQUATION x = ( - b /2 a ) 3) SET UP A CHART AND CHOOSE AT LEAST 5 VALUES FOR “ X” AND SOLVE FOR THE CORRESPONDING VALUES FOR “ y .“ ( use the value for “ X” at the axis of symmetry and also choose at least two values for “ X” on either side of the axis of symmetry. ) 4) PLOT THE ORDERED PAIRS FOUND IN STEP 2 AND GRAPH IN THE PARABOLA 5) THE “ X ” - INTERCEPTS ARE THE SOLUTIONS (also called the “roots” or “zeros”) TO THE EQUATION 6) CHECK EACH OF THE POSSIBLE SOLUTIONS FOUND IN STEP 5 BACK INTO THE ORIGINAL EQUATION TO MAKE SURE THAT IT IS AN ACTUAL SOLUTION. ( IF THERE ARE NO “ X ” - INTERCEPTS OR IF NONE OF THE POSSIBLE ANSWERS WORK WHEN CHECKED BACK INTO THE ORIGINAL EQUATION THEN THERE ARE “NO REAL ROOTS” OR THERE IS “ NO SOLUTION “ TO THE EQUATION. )


STEPS FOR WRITING A QUADRATIC FUNCTION INTO “STANDARD FORM” (THE FORM OF y = a ( x - h ) 2 + k ) . 1) CHECK TO MAKE SURE THE EQUATION IS WRITTEN IN “GOOD STANDARD QUADRATIC EQUATION FORM” ( y = ax2 + bx + c) INCLUDING ANY NEEDED SPACE HOLDERS 2) GROUP THE QUADRATIC AND LINEAR TERMS USING PARENTHESIS 3) FACTOR “ a “ ( THE COEFFICIENT OF THE QUADRATIC TERM ) OUT OF BOTH THE QUADRATIC AND LINEAR TERMS 4) COMPLETE THE SQUARE INSIDE THE PARENTHESIS 5) SUBTRACT THE PRODUCT OF “ a “ ( THE COEFFICIENT OF THE QUADRATIC TERM ) AND THE NUMBER ADDED INSIDE THE PARENTHESIS TO COMPLETE THE SQUARE FROM THE CONSTANT 6) FACTOR THE COMPLETED SQUARE AND SIMPLIFY THE CONSTANT 7) WRITE THE QUADRATIC IN THE FORM y = a ( x - h ) 2 + k .


Section: 3.2 Polynomial Functions and Graphs


Section: 3.3 Dividing Polynomials – Remainder and Factor Theorems

STEPS TO DIVIDE POLYNOMIALS BY POLYNOMIALS USING LONG DIVISION: 1) MAKE SURE THAT THE DIVIDEND AND THE DIVISOR ARE BOTH POLYNOMIALS IN DESCENDING ORDER WITH SPACE HOLDERS FOR EACH DEGREE 2) FIND THE MONOMIAL BY WHICH YOU WILL HAVE TO MULTIPLY THE “ LEADING MONOMIAL OF THE DIVISOR” BY TO GET THE “ LEADING MONOMIAL OF THE DIVIDEND “ 3) MULTIPLY THE ENTIRE POLYNOMIAL OF THE DIVISOR BY THE MONOMIAL FOUND IN STEP 2 AND SUBTRACT THE RESULTING POLYNOMIAL FROM THE DIVIDEND 4) BRING DOWN THE NEXT MONOMIAL TERM FROM YOUR ORIGINAL DIVIDEND AND REPEAT STEPS 2 AND 3 UNTIL THE DIVISOR WILL NO LONGER DIVIDE INTO THE REMAINDER 5) WRITE THE REMAINDER DIVIDED BY THE DIVISOR AND ADD THIS TERM TO THE POLYNOMIAL QUOTIENT 6) IF THE REMAINDER IS ZERO THEN THE QUOTIENT AND THE DIVISOR ARE BOTH FACTORS OF THE ORIGINAL DIVIDEND) IF THE REMAINDER IS ZERO THEN THE QUOTIENT IS A FACTOR OF THE ORIGINAL DIVIDEND


STEPS TO DIVIDE POLYNOMIALS BY POLYNOMIALS USING SYNTHETIC DIVISION: 1) MAKE SURE THAT THE DIVIDEND AND THE DIVISOR ARE BOTH POLYNOMIALS IN DESCENDING ORDER WITH SPACE HOLDERS FOR EACH DEGREE 2) MAKE SURE THAT THE DIVISOR IS IN THE FORM OF (OR CAN BE PUT IN THE FORM OF) x – a WHERE “ x “ IS A VARIABLE WHOSE COEFFICIENT AND EXPONENT ARE BOTH ONE AND “ a “ IS A REAL NUMBER. ( IF THIS CANNOT BE DONE THEN IT IS IMPOSSIBLE TO USE THIS METHOD AND YOU WILL HAVE TO USE LONG DIVISION TO DIVIDE THE POLYNOMIALS.) 3) WRITE “ a “ IN A BRACKET TO SET IT A PART 4) ON THE SAME LINE WRITE THE COEFFICIENTS OF THE DIVIDEND 5) LEAVE A ONE LINE SPACE IMMEDIATELY UNDER THESE COEFFICIENTS AND DRAW A N ADDITION LINE 6) CARRY THE FIRST COEFFICIENT DOWN IMMEDIATELY UNDER THE ADDITION 7) MULTIPLY THE NUMBER “ a “ BY THIS COEFFICIENT AND WRITE THE PRODUCT IMMEDIATELY UNDER THE SECOND COEFFICIENT 8) ADD THE SECOND COEFFICIENT AND THE PRODUCT RESULTING FROM STEP 7. WRITE YOUR SUM UNDER THE ADDITION LINE 9) REPEAT STEPS 7 AND 8 FOR ALL OF THE REMAINING COEFFICIENTS 10) THE LAST SUM TO BE BROUGHT DOWN UNDER THE ADDITION LINE IS ALWAYS YOUR REMAINDER AND SHOULD BE WRITTEN OVER THE DIVISOR 11) THE NEXT SUM IMMEDIATELY TO THE LEFT OF THE REMAINDER IS ALWAYS THE CONSTANT OF THE POLYNOMIAL QUOTIENT 12) THE NEXT SUM IMMEDIATELY TO THE LEFT OF THE CONSTANT IS ALWAYS THE COEFFICIENT OF THE FIRST DEGREE TERM OF THE POLYNOMIAL QUOTIENT 13) KEEP REPEATING THE PATTERN IN STEP 12 UNTIL ALL OF THE CONSTANT IS ALWAYS THE COEFFICIENT OF THE FIRST DEGREE TERM OF THE POLYNOMIAL QUOTIENT 13) KEEP REPEATING THE PATTERN IN STEP 12 UNTIL ALL OF THE SUMS ARE COEFFICIENTS OF ONE DEGREE HIGHER TERMS OF THE POLYNOMIAL QUOTIENT 14) THE DEGREE OF THE POLYNOMIAL QUOTIENT SHOULD ALWAYS BE ONE LESS THAN N THAT OF THE ORIGINAL DIVIDEND 15) IF THE REMAINDER IS ZERO THEN THE QUOTIENT AND THE DIVISOR ARE BOTH FACTORS OF THE ORIGINAL DIVIDEND


