Regents Review 2.0: Algebra I

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REGENTS REVIEW 2.0: THE NEXT GENERATION OF REGENTS EXAM PREP.

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[MUSIC] >> HI, MY NAME IS ANDREW SICKLES, I'M A HIGH SCHOOL MATH TEACHER AT COLONIE CENTRAL HIGH SCHOOL IN ALBANY, NEW YORK.

>> AND I'M ERIN BOTTA, ALSO A MATH TEACHER AT COLONIE CENTRAL HIGH SCHOOL IN ALBANY, NEW YORK.

>> AND THIS IS 5 THINGS YOU NEED TO KNOW ABOUT THE ALGEBRA 1 REGENTS EXAM.

SO WE JUST WANTED TO GO OVER A QUICK OVERVIEW OF THE REGENTS EXAM AND ITS STRUCTURE AND THEN TALK ABOUT WHERE WE CHOSE TO FOCUS OUR FIVE THINGS TO HIGHLIGHT IN THIS VIDEO TODAY.

SO I'M SURE YOU KNOW, BUT THE TEST IS BROKEN DOWN INTO FOUR SECTIONS.

PART ONE IS MULTIPLE CHOICE AND THERE'S 24 QUESTIONS; PART TWO IS SHORT ANSWER, THOSE ARE 8 QUESTIONS; PARTS THREE AND FOUR ARE MORE LONG ANSWERS, AND THOSE ARE FOUR AND ONE QUESTION RESPECTIVELY.

THE REGENTS HAS PUBLISHED ITS BREAKDOWN OF ITS TOPICS AND THE PERCENTAGE THAT THEY WEIGHT ON THE TEST, SO THE NUMBERS AND REASONING SECTION IS ROUGHLY 2-8% OF THE TEST.

ALGEBRA AND SOLVING ALGEBRAIC EQUATIONS ARE 50-56%.

FUNCTIONS MAKE UP 32-28% OF THE TEST, AND THEN STATISTICS AND PROBABILITY ANYWHERE FROM 5-10%.

BUT THAT'S A VERY VAGUE BREAKDOWN OF THE TOPICS, SO WHAT ERIN AND I DID IS WE WENT BACK AND LOOKED AT EVERY COMMON CORE REGENTS EXAM THAT HAS BEEN ADMINISTERED, AND WE LOOKED AT SPECIFIC TOPICS AND HOW FREQUENTLY THEY APPEARED IN THE PAST.

AND WHAT WE CAME UP WITH WAS THIS LIST, AND THERE'S DEFINITELY SOME HEAVY-HITTERS ON THERE, THOSE THAT WE CHOSE TO FOCUS ON, SO ON EVERY SINGLE TEST THE REGENTS HAS ASKED STUDENTS TO GRAPH A FUNCTION.

ON MOST TESTS, THE REGENTS IS ASKING TO PERFORM A LINEAR REGRESSION, A STATISTICAL ANALYSIS.

ON MOST EVERY TEST THEY'RE ASKING ABOUT A SYSTEM OF EQUATIONS OR INEQUALITIES, AND THEN MOST EVERY TEST THEY'RE ASKING TO INTERPRET FUNCTIONS OR GRAPHS THAT REPRESENT FUNCTIONS.

SO THOSE ARE THE FOUR THINGS THAT WE CHOSE TO HIGHLIGHT IN OUR FIVE THINGS TO TALK ABOUT IN ALGEBRA 1, AND OUR FIFTH SECTION IS GOING TO BE SOME CALCULATOR TRICKS TO HELP WITH THOSE FIRST PART QUESTIONS.

SO WITH THAT I WILL GIVE IT OVER TO MY COWORKER ERIN TO TALK ABOUT SYSTEM OF INEQUALITIES AND GRAPHING.

>> HELLO!

FOR OUR FIRST SECTION OF THINGS YOU NEED TO KNOW ABOUT THE ALGEBRA 1 EXAM, WE ARE GOING TO DIVE INTO GRAPHING A SYSTEM OF INEQUALITIES.

SO THIS WILL ALSO HIT THE TOPIC OF SOLVING A SYSTEM OF EQUATIONS GRAPHICALLY, IT'S A VERY, VERY POPULAR PART THREE QUESTION, SO THAT WOULD BE FOUR POINTS.

I CHOSE NUMBER 35 FROM THE JUNE 2021 EXAM, BECAUSE IT KIND OF HITS EVERYTHING THAT YOU'RE GOING TO POSSIBLY SEE WITH YOUR SYSTEM OF INEQUALITIES PROBLEM.

SO IF WE LOOK AT OUR TWO INEQUALITIES, I NOTICE THAT OUR FIRST INEQUALITY IS IN SLOPE/INTERCEPT FORM, SO THAT IS GREAT NEWS.

I CAN GO AHEAD AND JUMP RIGHT IN AND GRAPH THIS INEQUALITY.

MY Y-INTERCEPT IS 5, SO I START BY PUTTING A POINT AT 5 ON THE Y-AXIS.

AND THEN WE USE OUR SLOPE TO GATHER OUR NEW POINTS.

OUR SLOPE IS -3/4, SO THAT MEANS WE'RE GOING TO GO IN ONE NEGATIVE DIRECTION AND ONE POSITIVE DIRECTION.

SO WE CAN GO UP 3 UNITS, WHICH IS POSITIVE, LEFT 4, WHICH IS NEGATIVE, AND WE'RE OUT OF ROOM THERE.

SO THEN WE'RE GONNA GO DOWN 3 UNITS, TO THE RIGHT 4 UNITS.

ONE NEGATIVE DIRECTION, ONE POSITIVE DIRECTION.

NOW, BEFORE YOU GO AND GRAPH YOUR LINE, THERE'S SOMETHING I WANT TO REMIND YOU ABOUT INEQUALITIES: THERE ARE TWO DIFFERENT TYPES OF INEQUALITIES.

THE SYMBOL IN OUR FIRST INEQUALITY CONTAINS AN EQUALS SIGN - THAT MEANS THAT POINTS ON THAT LINE ARE SOLUTIONS.

OUR SECOND INEQUALITY DOES NOT HAVE AN EQUALS SIGN, WHICH MEANS THE POINT ON THAT LINE ARE NOT SOLUTIONS.

SO THE WAY WE DENOTE THE DIFFERENCE, THE WAY WE SHOW THAT THE POINTS ON THE LINE ARE OR ARE NOT SOLUTIONS, IS WITH THE TYPE OF LINE THAT WE DRAW.

IF WE HAVE AN EQUALS SIGN, WE ARE GOING TO MAKE THAT A SOLID LINE.

FOR OUR FIRST INEQUALITY, IT'S GOING TO BE A SOLID LINE.

OUR SECOND WILL BE A DASHED LINE.

SO I'M GOING TO GO AHEAD AND DRAW MY SOLID LINE.

AND WE WANNA MAKE SURE THAT IF WE'RE GRAPHING 2 FUNCTIONS ON THE SAME GRID, WE LABEL AT LEAST ONE, SO I'M GONNA GO AHEAD AND LABEL THIS LINE SO I DON'T FORGET.

OKAY, NOW WE NEED TO SHADE THE AREA OF THE COORDINATE PLANE THAT CONTAINS ALL OF THE SOLUTIONS, AND THAT WOULD BE ALL OF THE X AND Y VALUES THAT ARE LESS THAN OR EQUAL TO WHAT'S ON THAT LINE.

SO THIS BRINGS IN SHADING, AND THERE'S A COUPLE DIFFERENT WAYS YOU CAN LOOK AT SHADING, SOME STUDENTS FIND IT VERY EASY TO JUST FOLLOW WHAT THE SYMBOL SAYS - LESS THAN OR EQUAL TO MEANS I'M GONNA GO SHADE DOWN FROM MY LINE.