Section: 3.4 Zeros of Polynomial Functions

THE VALUES THAT SATISFY AN EQUATION ARE CALLED THE “ ROOTS “, THE “ SOLUTIONS “ OR THE “ ZEROS “ OF THE EQUATIONS

Steps for finding the roots of a polynomial function of degree three or higher Phase one (making the “ P N I “ chart)

1. Use Descartes rule of signs to find the number of possible positive real roots(remember that the number of possible positive real roots is equal to the number of sign changes in the coefficients in f(x) or that number decreased by an even number down to but not lower that zero) 2. use Descartes rule of signs to find the number of possible negative real roots (remember that the number of possible negative real roots is equal to the number of sign changes in the coefficients in f(-x) or that number decreased by an even number down to but not lower that zero) 3. make a “P N I” chart to find out the number of possible imaginary roots ( remember that each row in the “ P N I “ chart must add up the degree of the function f(x)

Phase two (Finding all of the “Possible Rational Zeros”)

1. make a “ P Q “ chart 2. list all of the integer factors of the constant of the polynomial function in the “P” column 3. list all of the integer factors of the leading coefficient of the polynomial function in the “Q”” column 4. List all of the possible rational zeros (all of the P/Q’s terms in simplified form)

Phase three (Finding all of the Rational Zeros) Use synthetic division to find which of the “Possible Rational Zeros” actually are Rational Zeros of the polynomial function (which of the “Possible Rational Zeros” when divided into the polynomial functional using synthetic division will have a remainder of zero) Phase four (Finding all of the zeros)

1. use synthetic division to find which of the “Possible Rational Zeros” actually are Rational Zeros of the polynomial function (which of the “Possible Rational Zeros” when divided into the polynomial functional using synthetic division will have a remainder of zero) 2. check to make sure that the roots you have found fit one of the possible scenarios from your “ P N I “ chart


Section: 3.5 Rational Functions and Graphs

Steps for Graphing Rational Functions 1. Try to simplify the rational function by factoring both the numerator and denominator and canceling any like terms. 2. Find any holes.

1 . Set any terms that were canceled out in step one that contains a variable equal to zero and solve them. These values are where there is a hole in the graph of the rational function.

3. Find the Vertical Asymptote(s) 1. If the simplified function still has a variable term left in the denominator set this denominator equal to zero and solve. The results are the vertical asymptote(s). 2. If the simplified function does not have a variable term left in the denominator then set the original functions denominator equal to zero and solve. The results are the vertical asymptote(s).

4. Find the Horizontal Asymptote - find out what happens to the function as “ x “ approaches infinity. 5. Find the Slant Asymptote if it exists

1. Check to see if the degree of the numerator of the rational expression id one degree greater than the degree of the denominator of the rational expression. If so, there will be a slant asymptote. 2. use long division to divide the numerator by the denominator and find the linear quotient. 3. Graph the linear quotient on the coordinate plane.

6. Choose some values for x and solve for the corresponding values of f(x) 1. choose at least five values for “ x “ that are less than the lowest vertical asymptote 2. if there are more than one vertical asymptote then choose at least five values for “ x “ that are between the lowest vertical asymptote and highest vertical asymptote 3. choose at least five values for “ x “ that are greater than the highest vertical asymptote

Coach Stones’ Expanded Standard Pre-Calculus Algorithm Packet 7. Graph the Rational Function

1. draw in the vertical asymptote(s) using dotted lines 2. draw in the horizontal asymptote (if it exists) using a dotted line 3. draw in the slant asymptote (if it exists) using a dotted line 4. plot the ordered pairs found in step 3. 5. draw in the graphs using solid lines and being careful not to allow the graphs to cross the asymptotes


Section: 3.6 Polynomial and Rational Inequalities

Section: 3.7 Modeling Variation

DIRECT VARIATION FORMULA

1. IF Y VARIES DIRECTLY WITH X THEN Y = K X or K = ( Y / X ) WHERE “ K “ IS THE CONSTANT OF VARIATION. 2. IF X VARIES DIRECTLY WITH Y THEN X = K Y OR K = ( X / Y ) WHERE “ K “ IS THE CONSTANT OF VARIATION


INVERSE VARIATION FORMULA 1. IF Y VARIES INVERSELY WITH X THEN X Y = K WHERE “ K “ IS THE CONSTANT OF VARIATION 2. IF X VARIES INVERSELY WITH Y THEN X Y = K WHERE “ K “ IS THE CONSTANT OF VARIATION