ANOTHER TRICK I'VE USED, IS YOU CAN LOOK AT YOUR Y-AXIS AND THE TOP OF YOUR Y-AXIS IS GOING TO BE GREATER THAN, THE BOTTOM OF YOUR Y-AXIS IS LESS THAN, AND THEN YOU SHADE FROM YOUR LINE TO THE APPROPRIATE SYMBOL.

IN THIS CASE, WE WOULD BE SHADING DOWN TOWARDS LESS THAN.

WHEN IN DOUBT, THERE IS ALWAYS ANOTHER METHOD YOU CAN USE TO DETERMINE WHERE TO SHADE, AND THAT WOULD BE TO PICK A TEST POINT.

AND THE TEST POINT THAT I LIKE TO USE IS 0,0.

IT MAKES DOING THE MATH REALLY EASY, SO WE SUBSTITUTE 0,0 IN, AS YOU CAN SEE I'VE ALREADY DONE HERE TO OUR ORIGINAL INEQUALITY.

AND IF YOU GET A TRUE STATEMENT, WHICH WE DO, YOU WANT TO SHADE WHERE 0,0 IS, YOU WANT THAT ORIGIN TO FALL IN YOUR SOLUTION SET, WHICH IT DOES.

MOVING ON TO OUR SECOND INEQUALITY, THIS ONE IS NOT IN SLOPE/INTERCEPT FORM, SO WE HAVE TO DO A LITTLE WORK IN ORDER TO DO THAT, AND I WANT TO REMIND YOU OF SOMETHING FIRST.

INEQUALITIES, IT'S VERY IMPORTANT TO REMEMBER, THAT WHEN YOU'RE SOLVING AN INEQUALITY, EITHER IF YOU'RE JUST SOLVING THE INEQUALITY FOR X, OR YOU'RE GRAPHING IT AND GETTING IT INTO SLOPE/INTERCEPT FORM, WHEN YOU MULTIPLY OR DIVIDE BY A NEGATIVE VALUE, YOU HAVE TO REMEMBER TO FLIP THE INEQUALITY SYMBOL.

AND I'LL TELL YOU THAT'S GONNA HAPPEN HERE, SO WATCH.

LET'S TAKE OUR ORIGINAL EQUATION AND JUST RE-WRITE IT OVER HERE TO THE RIGHT SO I CAN SHOW MY WORK.

AND I'M GONNA MOVE MY X-TERM TO THE OTHER SIDE • BRING MY -2Y IS GREATER THAN NEGATIVE 3X PLUS 4, AND NOW HERE WE ARE, WE ARE DIVIDING BY A NEGATIVE VALUE.

AND SO YOU WANNA BE REALLY CAREFUL AND MAKE SURE THAT YOU DO NOT FORGET TO FLIP THE SYMBOL, GIVING ME Y IS NOW LESS THAN -3/-2, IS POSITIVE 3/2X, AND 4 DIVIDED BY -2 IS -2.

AND NOW I'M GONNA GO AHEAD AND GRAPH THAT LINE.

PUT YOUR Y-INTERCEPT AT -2, OUR SLOPE IS +3/2, SO UP 3, TO THE RIGHT 2, NOTICE I'M TRAVELING IN BOTH POSITIVE DIRECTIONS.

OR YOU COULD GO IN BOTH NEGATIVE DIRECTIONS, BECAUSE A NEGATIVE DIVIDED BY A NEGATIVE IS POSITIVE.

REMEMBER THAT THIS NEEDS TO BE A DASHED LINE BECAUSE THE INEQUALITY SYMBOL IS LESS THAN, NOT LESS THAN OR EQUAL TO.

ALRIGHT, NOW TO SHADE, WE'RE AGAIN LESS THAN, SO YOU MIGHT SAY, OKAY I KNOW I'M GONNA GO DOWN OR I'M GONNA GO SHADE DOWN TOWARDS THAT LESS THAN SYMBOL.

OR, ONCE AGAIN, YOU CAN TEST THE POINT 0,0.

IF YOU GET A TRUE STATEMENT, YOU'RE GONNA SHADE WHERE 0,0 IS, IF YOU GET A FALSE STATEMENT, YOU'RE GONNA SHADE WHERE 0,0 IS NOT.

SO WHEN I TESTED 0,0 IN THAT SECOND INEQUALITY, I GOT 0 IS GREATER THAN 4, WHICH IS A FALSE STATEMENT, AND SO I'M GOING TO SHADE WHERE 0,0 IS NOT, WHICH IS IN THIS AREA OF THE GRAPH.

NOW THE SOLUTION SET LIES IN THE DOUBLE-SHADED AREA, SO TYPICALLY WE LABEL THAT AREA WITH A BIG S FOR SOLUTION SET, AND THAT MEANS ANY POINTS IN THAT AREA ARE SOLUTIONS, AND WILL WORK FOR BOTH INEQUALITIES.

NOW THEY'RE GONNA TYPICALLY, AFTER YOU COMPLETE GRAPHING THE SYSTEM, YOU'RE GONNA HAVE TO ANSWER A QUESTION, TYPICALLY ABOUT A POINT IN THE SOLUTION SET.

SOMETIMES THEY MIGHT ASK YOU TO GIVE A POINT IN THE SOLUTION SET, AND THEN IT'S UP TO YOU TO JUST PICK ANY POINT IN YOUR GRID WHERE YOUR SOLUTION SET IS.

SOMETIMES THEY GIVE YOU A POINT AND YOU HAVE TO SEE IF IT'S IN THE SOLUTION SET AND THEN EXPLAIN YOUR ANSWERS.

SO 6,3, THE POINT 6,3 IS WHAT THEY'RE ASKING ABOUT.

6 ON X, 3 ON Y, AND IT LIES RIGHT HERE.

NOTICE IT IS IN THE SHADED AREA OF THE BLUE, THE SECOND INEQUALITY, BUT NOT THE FIRST ONE, THE PINK INEQUALITY.

SO, IT IS NOT A SOLUTION TO THE SYSTEM BECAUSE IT DOES NOT LIE IN THE SOLUTION SET, AND THAT IS ENOUGH FOR YOU TO GET ALL OF YOUR POINTS FOR THAT ANSWER, IT DOES NOT LIE IN THE SOLUTION SET.

IF IT DID, YOU WOULD SAY IT DID.

I JUST WANNA ADD ANOTHER LITTLE NOTE TO THIS QUESTION, BECAUSE I'VE OFTEN SEEN THE SYSTEM WHERE THEY GIVE YOU A POINT THAT LIES ON THE DOTTED LINE, AND THEY WANNA KNOW WHETHER THAT IS IN THE SOLUTION SET, AND THAT CAN BE A COMMON MISTAKE FOR STUDENTS.

SO, FOR EXAMPLE, WHAT IF INSTEAD OF 6,3 THEY ASKED IF THE POINT -2,-5 WAS IN THE SOLUTION SET?

SO THE POINT -2,-5, IF I PLOT IT, IT FALLS RIGHT HERE, IT IS IN THE SOLUTION SET, THE SHADED AREA FOR THE FIRST INEQUALITY, BUT IT'S ON THE DASHED LINE OF THE SECOND INEQUALITY, SO IT IS ACTUALLY NOT A SOLUTION TO THE SYSTEM, BECAUSE IT IS A SOLUTION TO THE FIRST ONE, BUT NOT THE SECOND ONE, BECAUSE POINTS ON A DASHED LINE ARE NOT SOLUTIONS.

AND THAT'S YOUR INEQUALITY SYSTEM, I'M GONNA HAND IT OVER TO MR. TO MR. SICKLES, WHERE HE'S GONNA TALK TO YOU ABOUT FUNCTIONS.

>> HELLO AGAIN, SO THIS IS THE SECOND THING THAT WE DECIDED TO HIGHLIGHT IN THIS VIDEO, 5 THINGS YOU NEED TO KNOW ABOUT THE ALGEBRA 1 REGENTS, THIS IS INTERPRETING FUNCTIONS, AND WHAT WE MEAN BY THAT IS THE REGENTS EVERY YEAR LIKES TO GIVE YOU EITHER A GRAPH OR AN EQUATION THAT REPRESENTS SOME SITUATION IN OUR WORLD AND IT'S MODELED BY A FUNCTION.

SO I CHOSE TO HIGHLIGHT QUESTION 36 FROM THE JUNE 2022 REGENTS, WHICH IS A PROJECTILE MOTION QUESTION THAT'S BASED ON A FUNCTION EQUATION.

SO THIS QUESTION READS, "A BALL IS PROJECTED UP INTO THE AIR FROM THE SURFACE OF A PLATFORM TO THE GROUND BELOW.

"THE HEIGHT OF THE BALL ABOVE THE GROUND, IN FEET, IS MODELED BY THE FUNCTION F(T)>>-16T(SQUARED) PLUS 96T PLUS 112, WHERE T IS THE TIME, IN SECONDS, AFTER THE BALL IS PROJECTED."

SO THIS IS EXACTLY WHAT I MEAN, THEY ARE TRYING TO MODEL A REAL-LIFE SITUATION USING ALGEBRA, OR A FUNCTION - WELL, BOTH I GUESS - AND THEY'RE TRYING TO GET YOU TO UNDERSTAND OR MODEL THIS SITUATION USING YOUR ALGEBRAIC SKILLS AND ALSO YOUR COMPREHENSION SKILLS, AND THAT'S WHAT MAKES THESE QUESTIONS SO CHALLENGING, THERE'S A LOT OF VOCABULARY, A LOT OF TOPICS THAT THEY HIT IN THESE QUESTIONS, AND HOPEFULLY I'LL HIT ALL OF THEM WHEN I TALK ABOUT THIS PROBLEM.

SO THE FIRST QUESTION HERE SAYS "STATE THE HEIGHT OF THE PLATFORM, IN FEET," SO WHAT YOU HAVE TO UNDERSTAND IS, THE BALL IS BEING LAUNCHED FROM A CERTAIN ELEVATION AT A TIME OF 0 SECONDS.

SO WHAT THEY'RE REALLY ASKING IS WHAT IS THE HEIGHT OF THE BALL WHEN YOU HAVE AN INDEPENDENT VARIABLE OF T, 0 SECONDS.

SO WHAT THIS IS ASKING FOR IS EVALUATING THIS FUNCTION AT - SORRY I'M GONNA CHANGE MY FONT HERE - EVALUATING THIS FUNCTION AT A TIME OF 0 SECONDS.

SO WHAT YOU'RE GONNA DO IS YOU'RE GONNA PLUG 0 INTO T FOR THAT FUNCTION, -16 TIMES 0(SQUARED) PLUS 96 TIMES 0 PLUS 112.

SO THAT MEANS THAT AT A TIME OF 0, YOU'RE LAUNCHING A BALL FROM THAT PLATFORM.

THIS ONE'S NICE AND EASY BECAUSE THE 0S IN THE FIRST TWO TERMS NEGATE, THEY'RE CANCELED OUT, AND I AM LEFT WITH A HEIGHT OF 112 FEET, SO MY FUNCTION AT A TIME OF 0 IS 112 FEET.

WHAT THAT MEANS IS THE PLATFORM IS AT 112 FEET.

SO IT'S IMPORTANT TO REMEMBER THAT THE REGENTS DOES CARE A LOT ABOUT BEING SPECIFIC ABOUT YOUR UNITS, SO MAKE SURE YOU'RE SPECIFIC AND SAY THAT IT'S 112 FEET, NOT JUST 112, 112 FEET, BECAUSE THIS DOES MODEL A REAL-LIFE SITUATION.

THE SECOND PART SAYS, "STATE THE COORDINATES OF THE VERTEX; EXPLAIN WHAT THAT MEANS IN THE CONTEXT OF THE PROBLEM."

SO THIS IS ANOTHER VOCABULARY WORD, THE VERTEX, WHICH IS EITHER THE VERY TOP OR THE VERY BOTTOM OF A PARABOLA.

SO THIS SITUATION I'VE GOT A PARABOLA, IT'S A PROJECTILE MOTION, SO I'M LAUNCHING THE BALL FROM SOME POINT, AND THE VERTEX IS GOING TO BE THE VERY TIPPY-TOP OF THAT PARABOLA, RIGHT AT THE MAXIMUM VALUE.

SO THE COORDINATES OF THE VERTEX CAN BE CALCULATED WITH THE CALCULATOR; HOWEVER YOU DID LEARN AN ALGEBRAIC WAY THIS YEAR AS WELL USING THE AXIS OF SYMMETRY.

SO THE AXIS OF SYMMETRY FORMULA IS X = -B OVER 2 TIMES A, OF COURSE I DON'T HAVE X IN THIS PROBLEM, I DO HAVE T SO I'M GONNA USE T INSTEAD OF X.

AND THE B AND THE A ARE REFERENCING THE COEFFICIENTS IN A GENERIC QUADRATIC EQUATION - AX(SQUARED) PLUS BX PLUS C IS USUALLY EQUAL TO Y, OR RATHER, F(T).

SO I HAVE TO DETERMINE WHAT A IS AND WHAT B IS IN THE EQUATION THAT THEY'VE GIVEN ME.

SO I SEE MY A RIGHT THERE IS -16, SO A IS -16, B IS 96, AND C IS 112.

SO I'M JUST GONNA TAKE A AND B AND PLUG THEM INTO MY FORMULA TO FIGURE OUT WHAT THE AXIS OF SYMMETRY IS.

SO T >> -96 DIVIDED BY 2 TIMES SO T X -96 DIVIDED BY 2 TIMES -16, AND I GET AN AXIS OF SYMMETRY, T >> 3.

SO WHAT THAT AXIS OF SYMMETRY TELLS US IS THE X-COORDINATE, OR IN THIS CASE THE T-COORDINATE, OF THAT VERTEX.

SO I JUST FOUND THAT THAT VERTEX IS AT 3, SOMETHING, WHICH I HAVE TO USE MY FUNCTION EQUATION NOW TO FIGURE OUT.

SO NOW I HAVE THAT INDEPENDENT VARIABLE OF 3, NOW I HAVE TO TAKE IT AND SUBSTITUTE IT BACK INTO THE FUNCTION EQUATION TO GET MY DEPENDENT VARIABLE, Y OR F(T) IN THIS CASE.

SO MY FUNCTION AT A T VALUE OF 3, I'M JUST GONNA SUBSTITUTE IN 3 EVERY TIME I SEE T, PLUS 96, PLUS 112, AND THEN I CAN PLUG THAT RIGHT INTO THE CALCULATOR, YOU WILL GET A HEIGHT OF 256 FEET.

SO WHAT THAT MEANS IS MY VERTEXT, USING THE AXIS OF SYMMETRY, AND THEN PLUGGING IT BACK INTO THE EQUATION, IS 3,256.

THAT'S NOT ALL, THOUGH, THEY ARE ASKING WHAT THAT MEANS IN THE CONTEXT OF THE PROBLEM, SO YOU GOTTA REMEMBER YOU'RE LAUNCHING A BALL, THAT MEANS AT ITS MAXIMUM HEIGHT, AT ITS VERTEX, WHICH WOULD BE 256 FEET, IT WOULD TAKE 3 SECONDS, SO THE VERTEX IS 3,256, AND THAT REPRESENTS HOW AT 3 SECONDS, THE BALL REACHES ITS MAXIMUM HEIGHT OF 256 FEET.