JOINT VARIATION FORMULA 1. IF Y VARIES JOINTLY WITH X AND Z THEN Y = K X Z WHERE “ K “ IS THE CONSTANT OF VARIATION 2. IF X VARIES JOINTLY WITH Y AND Z THEN X = K Y Z WHERE “ K “ IS THE CONSTANT OF VARIATION 3. IF Z VARIES JOINTLY WITH X AND Y THEN Z = K X Y WHERE “ K “ IS THE CONSTANT OF VARIATION


Section: 4.2 Logarithmic Functions


Section: 4.3 Properties of Logarithms


Section: 4.4 Exponential and Logarithmic Equations


Section: 5.1 Angles and Radian Measure 3600 = 2 π Radians

To rewrite the radian measure of an angle in degrees, multiply the numbers of radians by (1800 /π Radians )

To rewrite the degree measure of an angle in radians, multiply the numbers of degrees by (π Radians / 1800 )


Standard position is an angle positioned so that the vertex is at the origin and its initial side is along the positive x-axis.

Coterminal Angles - When two angles in standard position have the same terminal sides, they are called coterminal angles. To find an angle that is coterminal add or subtract a multiple of 3600 or 2π Radians.

Section: 5.2 Right Triangle Trigonometry

Sum of the interior angles of any triangle is 180O

30O - 60O -90O Right Triangles a = x b = x √ 3 c = 2x

45O - 45O -90O Right Triangles a = x b = x c = x √ 2


Reciprocal Identities

sin x = ( 1 / csc x ) csc x = ( 1 / sin x )

cos x = ( 1 / sec x ) sec x = ( 1 / cos x ) tan x = ( 1 / cot x ) cot x = ( 1 / tan x )

Quotient Identities

tan x = ( sin x / cos x )

cot x = ( cos x / sin x ) Pythagorean Identities

sin2 x + cos2 x = 1 or 1+ tan2 x = sec2 x or 1+ cot2 x = csc2 x or sin2 x = 1 – cos2 x or 1 = sec2 x – tan2 x or 1 = csc2 x – cot2 x or cos2 x = 1 – sin2 x tan2 x = sec2 x – 1 cot2 x = csc2 x – 1


Cofunction Identities ( If θ is in radians, replace 90o with /2. ) sin θ = cos ( 90o – θ ) csc θ = sec ( 90o – θ ) cos θ = sin ( 90o – θ ) sec θ = csc ( 90o – θ ) tan θ = cot ( 90o – θ ) cot θ = tan ( 90o – θ )

Section: 5.3 Trigonometric Functions of Any Angle

Trigonometric Functions, X in Standard Position given a point ( x , y ) r = √ x2 + y2 Where r = hypotenuse x = adjacent and y = opposite 1. sin Xo = y / r 4. csc Xo = r / y 2. cos Xo = x / r 5. sec Xo = r / x 3. tan Xo = y / x 6. cot Xo = x / y

QUADRANT SIGNS OF TRIG FUNCTIONS I ALL (sin, cos, tan, csc, sec, and Cot are all Positive in the 1st Quadrant) II STUDENTS (only sin and csc are Positive in the 2nd Quadrant) III TAKE (only tan and cot are Positive in the 3rd Quadrant) IV CALCULUS (only cos and Sec are Positive in the 4th Quadrant)

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Steps to find the value of a Trigonometric function of any angle “X” 1. Find the reference angle “ X’ “ 2. Find the value of the Trigonometric function for “ X’ “ 3. Using the quadrant in which the terminal side of “X ” lies, determine the sign of the trigonometric function of “X ”


Section: 5.4 Trigonometric Functions of Real Numbers, Periodic Functions


Section: 5.5 Graphs of Sine and Cosine

Graphing Variations of the y = A sin (Bx - C) + D and the y = A cos (Bx - C) + D functions.

1. Write the function in the form of y = A sin (Bx - C) + D or y = A cos (Bx - C) + D filling in the

appropriate values for any constants “A”, “B”, “C”, or “D” not already listed in the original function. 2. Identify the amplitude using the formula - “amplitude” = | A |. 3. Identify the period using the formula - “period” = 2 / B. 4. Identify the phase shift, if any, using the formula - “phase shift” = C/B. 5. Identify the vertical shift, if any, using the formula y = D. 6. Find (period / 4 ) which is the value which you will add to each “x” value to make your table. 7. Make a Table: Start with the value of “x1” where x1 = phase shift. To find each subsequent value for “x”

add the value solved for in step 6 to the previous “x”. (For example: x2 = x1 + (period / 4 ), x3 = x2 + (period / 4 ), etc.).

8. Find the corresponding values of “y” for the five key points by evaluating the function at each of the

values of “x” from step 7. 9. Connect the five key points with a smooth curve and graph one complete cycle of the function. 10. Extend the graph in step 9 to the left or right as desired.

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Section: 5.6 Graphs of Other Trigonometric Functions

Graphing Variations of the y = A tan (Bx - C) + D function. 1. Write the function in the form of y = A tan (Bx - C) + D filling in the appropriate values for any

constants “A”, “B”, “C”, or “D” not already listed in the original function. 2. Find two consecutive asymptotes by finding an interval containing one period using: Bx - C = - ( / 2)

call it Asymptote1 and Bx - C = ( / 2) call it Asymptote2 3. Identify an x-intercept midway between the two consecutive asymptotes using: midpoint1 = ( Asymptote1 + Asymptote2 ) / 2 4. Find the two points on the graph ( 1 / 4 ) and ( 3 / 4 ) of the way between the consecutive asymptotes

using:

( 1 / 4 ) midpoint1 = ( Asymptote1 + x-intercept1 ) / 2 and the ( 3 / 4 ) midpoint2 = ( Asymptote2 + x-intercept1 ) / 2 5. Make a table using the following values for: x y x1 = x value at Asymptote1 y1 = undefined - which creates an Asymptote x2 = ( 1 / 4 ) midpoint1 y2 = - A + D

x3 = x-midpoint1 y3 = D x4 = ( 3 / 4 ) midpoint2 y4 = A + D x5 = x value at Asymptote2 y5 = undefined - which creates an Asymptote 6. Connect the five key points with a smooth curve and graph one complete cycle of the function. 7. Extend the graph in step 6 to the left or right as desired.

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Graphing Variations of the y = A cot (Bx - C) + D function. 1. Write the function in the form of y = A cot (Bx - C) + D filling in the appropriate values for any

constants “A”, “B”, “C”, or “D” not already listed in the original function. 2. Find two consecutive asymptotes by finding an interval containing one period using:

Bx - C = 0 call it Asymptote1 and Bx - C = call it Asymptote2

3. Identify an x-intercept midway between the two consecutive asymptotes using: midpoint1 = ( Asymptote1 + Asymptote2 ) / 2 4. Find the two points on the graph ( 1 / 4 ) and ( 3 / 4 ) of the way between the consecutive asymptotes

using:

( 1 / 4 ) midpoint1 = ( Asymptote1 + x-intercept1 ) / 2 and the ( 3 / 4 ) midpoint2 = ( Asymptote2 + x-intercept1 ) / 2 5. Make a table using the following values for: x y x1 = x value at Asymptote1 y1 = undefined - which creates an Asymptote x2 = ( 1 / 4 ) midpoint1 y2 = A + D x3 = x-midpoint1 y3 = D x4 = ( 3 / 4 ) midpoint2 y4 = - A + D x5 = x value at Asymptote2 y5 = undefined - which creates an Asymptote 6. Connect the five key points with a smooth curve and graph one complete cycle of the function. 7. Extend the graph in step 6 to the left or right as desired.

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Section: 5.7 Inverse Trigonometric Functions

Inverse Trigonometric Ratios 1. sin - 1 (sin θ) = θ 4. csc - 1 (csc θ) = θ 2. cos - 1 (cos θ) = θ 5. sec - 1 (sec θ) = θ 3. tan - 1 (tan θ) = θ 6. cot - 1 (cot θ) = θ


Section: 5.8 Applications of Trigonometric Functions

Strategies for solving trigonometric word problems.

1. Read the word problem twice. The first time you read to get a basic understanding of the situation described and the question that is asked. The second time you read looking for the details described in the word problem.

2. Using the information in the details of the word problem make a drawing and develop an equation to solve.

3. Solve the equation and check your answer.


Section: 6.1 Verifying Trigonometric Identities

Guidelines for verifying Trigonometric Identities 1. Work with each side of the equation independently of the other side. Start with the more complicated

side and transform it in a step-by-step fashion until it looks exactly like the other side. 2. Analyze the identity and look for opportunities to apply the fundamental identities. 3. Try using one or more of the following techniques: A. Rewrite the more complicated side in terms of sine and cosine. B. Factor out the greatest common factor. C. Separate a single-term quotient into two terms: a + b = a + b or a - b = a - b c c c c c c D. Combine fractional expressions using the least common denominator. E. Multiply the numerator and the denominator by a binomial factor that appears on the other side of the identity. 4. Don’t be afraid to stop and start over if you are not getting anywhere. Creative puzzle solvers know that

strategies leading to dead ends often provide good problem-solving ideas. Fundamental Identities: 1. Reciprocal Identities sin x = ( 1 / csc x ) csc x = ( 1 / sin x ) cos x = ( 1 / sec x ) sec x = ( 1 / cos x ) tan x = ( 1 / cot x ) cot x = ( 1 / tan x ) 2. Quotient Identities

tan x = ( sin x / cos x ) cot x = ( cos x / sin x )


3. Pythagorean Identities

sin2 x + cos2 x = 1 or sin2 x = 1 – cos2 x or cos2 x = 1 – sin2 x

1+ tan2 x = sec2 x or 1 = sec2 x – tan2 x or tan2 x = sec2 x – 1

1+ cot2 x = csc2 x or 1 = csc2 x – cot2 x or cot2 x = csc2 x – 1

4. Even-Odd Identities ( Negative Formulas) cos ( -x ) = cos x sec ( -x ) = sec x sin ( -x ) = - sin x csc ( -x ) = - csc x tan ( -x ) = - tan x cot ( -x ) = - cot x 5. Cofunction Identities ( If θ is in radians, replace 90o with /2. ) sin θ = cos ( 90o – θ ) csc θ = sec ( 90o – θ ) cos θ = sin ( 90o – θ ) sec θ = csc ( 90o – θ ) tan θ = cot ( 90o – θ ) cot θ = tan ( 90o – θ ) 6. Periodic Properties of Sine and Cosine Functions Sin (θ + 2 n ) Cos (θ + 2 n ) 7. Periodic Properties of Tangent and Cotangent Functions

Tan (θ + n ) Cot (θ + n )

Section: 6.2 Sum and Difference Formulas

Sum and Difference Formulas for Cosines and Sines cos( + ) = cos cos - sin sin cos( - ) = cos cos + sin sin sin ( + ) = sin cos + cos sin sin ( - ) = sin cos - cos sin Sum and Difference Formulas for Tangents tan ( + ) = ( tan + tan ) / ( 1 - tan tan ) tan ( - ) = ( tan - tan ) / ( 1 + tan tan )

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Section: 6.3 Double Angle/Half Angle Formulas, Power Reducing

Double-Angle Formulas sin (2 ) = 2 sin cos cos (2 ) = cos2 - sin2 or cos (2 ) = 2 cos2 - 1 or cos (2 ) = 1 - 2 sin2 tan (2 ) = ( 2 tan ) / (1 - tan2 ) Half-Angle Formulas where the sign is determined by the quadrant in which ( /2) lies. sin ( /2) = ± √ ( (1 - cos ) / 2 ) cos ( /2) = ± √ ( (1 + cos ) / 2 ) tan ( /2) = ± √ ( (1 - cos ) / (1 + cos ) ) or tan ( /2) = ( 1 - cos ) / ( sin ) or tan ( /2) = ( sin ) / ( 1 + cos ) Power Reducing Formulas sin2 = 1 - cos2 cos2 = 1 + cos2 tan2 = 1 - cos2