SO A LOT OF VOCABULARY THERE, VERTEX, MAXIMUM, MINIMUM, AND THEN UNDERSTANDING THE AXIS OF SYMMETRY TO SOLVE THE PROBLEM THAT WAY.

OKAY, THE LAST PART SAYS "STATE THE ENTIRE INTERVAL OVER WHICH THE BALL'S HEIGHT IS DECREASING."

SO THERE'S REALLY TWO VOCABULARY WORDS THERE: INTERVAL AND DECREASING, DECREASING'S A LITTLE BIT EASIER, THAT'S WHEN THE HEIGHT OF THE BALL IS GOING DOWN.

AND INTERVAL IS REFERENCING X-VALUES USUALLY, IN THIS CASE IT'S T-VALUES.

SO AT WHAT TIMES IS THE BALL GOING DOWN, ESSENTIALLY.

SO WHAT IT'S REALLY ASKING FOR IS THE 0S, BECAUSE I ALREADY KNOW AT ITS MAXIMUM THE HEIGH OF OF THE BALL WAS 256 FEET BUT IT TOOK 3 SECONDS TO GET THERE - I NEED TO KNOW HOW LONG IT TAKES FROM THOSE 3 SECONDS TO REACH ITS MINIMUM, OR REACH THE GROUND, AT 0 FEET, SO IT'S REALLY LOOKING FOR THE 0S OF THAT FUNCTION.

IN THAT CASE, I'M GONNA SUBSTITUTE 0 IN FOR F(T) AND THEN SOLVE FOR T: -16T(SQUARED), PLUS 96T, PLUS 112, AND THIS IS A QUADRATIC AND YOU CAN USE YOUR SKILLS TO SOLVE A QUADRATIC HERE, AND IN THIS CASE THIS ONE IS FACTORABLE, ALL OF THESE COEFFICIENTS ARE DIVISIBLE BY 16, SO I'M JUST GOING TO DIVIDE EVERYTHING IN THIS CASE BY -16 BECAUSE I REALLY LIKE TO HAVE THAT FIRST TERM TO BE POSITIVE, THAT A VALUE TO BE POSITIVE WHEN I'M FACTORING.

AND OF COURSE THE 0 IS DIVIDED BY -16, AND I GET 0= T(SQUARED) -6T.

THIS ONE IS FACTORABLE, SO I'M GONNA SAY 2 NUMBERS THAT MULTIPLY TO -7 AND ADD TO -6 ARE OF COURSE -7 AND +1, STILL EQUAL TO 0, AND NOW WE CAN USE OUR ZERO PRODUCT LAW TO SPLIT THOSE TWO UP AND THEN SOLVE EACH INDIVIDUAL 0.

T MINUS 7 = 0, ALSO T PLUS 1= 0, SO I GET A TIME OF 7 AND A TIME OF -1.

SO NOW YOU'VE GOT THOSE TWO 0S, THOSE ARE BOTH COMPLETELY RIGHT, BUT YOU'VE GOTTA THINK ABOUT IN THE CONTEXT OF THE PROBLEM WHICH ONE MAKES MORE SENSE.

SO YOU KNOW THAT THE BALL IS NOT GOING TO LAND ON THE GROUND AT -1 SECONDS, THAT DOESN'T MAKE MUCH SENSE, SO REALLY IT'S AT 7 SECONDS WHEN THE BALL WOULD REACH THE GROUND.

SO TRY TO ALWAYS WRITE A CONCLUSION AT THESE PROBLEMS, RIGHT, SO I USED THE 0S TO FIGURE OUT THAT THE BALL LANDED AT A TIME OF 7 SECONDS, AND THE MAXIMUM HEIGHT OCCURRED AT 3 SECONDS, SO REALLY THE INTERVAL THAT THE BALL IS DECREASING, IS GOING DOWN, IS FROM THOSE 3 SECONDS TO THE 7 SECONDS.

I CAN ALSO USE INEQUALITY NOTATION TO WRITE THIS, SO REALLY MY INTERVAL IS 3 IS LESS THAN T WHICH IS LESS THAN 7.

SO AGAIN, THESE PROBLEMS ARE ALL MODELING SITUATIONS IN OUR REAL WORLD, THERE'S A LOT OF VOCABULARY, BUT IT TENDS TO BE CONSISTENT, THEY CONSISTENTLY COME UP OVER AND OVER AND OVER, LIKE VERTEX IF IT'S A QUADRATIC, MAXIMUM OR MINIMUM, INTERVALS COME UP A LOT, INCREASING AND DECREASING AND CONSTANT, SO TRY A LOT OF THESE PROBLEMS.

WHAT I DID HERE IS I WENT THROUGH A COUPLE OF THE RECENT REGENTS AND JUST GAVE SOME KEYS TO PROBLEMS THAT LOOK LIKE THIS, IF YOU NEED TO STOP THE VIDEO AND TRY THEM OUT IT'S A GREAT OPPORTUNITY TO DO SO.

ONE OTHER THING I DID WANT TO TALK ABOUT WAS AVERAGE RATE OF CHANGE, BECAUSE A LOT OF THESE PROBLEMS DEAL WITH AVERAGE RATE OF CHANGE.

SO AVERAGE RATE OF CHANGE FOR A FUNCTION PROBLEM IS REALLY ASKING FOR A SLOPE.

SO IF YOU'RE ASKED TO FIND AN AVERAGE RATE OF CHANGE, IT'S REALLY THE SAME AS FINDING A SLOPE.

IN THIS PROBLEM HERE IT'S JUST MODELING THE POPULATION OF A TOWN OVER TIME, AND THEY'RE ASKING FOR THE AVERAGE RATE OF CHANGE IN THE POPULATION BETWEEN AN INTERVAL OF, IN THIS CASE, 30 TO 40 YEARS, SO IT'S REALLY JUST CALCULATING A SLOPE BETWEEN YEARS 30 AND 40.

SO TRY A BUNCH OF THOSE PROBLEMS, TYPICALLY LIKE WE SAID IT'S ALWAYS THE FIRST PROBLEM IN PART THREE, THERE'S BEEN VERY FEW REGENTS WHERE THERE HASN'T BEEN A FUNCTION THAT'S BEEN MODELED, SO I THINK IT'S A VALUABLE EXERCISE TO TRY A BUNCH OF THOSE OUT.

THANK YOU.

>> OKAY, SO FOR OUR NEXT TOPIC WE'RE GOING TO DIVE INTO STATISTICS HERE WITH A LINEAR REGRESSION ANALYSIS.

THIS IS A VERY POPULAR QUESTION AND IT'S VERY FORMULAIC, THEY BASICALLY ASK YOU TO DO THE SAME THING EVERY TIME, SO IF YOU FEEL PREPARED AFTER GOING THROUGH THIS EXAMPLE, YOU'LL BE READY TO TACKLE THIS QUESTION.

SO THEY'RE GOING TO GIVE YOU A SET OF DATA THAT'S GOING TO INVOLVE A BI-VARIATE RELATIONSHIP, MEANING YOU'VE GOT AN X VARIABLE AND YOU'VE GOT A Y VARIABLE, AND THERE'S SOME SORT OF RELATIONSHIP BETWEEN THE TWO.

THIS TABLE HERE SHOWS THE NUMBER OF MATH CLASSES MISSED DURING A SCHOOL YEAR AND FINAL EXAM SCORES FOR 9 STUDENTS.