2 2 1 + cos2

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Section: 6.5 Trigonometric Equations

Strategies for solving Trigonometric Functions

1. Isolate the trigonometric function on one side of the equation. 2. Solve for the variable. (Remember, if the equation includes more than one trigonometric function or is in

quadratic form you may need to use factoring techniques in order to solve for the variable.) 3. Add the value to include the periocity of the function. (for Sine and cosine add 2πn and for tangent

add πn) 4. If the domain is limited to [ 0 , 2 π ) then make a table. Choose values for n and solve for values of the

variable. Include in your final answer only those solution which are in the domain of [ 0 , 2 π )

Section: 7.1 Law of Sines

Law of Sines: a = b = c or sin A sin B sin C sin A sin B sin C a = b = c Formula to find the Altitude (height) of a Triangle: h = b sin A or h = a sin B Steps for solving Oblique Triangles using the Law of Sines If you are given SAA or ASA

1. Find the unknown angle using “180o – the sum of the two known angles.” 2. Apply the Law of Sines to find the unknown side lengths

If you are given SSA (the “Ambiguous Case”)

1. Using the given angle, the side length opposite this given angle and the other given side, apply the Law of Sines to find the unknown angle. (If this angle measure is “not” possible in a triangle then there is “No Solution” and the triangle cannot exist.)

2. Find the third angle by using “180O – the given angle – the angle solved for in step 1”. 3. Find the remaining side length using the Law of Sines. 4. To check to see if there is a second triangle that could be formed using the originally given information.

Find the second possibility for the angle found in step 1 using “1800 – the angle solved for in step 1”. 5. Find the second possibility for the angle solved for in step two using “1800 – the originally given angle

and the angle found in step 4”. If this angle is negative then there is not a second triangle which can be formed using the original information. If this angle is positive there is a second triangle which can be formed using the original information so proceed to step 6.

6. Apply the Law of Sines using the originally given angle measure, the originally give side length opposite the given angle and the angle solved for in step 5 to solve for the still unknown side length.


Area of Oblique Triangles: Area = (1 / 2) b c sin A = (1 / 2) a b sin C = (1 / 2) a c sin B

Section: 7.2 Law of Cosines

Law of Cosines: a2 = b2 + c2 - 2 b c cos A b2 = a2 + c2 - 2 a c cos B c2 = a2 + b2 - 2 a b cos C

Solving SAS Triangle:

1. Use the Law of Cosines to find the side opposite the given angle. 2. Use the Law of Sines to find the angle opposite the shorter of the two given sides. This angle is always

acute. 3. Find the third angle by subtracting the measure of the given angle and the angle found in step 2 from

180O Solving SSS Triangle:

1. Use the Law of Cosines to find the angle opposite the longest side. 2. Use the Law of Sines to find the angle opposite the shorter of the two given sides. (This angle is always

acute.) 3. Find the third angle by subtracting the measure of the given angle and the angle found in step 2 from

180O Heron’s Formula for the Area of a Triangle: Area = √ s ( s - a ) ( s - b ) ( s - c ) where s = (1 / 2) ( a + b + c )


Section: 7.3 Polar Coordinates

Relationships between Polar and Rectangular Coordinates: x = r cos y = r sin x2 + y2 = r2 or r2 = x2 + y2 r = √ x + y

tan = ( y / x )

€

θ

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θ

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θ


Section: 9.3 Matrix Operations

A “ MATRIX” IS A SYSTEM OF ROWS AND COLUMNS AND IS USUALLY NAMED USING AN UPPERCASE LETTER FOLLOWED BY A SUBSCRIPTED DESCRIPTION OF THE NUMBER OF ROWS AND THE NUMBER OF COLUMNS IN THE MATRIX. (FOR EXAMPLE A2 x 3 MEANS MATRIX “ A “ HAS TWO ROWS AND THREE COLUMNS) MATRICES ARE EQUAL IF AND ONLY IF 1) THEY BOTH HAVE THE SAME DIMENSIONS (ORDER) 2) AND CORRESPONDING ELEMENTS ARE EQUAL

THE STEPS TO ADD OR SUBTRACT MATRICES 1) IN ORDER FOR THEM TO BE ABLE TO BE ADDED OR SUBTRACTED THEY MUST BOTH HAVE THE SAME DIMENSIONS (ORDER) 2) THEN ADD OR SUBTRACT THE CORRESPONDING ELEMENTS 3) THE ANSWER WILL BE A MATRIX WITH THE SAME DIMENSIONS AS THE ORIGINAL MATRICES THAT WERE ADDED OR SUBTRACTED BUT THE NEW ELEMENTS WILL BE THE SUMS OR DIFFERENCES OF THE CORRESPONDING ELEMENTS


SCALAR MULTIPLICATION OF A MATRIX (A NUMBER) - JUST MULTIPLY EVERY ELEMENT IN THE MATRIX BY THAT SCALAR. SCALAR MULTIPLICATION IS USED IN THE DILATION (THE REDUCTION OR ENLARGEMENT) OF A GEOMETRIC FIGURE


MATRIX MULTIPLICATION IS “ NOT “ COMMUTATIVE MATRIX MULTIPLICATION (MULTIPLYING A MATRIX BY A MATRIX) 1) TO BE ABLE TO MULTIPLY TWO MATRICES YOU MUST HAVE THE SAME NUMBER OF COLUMNS IN THE “LEFT HAND” MATRIX AS YOU HAVE ROWS IN THE “RIGHT HAND” MATRIX ( IF NOT THE MULTIPLICATION CANNOT BE PERFORMED AND IT IS SAID TO BE “UNDEFINED” FOR THE TWO MATRICES 2) IF THE MULTIPLICATION CAN BE DONE IT IS A “ROW BY COLUMN” MULTIPLICATION


Section: 9.4 Matrix Equations


Section: 9.5 Determinants and Cramer’s Rule


Section: 10.1 The Ellipse


STEPS FOR WRITING AN EQUATION FOR A HORIZONTAL ELLIPSE INTO STANDARD FORM [(x - h)2/ a2] + [(y - k)2/ b2] =1 1) ISOLATE THE CONSTANT 2) USING PARENTHESES, GROUP THE x2 AND THE x TERMS 3) USING PARENTHESES, GROUP THE y2 AND THE y TERMS 4) FACTOR OUT THE COEFFICIENT OF THE x2 TERM FROM BOTH THE x2 AND THE x TERMS AND COMPLETE THE SQUARE INSIDE THE PARENTHESIS 5) FACTOR OUT THE COEFFICIENT OF THE y2 TERM FROM BOTH THE y2 AND THE y TERMS AND COMPLETE THE SQUARE INSIDE THE PARENTHESIS 6) ADD TO THE CONSTANT THE PRODUCT OF THE VALUE FACTORED OUT IN STEP 4 AND THE NUMBER YOU ADDED IN INSIDE THE “ x “ PARENTHESIS TO COMPLETE THE SQUARE. 7) ADD TO THE CONSTANT THE PRODUCT OF THE VALUE FACTORED OUT IN STEP 5 AND THE NUMBER YOU ADDED IN INSIDE THE “ y “ PARENTHESIS TO COMPLETE THE SQUARE. 8) WRITE INTO THE STANDARD FORM OF A HORIZONTAL ELLIPSE [(x - h)2/ a2] + [(y - k)2/ b2] =1 MAKING SURE THAT THE CONSTANT IS 1 IF NOT USE DIVISION TO MAKE IT 1

STEPS FOR WRITING AN EQUATION FOR A VERTICAL ELLIPSE INTO STANDARD FORM [(x - h)2/ b2] + [(y - k)2/ a2] =1 1) ISOLATE THE CONSTANT 2) USING PARENTHESES, GROUP THE x2 AND THE x TERMS 3) USING PARENTHESES, GROUP THE y2 AND THE y TERMS 4) FACTOR OUT THE COEFFICIENT OF THE x2 TERM FROM BOTH THE x2 AND THE x TERMS AND COMPLETE THE SQUARE INSIDE THE PARENTHESIS 5) FACTOR OUT THE COEFFICIENT OF THE y2 TERM FROM BOTH THE y2 AND THE y TERMS AND COMPLETE THE SQUARE INSIDE THE PARENTHESIS 6) ADD TO THE CONSTANT THE PRODUCT OF THE VALUE FACTORED OUT IN STEP 4 AND THE NUMBER YOU ADDED IN INSIDE THE “ x “ PARENTHESIS TO COMPLETE THE SQUARE. 7) ADD TO THE CONSTANT THE PRODUCT OF THE VALUE FACTORED OUT IN STEP 5 AND THE NUMBER YOU ADDED IN INSIDE THE “ y “ PARENTHESIS TO COMPLETE THE SQUARE. 8) WRITE INTO THE STANDARD FORM OF A VERTICAL ELLIPSE [(x - h)2/ b2] + [(y - k)2/ a2] =1 MAKING SURE THAT THE CONSTANT IS 1 IF NOT USE DIVISION TO MAKE IT 1

INFORMATION ABOUT ELLIPSES

TYPE OF ELLIPSE HORIZONTAL ELLIPSE VERTICAL ELLIPSE FORM OF EQUATION [(x - h)2/ a2] + [(y - k)2/ b2] =1 [(x - h)2/ b2] + [(y - k)2/ a2] =1 CENTER ( h , k ) ( h , k ) FOCI ( h +/- c , k ) ( h , k +/- c ) LENGTH OF MAJOR AXIS 2 a 2 a ENDPOINTS OF MAJOR AXIS ( h +/- a , k ) ( h , k +/- a ) LENGTH OF MINOR AXIS 2 b 2 b ENDPOINTS OF MINOR AXIS ( h , k +/- b ) ( h +/- b , k ) HELPFUL EQUATION a2 - b2 = c2 a2 - b2 = c2


Section: 10.2 The Hyperbola


STEPS FOR WRITING AN EQUATION FOR A RIGHT/LEFT HYPERBOLA INTO STANDARD FORM [(x - h)2/ a2] - [(y - k)2/ b2] =1 1) ISOLATE THE CONSTANT 2) USING PARENTHESES, GROUP THE x2 AND THE x TERMS 3) USING PARENTHESES, GROUP THE y2 AND THE y TERMS 4) FACTOR OUT THE COEFFICIENT OF THE x2 TERM FROM BOTH THE x2 AND THE x TERMS AND COMPLETE THE SQUARE INSIDE THE PARENTHESIS 5) FACTOR OUT THE COEFFICIENT OF THE y2 TERM FROM BOTH THE y2 AND THE y TERMS AND COMPLETE THE SQUARE INSIDE THE PARENTHESIS 6) ADD TO THE CONSTANT THE PRODUCT OF THE VALUE FACTORED OUT IN STEP 4 AND THE NUMBER YOU ADDED IN INSIDE THE “ x “ PARENTHESIS TO COMPLETE THE SQUARE. 7) ADD TO THE CONSTANT THE PRODUCT OF THE VALUE FACTORED OUT IN STEP 5 AND THE NUMBER YOU ADDED IN INSIDE THE “ y “ PARENTHESIS TO COMPLETE THE SQUARE. 8) WRITE INTO THE STANDARD FORM OF A RIGHT/LEFT HYPERBOLA [(x - h) 2/ a2] - [(y - k) 2/ b2] =1 MAKING SURE THAT THE CONSTANT IS 1 IF NOT USE DIVISION TO MAKE IT 1