SO THE NUMBER OF CLASSES MISSED IS YOUR X, THAT'S YOUR INDEPENDENT VARIABLE, AND THAT'S GOING TO AFFECT YOUR Y, YOUR FINAL EXAM SCORE, YOUR DEPENDENT VARIABLE.

SO WHAT WE NEED TO DO IS WE NEED TO GET THIS DATA INTO OUR CALCULATOR TO ASSESS THE RELATIONSHIP AND CREATE AN EQUATION TO MODEL THE RELATIONSHIP SO THAT WE CAN MAKE PREDICTIONS ON WHAT YOUR FINAL EXAM SCORE WOULD BE BASED ON THE NUMBER OF MATH CLASSES YOU MISSED.

THIS IS A VERY CALCULATOR-HEAVY SECTION, SO IF YOU WANT TO PAUSE AND WRITE DOWN THE CALCULATOR STEPS, OR JUST MAKE SURE YOU HAVE YOUR CALCULATOR WITH YOU, THAT MIGHT BE HELPFUL FOR THIS SECTION.

JUST A QUICK NOTE, YOU DO HAVE TO HAVE A CERTAIN MODE TURNED ON, YOU NEED TO HAVE YOUR STAT DIAGNOSTICS ON, SO THE WAY YOU WOULD DO THAT IS YOU HIT THE MODE BUTTON AND USE YOUR DOWN ARROW UNTIL YOU GET TO STAT DIAGNOSTICS, AND THEN YOU TOGGLE OVER THE WORD ON, MAKE SURE WE'RE HIGHLIGHTED ON THE WORD ON, AND HIT ENTER.

SO ONCE YOU DO THAT YOU'RE READY TO GO.

THE FIRST THING YOU NEED TO DO IS YOU NEED TO TELL YOUR CALCULATOR WHAT THE DATA IS.

SO WE'RE GONNA ENTER OUR DATA INTO TWO LISTS.

YOU FIND THOSE BY GOING TO STAT, AND THE FIRST OPTION IN YOUR STAT APPLICATION IS EDIT, AND IT'S GONNA BE CHOICE ONE, EDIT.

WHAT WILL POP UP IS A SET, A TABLE OF LISTS.

IF YOU ALREADY HAVE DATA IN YOUR LISTS, YOU'RE GONNA WANT TO CLEAR IT SO YOU CAN ENTER YOUR NEW DATA IN, SO YOU HIT THE CARROT BUTTON TO GET UP INTO THE LIST, AND THEN YOU HIT CLEAR AND THEN ENTER AND YOUR LIST WILL BE EMPTY.

NEXT, YOU'RE GOING TO ENTER YOUR DATA WITH YOUR X VALUES INTO L1, TYPE THE NUMBER, HIT ENTER, AND THEN YOUR Y VALUES INTO L2, TYPE THE NUMBER AND HIT ENTER.

BE CAREFUL, MAKE SURE YOU KEEP YOUR PAIRS TOGETHER, AND MAKE SURE YOU ARE TYPING EVERYTHING IN CAREFULLY, BECAUSE JUST ONE LITTLE TYPO WILL CHANGE YOUR ENTIRE ANSWER.

NOW THAT YOU'VE ENTERED THE DATA IN YOU'RE GOING TO ASK YOUR CALCULATOR TO ANALYZE IT.

SO WE ARE GOING TO GO TO STAT AND THEN ARROW OVER TO CALC.

WHAT WE'RE GOING TO LOOK FOR IS OUR REGRESSION EQUATION AND THE CORRELATION COEFFICIENT.

SO I HIT STAT, ARROW OVER TO CALC, THEN ARROW DOWN TO CHOICE 4, LINREG(AX PLUS B).

THIS IS WHAT WILL POP UP ON YOUR SCREEN.

TAKE A QUICK MOMENT TO MAKE SURE THE LISTS ARE CORRECT: X IS L1, Y IS L2, YOU DON'T REALLY USE FREQUENCY LIST OR STORE REGULAR EQUATION, SO WE'RE GONNA GO DOWN TO CALCULATE AND HIT ENTER.

IF YOU'VE ENTERED THIS DATA INTO YOUR CALCULATOR, YOU SHOULD GET THIS INFORMATION FROM CHOICE 4.

NOW, IF YOU DIDN'T HAVE YOUR STAT DIAGNOSTICS TURNED ON LIKE I TOLD YOU TO DO IN THE BEGINNING, THESE TWO PIECES OF INFORMATION WOULDN'T BE HERE AND YOU WOULDN'T BE ABLE TO ANSWER THE SECOND PART OF THE QUESTION.

SO YOU GO BACK, MAKE SURE THOSE DIAGNOSTICS ARE ON.

THIS IS ALL THE INFORMATION THAT WE NEED TO ANSWER THE QUESTION.

FOR OUR FIRST PART, WE ARE WRITING OUR LINEAR REGRESSION EQUATION, AND WE'RE ROUNDING ALL OUR VALUES TO THE NEAREST HUNDREDTH.

SO Y >> AX PLUS B, A IS OUR SLOPE, ROUNDED TO THE NEAREST HUNDREDTH, THAT IS GOING TO BE -2.81, AND B, OUR Y-INTERCEPT, IS 97.55, THEREFORE YOUR EQUATION IS Y >> 2.81X PLUS 97.55.

NOW THEY'RE GOING TO ASK YOU A FOLLOW-UP QUESTION, AND IT'S GOING TO BE SOMETHING THAT YOU'RE GOING TO HAVE TO ANSWER AND THEN INTERPRET IN CONTEXT OF THE PROBLEM.

ONE OF THE THINGS THAT THEY COULD POSSIBLY ASK YOU TO FIND, AS THEY DO IN THIS EXAMPLE, IS THE CORRELATION COEFFICIENT.

SO THE CORRELATION COEFFICIENT IS THE R VALUE.

WE DON'T USE R(SQUARED) IN ALGEBRA 1, SO WE DON'T NEED TO WORRY ABOUT THAT VALUE, BUT R TELLS US THE DIRECTION AND THE STRENGTH OF THE RELATIONSHIP BETWEEN THE TWO VARIABLES.

SO OUR R VALUE HERE, ROUNDED TO THE NEAREST HUNDREDTH, IS -.97.

SOMETIMES THEY ASK YOU TO EXPLAIN WHAT THAT MEANS WITHIN THIS PART OF THE QUESTION, AND SOMETIMES THEY STATE IT AS A SEPARATE PART, WHICH THEY DO HERE.

HERE THEY SAY, "STATE WHAT THE CORRELATION COEFFICIENT INDICATES ABOUT THE LINEAR FIT OF THE DATA," AND THERE ARE TWO THINGS THAT YOU NEED TO STATE.

ONE, YOU NEED TO STATE THE DIRECTION, AND THAT'S GOING TO BE POSITIVE OR NEGATIVE, THIS IS A NEGATIVE VALUE SO IT IS A NEGATIVE RELATIONSHIP, WHICH MEANS THAT AS YOUR X INCREASES, NUMBER OF MATH CLASSES MISSED, YOUR Y DECREASES, FINAL EXAM SCORE, SO THERE'S A NEGATIVE RELATIONSHIP.

THE R VALUE ALSO TELLS US HOW STRONG THE RELATIONSHIP IS, AND IF YOU WERE TO PLOT YOUR POINTS, HOW CLOSE THEY ARE TO THE LINE THAT YOU CREATED IN THE FIRST PART.

YOUR R VALUE CAN RANGE FROM -1, WHICH IS A VERY STRONG NEGATIVE RELATIONSHIP, TO 0, WHICH IS A WEAK OR NO RELATIONSHIP, ALL THE WAY TO +1, WHICH IS A VERY STRONG, POSITIVE RELATIONSHIP.