STEPS FOR WRITING AN EQUATION FOR AN UP/DOWN HYPERBOLA INTO STANDARD FORM [(y - k)2/ a2] - [(x - h)2/ b2] =1 1) ISOLATE THE CONSTANT 2) USING PARENTHESES, GROUP THE x2 AND THE x TERMS 3) USING PARENTHESES, GROUP THE y2 AND THE y TERMS 4) FACTOR OUT THE COEFFICIENT OF THE x2 TERM FROM BOTH THE x2 AND THE x TERMS AND COMPLETE THE SQUARE INSIDE THE PARENTHESIS 5) FACTOR OUT THE COEFFICIENT OF THE y2 TERM FROM BOTH THE y2 AND THE y TERMS AND COMPLETE THE SQUARE INSIDE THE PARENTHESIS 6) ADD TO THE CONSTANT THE PRODUCT OF THE VALUE FACTORED OUT IN STEP 4 AND THE NUMBER YOU ADDED IN INSIDE THE “ x “ PARENTHESIS TO COMPLETE THE SQUARE. 7) ADD TO THE CONSTANT THE PRODUCT OF THE VALUE FACTORED OUT IN STEP 5 AND THE NUMBER YOU ADDED IN INSIDE THE “ y “ PARENTHESIS TO COMPLETE THE SQUARE. 8) WRITE INTO THE STANDARD FORM OF AN UP/DOWN HYPERBOLA [(y - k) 2/ a2] - [(x - h) 2/ b2] =1 MAKING SURE THAT THE CONSTANT IS 1 IF NOT USE DIVISION TO MAKE IT 1


INFORMATION ABOUT HYPERBOLAS TYPE OF HYPERBOLA RIGHT AND LEFT UP AND DOWN FORM OF EQUATION [(x - h)2/ a2] - [(y - k)2/ b2] =1 [(y - k)2/ a2] - [(x - h)2/ b2] =1 where “ a “ is always under the positive term EQUATION OF ASYMPTOTES y = +/- (b/a) x y = +/- (a/b) x POINT OF INTERSECTION OF ( h , k ) ( h , k ) VERTICES ( h +/- a , k ) ( h , k +/- a ) FOCI ( h +/- c , k ) ( h , k +/- c ) TRANSVERSE AXIS h +/- a k +/- a CONJUGATE AXIS k +/- b h +/- b LENGTH OF TRANSVERSE AXIS 2 a 2 a LENGTH OF CONJUGATE AXIS 2 b 2 b HELPFUL EQUATION a2 + b2 = c2 a2 + b2 = c2

Section: 10.3 The Parabola


Steps for finding the important features of a parabola and graphing the parabola:

1. Write the equation in “Standard Form” depending on which variable is squared. a. Isolate both the quadratic and linear terms for the variable that is squared on the left side of the

equal sign. b. Complete the square on the left side of the equal sign (remember to add the same value to both

sides of the equal sign). c. Factor the perfect square trinomial on the left hand side of the equal sign and simplify the right

hand side of the equal sign. d. Factor out the “Greatest Common Factor” from the binomial on the right hand side of the equal

sign. 2. Use the formula sheet to find the values for “p”, “h”, and “k”. 3. Use the formula sheet to find the direction of the opening of the parabola. 4. Use the formula sheet and the values of “h” and “k” to find the Vertex. 5. Use the formula sheet and the values of “p” and either “h” or “k” to find the Focus. 6. Use the formula sheet and the values of “p” and either “h” or “k” to find the Directrix. 7. Use the formula sheet and the values of “p” and either “h” or “k” to find the Axis of Symmetry. 8. Use the formula sheet and the value of “p” to find the length of the Latus Rectum. 9. Plot the features of the parabola found in steps 4 thru 8 and graph the parabola.


INFORMATION ABOUT PARABOLAS Vertical Parabola (Up/Down) Horizontal Parabola (Right/Left) ( y – k )2 = 4 p ( x – h ) ( x – h )2 = 4 p ( y – k ) Vertex Vertex ( h , k ) ( h , k ) Axis of Symmetry Axis of Symmetry x = h y = k Focus Focus ( h , k + p ) ( h + p , k ) Directrix Directrix y = k – p x = h – p Direction of the opening Direction of the opening If p > 0 the parabola opens up If p > 0 the parabola opens right If p < 0 the parabola opens down If p < 0 the parabola opens left Length of the Latus Rectum Length of the Latus Rectum | 4 p | | 4 p |

Section: 8.4 Systems of Non-Linear Equations in Two Variables

PARALLEL LINES - IF THE SLOPES OF TWO LINES ARE EQUAL BUT THE “ y “ - INTERCEPTS OF THE TWO LINES ARE DIFFERENT THEN THE LINES ARE PARALLEL. THE LINES THEN HAVE “NO SOLUTIONS” AND THE LINES ARE SAID TO BE “ INCONSISTENT”

COLLINEAR LINES - IF THE SLOPES OF TWO LINES ARE EQUAL AND THE “ y “ - INTERCEPTS OF THE TWO LINES ARE THE SAME THEN THE LINES ARE COLLINEAR (THEY ARE SAME LINE). THESE LINES HAVE “INFINITELY MANY SOLUTIONS” AND ARE SAID TO BE “CONSISTENT AND DEPENDENT”

LINES THAT INTERSECT AT ONE POINT - IF THE SLOPES OF TWO LINES ARE DIFFERENT THEN THE LINES INTERSECT AT EXACTLY ONE POINT AND THE LINES ARE SAID TO BE “CONSISTENT AND INDEPENDENT ”


STEPS FOR SOLVING A SYSTEM OF EQUATIONS USING THE “SUBSTITUTION METHOD” - 1) CHOOSE ONE OF THE EQUATIONS. THEN CHOOSE ONE OF THE VARIABLES IN THAT EQUATION AND SOLVE FOR IT. (HINT: IF POSSIBLE, CHOOSE THE EQUATION THAT HAS A VARIABLE WITH A COEFFICIENT OF “ONE”) 2) SUBSTITUTE THE VALUE FOUND IN STEP 1 INTO THE OTHER EQUATION AND SOLVE FOR THE STILL UNKNOWN VARIABLE. 3) SUBSTITUTE THE VALUE FOUND IN STEP 2 BACK INTO THE ORIGINAL EQUATION AND SOLVE FOR THE STILL UNKNOWN VARIABLE 4) CHECK YOUR ANSWERS BACK INTO BOTH OF THE ORIGINAL EQUATIONS TO MAKE SURE THEY MAKE BOTH STATEMENTS TRUE AT THE SAME TIME. STEPS FOR SOLVING A SYSTEM OF EQUATIONS USING THE “ELIMINATION METHOD” 1)MAKE SURE ALL EQUATIONS ARE WRITTEN IN GOOD STANDARD LINEAR EQUATION FORM THEN LINE THE EQUATIONS UP DIRECTLY UNDER ONE ANOTHER SO THAT TERMS WITH THE SAME VARIABLES ARE DIRECTLY UNDER EACH OTHER. 2) CHECK TO SEE IF YOU CAN SIMPLY ADD THE TWO EQUATIONS AND HAVE ONE OF THE VARIABLES ELIMINATE. (IF NOT, YOU WILL HAVE TO CHOOSE WHICH VARIABLE YOU WISH TO ELIMINATE AND THEN DECIDE WHAT NUMBER OR NUMBERS TO MULTIPLY ONE OR BOTH OF THE EQUATIONS BY IN ORDER TO COMPLETE STEP 2. 3) ONCE ONE OF THE VARIABLES HAS BEEN ELIMINATED SOLVE FOR THE OTHER VARIABLE. 4) THEN SUBSTITUTE THE VALUE SOLVED FOUND IN STEP 3 BACK INTO ONE OF THE ORIGINAL EQUATIONS AND SOLVE FOR THE OTHER VARIABLE. 5) CHECK YOUR ANSWERS BACK INTO BOTH OF THE ORIGINAL EQUATIONS TO MAKE SURE THEY MAKE BOTH STATEMENTS TRUE AT THE SAME TIME.

Section: 8.5 Systems of Inequalities

STEPS FOR GRAPHING SYSTEMS OF LINEAR INEQUALITIES 1) CHECK TO SEE IF ALL OF THE INEQUALITIES ARE WRITTEN IN GOOD SLOPE INTERCEPT FORM. (IF NOT REWRITE INTO THIS FORM.) 2) PLOT THE Y INTERCEPT OF THE FIRST INEQUALITY ON YOUR COORDINATE PLANE 3) USE THE SLOPE OF THE FIRST INEQUALITY TO LOCATE AND PLOT ANOTHER POINT ON THE LINE 4) IF THE INEQUALITY SYMBOL IS STRICTLY LESS THAN “ < “ OR STRICTLY GREATER THAN “ > ” THEN CONNECT THE TWO PLOTTED POINTS WITH A DOTTED LINE 5) IF THE INEQUALITY SYMBOL IS LESS THAN OR EQUAL TO “ ≤ “ OR GREATER THAN OR EQUAL TO “ ≥ ” THEN CONNECT THE TWO PLOTTED POINTS WITH A SOLID LINE 6) IF THE INEQUALITY SYMBOL IS STRICTLY LESS THAN “ < “ OR IS LESS THAN OR EQUAL TO “ ≤ “ SHADE THE AREA UNDER THE LINE 7) IF THE INEQUALITY IS STRICTLY GREATER THAN “ > ” OR GREATER THAN OR EQUAL TO “ ≥ ” SHADE THE AREA ABOVE THE LINE 8) REPEAT STEPS 2 THROUGH 7 FOR ALL OTHER INEQUALITIES IN THE SYSTEM 9) THE AREA WHICH IS SHADED IN THE GRAPH OF ALL INEQUALITIES OF THE SYSTEM AT THE SAME TIME IS YOUR ANSWER


Section: 11.1 Sequences and Summation Notation


nth term of an Arithmetic Sequence an = a1 + ( n - 1 ) d Where d = ( a2 ) - ( a1 ) Sum of an Arithmetic Series sn = (n/2) [ 2 * a1 + ( n - 1 ) d ] Sum of an Arithmetic Series sn = (n/2) [ a1 + an ]

nn Sum of an Arithmetic Series in Sigma Notation E = a * b

n1

Arithmetic Means d = [ an - a1 ] / (n - 1)


Section: 11.2 Arithmetic Sequences


nth term of an Arithmetic Sequence an = a1 + ( n - 1 ) d Where d = ( a2 ) - ( a1 ) Sum of an Arithmetic Series sn = (n/2) [ 2 * a1 + ( n - 1 ) d ] Sum of an Arithmetic Series sn = (n/2) [ a1 + an ]

nn

Sum of an Arithmetic Series in Sigma Notation E = a * b n1

Arithmetic Means d = [ an - a1 ] / (n - 1)


Section: 11.3 Geometric Sequences and Series


nth term of an Geometric Sequence an = a1 * r( n - 1 ) Where r = ( a2 ) / ( a1 ) Sum of an Geometric Series sn = ( a1 - a1 r n ) and “r” does not = 1

1 - r Sum of an Geometric Series sn = a1 ( 1 – r n ) and “r” does not = 1

1 – r

nn Sum of an Geometric Series in Sigma Notation E = a1 * r n -1

n=1 Alternate formula for finding the Sum of an Geometric Series sn = a1 - an * r

1 - r Geometric Means r = the (n - 1)th root of [ an / a1 ] Sum of an Infinite Geometric Series s = a1 with -1 < r < 1

1 - r

infinity Sum of an Infinite Geometric Series in Sigma Notation E = a1 * r n -1

n=1

coach stones’ expanded standard pre-calculus algorithm ... · coach stones’ expanded standard...

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