SO THE CLOSER YOU ARE OUT TO THOSE ENDS, THE STRONGER THE RELATIONSHIP.

SO AN R VALUE OF -.97 IS GOING TO INDICATE THAT THERE IS A STRONG, NEGATIVE RELATIONSHIP BETWEEN - AND THIS IS IMPORTANT, GIVE THE CONTEXT OF THE PROBLEM, WHAT IS THE X VALUE?

NUMBER OF CLASSES MISSED.

AND WHAT'S BEING AFFECTED, THE Y VALUE; FINAL EXAM SCORE.

NOW THAT'S IT FOR THIS PARTICULAR QUESTION, THERE'S YOUR FOUR POINTS, BUT OFTENTIMES THEY DON'T ASK ABOUT THE CORRELATION COEFFICIENT OR THEY DO BUT IN ADDITION TO ASKING ABOUT THAT, YOU COULD BE ASKED TO EXPLAIN WHAT THE SLOPE AND Y-INTERCEPT MEAN IN CONTEXT OF THE PROBLEM.

SO I WANNA MAKE SURE I SHOW YOU THAT, JUST IN CASE YOU ENCOUNTER THAT ON YOUR EXAM.

SO IF I ADD A LITTLE EXTRA PART HERE, "INTERPRET THE SLOPE AND Y-INTERCEPT IN CONTEXT."

A LOT OF TIMES WE REFER TO THEM AS PARAMETERS OF THE EQUATION AS WELL.

SO OUR SLOPE, THAT'S GONNA BE THE COEFFICIENT OF X, THAT -2.81, THAT IS GOING TO BE THE CHANGE IN Y FOR EACH UNIT CHANGE IN X.

PUT IT TO THE CONTEXT OF THE PROBLEM, WHAT DOES THAT MEAN?

FOR EACH CLASS MISSED, A STUDENT'S FINAL EXAM SCORE, THEIR Y, WILL BE PREDICTED TO DECREASE BY 2.81 POINTS.

MORAL OF THE STORY: DON'T SKIP MATH.

OR ANY CLASS.

ALRIGHT, THEN OUR Y-INTERCEPT IS ALWAYS YOUR STARTING POINT, IT'S THE Y VALUE WHEN X IS 0.

SOMETIMES IT DOESN'T MAKE SENSE IN THE CONTEXT OF THE PROBLEM, BUT IN THIS CASE IT DOES, AND IT WOULD MEAN, A Y-INTERCEPT OF 97.55, WOULD MEAN A STUDENT WHO HAS MISSED 0 CLASSES WOULD BE PREDICTED TO HAVE A FINAL EXAM SCORE OF 97.55.

AND THAT'S EVERYTHING YOU NEED TO KNOW ABOUT LINEAR REGRESSION; I'LL BE BACK WITH OUR NEXT SECTION.

>> IN OUR NEXT SECTION, WE ARE GONNA TACKLE THE LAST QUESTION OF THE EXAM, PART FOUR, THE SIX-POINTER, WHICH IS ALMOST ALWAYS SOME SORT OF SYSTEM OF EQUATIONS.

NOW THE THING WITH THIS QUESTION IS IT IS THE LAST QUESTION IN THE TEST, PROBABLY THE LAST QUESTION YOU'LL DO, AND SO BY THAT POINT YOU'RE MENTALLY AND PHYSICALLY EXHAUSTED.

SO IT'S GOOD TO GO IN PREPARED FOR THIS QUESTION.

IT'S USUALLY A SYSTEM OF EQUATIONS WHERE YOU'RE GIVEN A WORD PROBLEM, AN APPLICATION YOU NEED TO MODEL WITH TWO EQUATIONS, AND THEN YOU MIGHT NEED TO SOLVE IT ALGEBRAICALLY, GRAPHICALLY, OR EITHER WAY.

BECAUSE WE ALREADY WORKED THROUGH A GRAPHING SYSTEM WITH OUR INEQUALITY EXAMPLE EARLIER, I'M GOING TO STICK TO SOLVING OUR SYSTEMS ALGEBRAICALLY, AND I'M GOING TO SHOW YOU TWO DIFFERENT METHODS THAT COULD POP UP.

HERE IS OUR FIRST EXAMPLE.

"AT A LOCAL GARDEN SHOP, THE PRICE OF PLANTS INCLUDES SALES TAX."

NOW THAT JUST MEANS YOU DON'T HAVE TO WORRY ABOUT SALES TAX FOR THIS PROBLEM.

"THE COST OF 4 LARGE PLANTS AND 8 MEDIUM PLANTS IS $40.

"THE COST OF 5 LARGE PLANTS AND 2 MEDIUM PLANTS IS $28."

THEY TELL US TO USE L FOR THE COST OF A LARGE PLANT, AND M FOR THE COST OF A MEDIUM PLANT, AND WRITE A SYSTEM OF EQUATIONS THAT MODELS THIS SITUATION.

I LIKE TO CALL THIS TYPE OF PROBLEM, WHERE YOU HAVE TWO SIMILAR EQUATIONS, IF YOU CAN WRITE THE FIRST EQUATION, THE SECOND EQUATION IS GONNA BE A LOT LIKE IT.

SO OUR FIRST EQUATION INCLUDES THE LARGE PLANTS, THE MEDIUM PLANTS, AND THE COST OF THE QUANTITY THAT WE BOUGHT, WHICH WOULD BE 4L PLUS 8M >> 40.

WOULD BE 4L PLUS 8M = 40.

THEN OUR SECOND EQUATION IS GOING TO SHOW THAT 5 LARGE PLANTS PLUS 2 MEDIUM PLANTS COST $28.

SO AS YOU CAN SEE, WE'VE GOT TWO SIMILAR EQUATIONS, WHICH MEANS THAT WHEN I GO AHEAD AND SOLVE IT, I'M PROBABLY GOING TO USE THE METHOD OF ELIMINATION.

BUT LET'S SEE WHAT THEY'RE ASKING US TO DO FIRST.

"COULD THE COST OF ONE LARGE PLANT BE $5.50 AND THE COST OF ONE MEDIUM PLANT BE $2.25?

"JUSTIFY YOUR ANSWER."

SO BEFORE THEY'RE ASKING YOU TO SOLVE, THEY'RE ASKING YOU TO TEST A POINT, TEST THE POINT 5.50 FOR L AND 2.25 FOR M, AND YOU NEED IT TO WORK IN BOTH EQUATIONS IN ORDER FOR IT TO BE A SOLUTION.

SO I START WITH MY FIRST EQUATION AND I SUBSTITUTE 5.50 AND 2.25 IN RESPECTIVELY, DO MY MATH AND I DO GET A TRUE STATEMENT, 40 = 40.

SO YOU WOULD SPEND $40 ON 4 LARGE PLANTS AND 8 MEDIUM PLANTS IF THEY COST 5.50 AND 2.25.

BUT DOES THIS WORK FOR THE OTHER SCENARIO, DOES THIS WORK FOR THE OTHER EQUATION?

LET'S SEE.

WHEN WE WORK IT OUT, IT DOES NOT, WE GET A FALSE STATEMENT.

SO ALTHOUGH IT WORKED FOR OUR FIRST EQUATION, IN ORDER TO BE A SOLUTION IT HAS TO WORK IN BOTH, SO THE ANSWER HERE IS NO, THE COST COULD NOT BE 5.50 AND 2.25.

IN OUR NEXT PART, WE ARE GOING TO FIND THE COST OF THE PLANTS, AND WE'RE GOING TO SOLVE THIS USING THE PROCESS OF ELIMINATION.

WHAT I DO, IS I TAKE MY TWO EQUATIONS AND WE LINE THEM UP VERTICALLY, I KNOW YOU CAN'T REALLY SEE MY SECOND EQUATION THERE.

AND THEN WHAT WE - I'LL DO THAT.

WHAT WE DO, IS WE LOOK AT OUR PAIRS OF VARIABLES AND SEE IF WE CAN MAKE THEM INTO OPPOSITE COEFFICIENTS, SO THE SAME VALUE BUT OPPOSITE SIDES, SO THAT WOULD MEAN WE WANT TO GET 4L AND 5L TO BE OPPOSITE COEFFICIENTS.

OR, 8M AND 2M, AND THE EASIEST STRATEGY IS GOING TO BE TO WORK WITH YOUR MS, BECAUSE IF I MULTIPLY 2 BY -4, I GET -8 AND WE WILL HAVE OPPOSITE COEFFICIENTS FOR OUR M VARIABLE.

SO I HAVE TO MULTIPLY MY ENTIRE SECOND EQUATION BY -4.

BRING YOUR FIRST EQUATION OVER, RE-WRITE YOUR SECOND EQUATION, -20L MINUS -8M = -112, AND THEN ADD THOSE EQUATIONS TOGETHER.

WHEN WE DO THAT, OUR MS ARE ELIMINATED, BECAUSE 8M AND -8M EQUALS 0M.

LAST STEP IS WE DIVIDE BY -16 TO GET L= 4.5, WHICH MEANS THAT THE COST OF A LARGE PLANT IS $4.50.

NOW WE NEED TO GO BACK AND FIND THE COST OF A MEDIUM PLANT BY SUBSTITUTING L INTO ONE OF OUR ORIGINAL EQUATIONS, AND IT DOESN'T MATTER WHICH ONE, I USED THE SECOND ONE.

SO 5 TIME 4.5 PLUS 2M= 28, GIVES US THIS, CONTINUE TO SOLVE, AND WE END UP GETTING 2.75 FOR M. SO IN ANSWER TO THE QUESTION, THE COST OF ONE LARGE PLANT IS $4.50 AND THE COST OF ONE MEDIUM PLANT IS $2.75.

SO THAT WAS AN EXAMPLE WHERE WE HAD SIMILAR EQUATIONS BUT DIFFERENT SCENARIOS.

NOW WE'RE GONNA LOOK AT A DIFFERENT TYPE OF SYSTEM THAT WE'RE ALSO GOING TO SOLVE ALGEBRAICALLY.

SO JUMPING INTO THE SECOND EXAMPLE, "ALLYSA SPENT $35 TO PURCHASE 12 CHICKENS.

"SHE BOUGHT TWO DIFFERENT TYPES OF CHICKENS: AMERICANA CHICKENS, WHICH COST $3.75 EACH, AND DELAWARE CHICKENS, WHICH COST $2.50 EACH.

"WRITE A SYSTEM OF EQUATIONS THAT CAN BE USED TO DETERMINE THE NUMBER OF AMERICANA CHICKENS, A, AND THE NUMBER OF DELAWARE CHICKENS, D, SHE PURCHASED."

SO NOW WE HAVE TWO DIFFERENT EQUATIONS HERE TALKING ABOUT TWO DIFFERENT THINGS INVOLVING THESE CHICKENS.

YOU HAVE HOW MANY CHICKENS SHE HAS, WHICH HAS NOTHING TO DO WITH WHAT THEY COST TO PURCHASE, AND THEN YOU HAVE HOW MUCH THEY COST TO PURCHASE AND THE AMOUNT OF MONEY THAT SHE SPENT.

SO OUR FIRST EQUATION IS OUR TOTAL QUANTITY, WHICH IS THE HOW MANY, THE ANSWER TO HOW MANY.

THE NUMBER OF AMERICANA CHICKENS, A, PLUS THE NUMBER OF DELAWARE CHICKENS, D, WAS A TOTAL OF 12, SO THERE'S MY FIRST EQUATION.

MY SECOND EQUATION IS GOING TO TAKE IN THE VALUE OF IT ALL, THE AMOUNT OF MONEY SHE SPENT ON THE TWO DIFFERENT PRICES FOR THE TWO DIFFERENT TYPES OF CHICKENS, SO THAT I LIKE TO CALL OUR TOTAL VALUE EQUATION.

AND HERE'S WHERE WE TAKE INTO ACCOUNT THAT THE AMERICANA CHICKEN COST 3.75 PLUS THE DELAWARE CHICKENS COST 2.50, AND SHE SPENT A TOTAL OF $35.

NOW THESE QUESTIONS CAN SHOW UP IN A LOT OF DIFFERENT WAYS, A LOT OF TIMES THEY SHOW UP AS MONEY, WHERE YOU HAVE DIMES AND QUARTERS, AND YOU HAVE A CERTAIN AMOUNT OF DIMES AND QUARTERS, AND THEN YOU HAVE A CERTAIN VALUE THAT THAT MONEY'S WORTH.

I'VE ALSO SEEN AN EXAMPLE WHERE THEY TALKED ABOUT NUMBERS OF BICYCLES AND TRICYCLES.

YOU HAVE A CERTAIN AMOUNT OF BOTH, AND THEN YOU HAVE TO TAKE INTO ACCOUNT HOW MANY WHEELS YOU HAVE IN TOTAL, WHICH IS DIFFERENT FOR BICYCLES AND TRICYCLES.

THIS EXAMPLE IS TALKING ABOUT PRICES OF CHICKENS.

FIRST STEP HERE, THEY WANT US TO DETERMINE ALGEBRAICALLY HOW MANY OF EACH TYPE OF CHICKEN SHE PURCHASED.

SO WHAT I WANT TO SHOW YOU THIS EXAMPLE - OR WHY I WANT TO SHOW YOU THIS EXAMPLE, IS BECAUSE IT LENDS ITSELF TO A DIFFERENT METHOD OF SOLVING YOUR SYSTEM.

IN THIS CASE, I WOULD USE THE METHOD OF SUBSTITUTION.

WE TAKE OUR REALLY SIMPLE EQUATION, THAT A PLUS D= 12, AND WE'RE GOING TO SOLVE FOR EITHER ONE OF THOSE VARIABLES.

AND I'M GOING TO CHOOSE TO SUBTRACT A ON BOTH SIDES, SO THAT I SOLVE FOR D, AND D IS EQUAL TO THE EXPRESSION OF 12 MINUS A. I GO TO MY OTHER EQUATION AND I REPLACE THE VARIABLE D WITH THE EXPRESSION I FOUND FOR D, 12 MINUS A, AND THAT LOOKS LIKE THIS.

AND THEN WE'RE GOING TO DISTRIBUTE OUR $2.50, SIMPLIFY, AND SOLVE • TO GET A= 4, AND THEN WE JUST TAKE THAT VALUE, THAT MEANS WE HAVE 4 AMERICANA CHICKENS, WHICH MEANS 12 MINUS 4 GIVES ME 8 DELAWARE CHICKENS.

SO IN AN ANSWER TO THAT QUESTION, ALLYSA PURCHASED 4 AMERICANA CHICKENS AND 8 DELAWARE CHICKENS.

NOW THEY TYPICALLY ASK YOU SOME SORT OF EXTRA QUESTION, LIKE IN OUR LAST EXAMPLE BEFORE WE SOLVED, THEY ASKED US IF A CERTAIN PRICE COULD WORK.

NOW, WE SOLVED AND THEY'RE ASKING US ANOTHER QUESTION, SO I'M JUST GONNA GO THROUGH THIS PART QUICKLY, BECAUSE THEY COULD CHANGE WHAT TYPE OF QUESTION THEY ASK FOR EACH EXAMPLE.

WHAT THEY'RE TELLING US HERE IS THAT EACH AMERICANA CHICKEN LAYS 2 EGGS PER DAY AND EACH DELAWARE CHICKEN LAYS 1 EGG PER DAY.

SHE SELLS EGGS BY THE FULL DOZEN FOR $2.50, HOW MUCH MONEY WILL SHE MAKE AT THE END OF HER FIRST WEEK WITH HER CHICKENS?

SO, HOW MANY EGGS IS AN AMERICANA CHICKEN GONNA LAY?

WE HAVE FOUR OF THEM, THEY'RE EACH GONNA LAY 2 EGGS PER DAY FOR 7 DAYS, GIVING HER 56 EGGS.

OUR DELAWARE CHICKENS, WE HAVE 8 OF THEM, THEY ARE GOING TO LAY 1 EGG PER DAY FOR THOSE 7 DAYS AND GIVE US ALSO 56 EGGS.

WHICH IS A TOTAL OF 112 EGGS, AND AS THEY TOLD US, SHE SELLS THEM BY THE FULL DOZEN, HOW MANY FULL DOZENS ARE IN 112?

9.3(REPEATING) DOZENS, SO THERE ARE 9 FULL DOZENS THAT SHE HAS, AT $2.50 PER DOZEN, WHICH WOULD GIVE HER $22.50 AT THE END OF THE FIRST WEEK.

HOPE THAT WAS HELPFUL FOR YOUR PART FOUR QUESTION.

MR. SICKLES IS GONNA TAKE OVER WITH SOME CALCULATOR TIPS AND TRICKS.

>> OKAY, SO THE LAST THING WE WANNA TALK ABOUT IS CALCULATOR TIPS AND TRICKS, AND USUALLY THESE ARE BEST USED FOR MULTIPLE CHOICE QUESTIONS BECAUSE YOU CAN USE SOME CALCULATOR TRICKS EITHER TO CHECK THE WORK THAT YOU'VE DONE, TO CHECK YOUR ALGEBRA, OR, IF YOU'RE TRULY STUMPED, YOU CAN USE THE CALCULATOR TO HOPEFULLY GIVE YOU SOME INSIGHT ONTO WHAT THE ANSWER IS.

SO WHAT I DID HERE IS I JUST HIGHLIGHTED OR I JUST PICKED AND CHOOSED A BUNCH OF MULTIPLE CHOICE QUESTIONS THAT LEND ITSELF WELL TO USING THE CALCULATOR TO SOLVE.

SO THE FIRST EXAMPLE OF THIS IS ASKING FOR AN EQUIVALENT EXPRESSION.

SO WHAT I'VE GOT HERE IS I'VE GOT THE ALGEBRA DONE OUT, THEY'RE ASKING WHICH EXPRESSION IS EQUIVALENT TO THE QUADRATIC X(SQUARED) PLUS 5X MINUS 6.

SO THE ALGEBRA HERE YOU WOULD WANT TO FACTOR X(SQUARED) PLUS 5X MINUS 6 INTO X PLUS 6, X MINUS 1, BUT IF YOU'RE TRULY STUMPED OR YOU NEED TO CHECK YOUR WORK, YOU CAN ABSOLUTELY USE YOUR CALCULATOR AS A WAY TO CHECK.

AND WHAT YOU'LL DO IS YOU'LL GO INTO Y EQUALS IN THE CALCULATOR, YOU WANT TO TYPE IN THE FUNCTION THAT YOU'RE TRYING TO GET THE EQUIVALENT VALUE FOR, SO X(SQUARED) PLUS 5X MINUS 6, AND WHAT YOU CAN DO IS IN Y2 YOU CAN TEST ALL OF YOUR OPTIONS AND SEE WHICH ONE MATCHES THE FUNCTION THAT YOU'RE TRYING TO TEST.

SO I'M JUST GOING TO PLUG IN CHOICE A HERE.

X PLUS 3 TIMES X MINUS 2 I KNOW FROM THE ALGEBRA THAT I'VE DONE OUT THAT THIS ISN'T THE ANSWER, BUT I CAN PROVE THAT BY EITHER GOING TO THE GRAPH AND SEEING THAT I'VE GOT TWO VERY DIFFERENT QUADRATICS HERE, OR IN THE TABLE I CAN SEE I'VE GOT DIFFERENT VALUES FOR Y FOR EACH X VALUE.

SO I WOULD KNOW THAT A IS CERTAINLY NOT THE ANSWER.

WHAT IT WOULD LOOK LIKE IF YOU'VE ACTUALLY GOT THE RIGHT ANSWER, AND I'M GOING TO PLUG IN CHOICE D, WHICH IS THE RIGHT ANSWER, X PLUS 6 TIMES X MINUS 1.

IN MY TABLE YOU'LL NOTICE THAT FOR EVERY X VALUE THE Y VALUE MATCHES, OR IF I GO TO MY GRAPH I REALLY ONLY SEE ONE QUADRATIC, BECAUSE IT'S JUST ONE SUPERIMPOSED ON TOP OF THE OTHER.

SO ANY QUESTION THAT'S ASKING ABOUT EQUIVALENT EXPRESSIONS YOU CAN USE THE CALCULATOR IN Y1 AND Y2 TO EITHER CHECK YOUR WORK OR SOLVE, IF YOU'RE TRULY LOST.

SO THAT DOESN'T JUST WORK WITH QUADRATICS, IT WILL WORK WITH EXPONENTIALS AS WELL, I'M NOT GONNA SHOW THIS ONE IN THE CALCULATOR, SO EXPONENTIALS WORK GREAT, YOU WOULD PLUG EACH OF THOSE CHOICES IN AS Y2 AND SEE IN THIS CASE WHICH ONE DOESN'T MATCH, BECAUSE THEY'RE ASKING WHICH ONE IS NOT EQUIVALENT.

OR LINEAR EQUATIONS AS WELL.

I'VE GOT THE LINEAR EQUATION 3 TIMES THE QUANTITY OF X PLUS FOUR, MINUS THE QUANTITY OF 2X PLUS 7, AND I'VE GOT FOUR DIFFERENT LINEAR OPTIONS I COULD CHECK, SO YOU COULD TRY ALL OF THOSE IN THE CALCULATOR, EITHER AS A WAY TO CHECK YOUR WORK OR TO FIGURE OUT THE RIGHT ANSWER USING YOUR TABLE AND Y1 AND Y2.

IT ALSO WORKS REALLY WELL WITH GRAPHING, SO I'VE GOT A GRAPH HERE, A POLYNOMIAL FUNCTION THAT THEY'VE GIVEN US AND IT'S A, LOOKS LIKE A CUBIC FUNCTION TO ME, I'VE GOT KIND OF TWO TURNING POINTS AND I'VE GOT THREE ROOTS.

THE STRATEGY THAT IIKE TO PREACH IS, LABEL THE POINTS THAT YOU CAN LABEL ON THE GRAPH AND THEN YOU CAN USE THE TABLE AND THE CALCULATOR TO FIGURE OUT IF THOSE POINTS MATCH THE GRAPHS WHEN YOU TEST THE OTHER FUNCTIONS, OR ALL THE OPTIONS.

RIGHT, SO I JUST WENT THROUGH AND I LABELED MY THREE ZEROS, -2,0, 1,0, AND 2,0, AND I ALSO SEE I'VE GOT ANOTHER VALUE THERE OF -1,6 THAT IS PRETTY EASY FOR ME TO SPOT.

SO WHAT WE CAN DO IS GRAPH EACH OF THESE OPTIONS AND SEE IF THE TABLE MATCHES, RIGHT, SO I'M GONNA GO THROUGH AND TEST THE FIRST ONE AND YOU'LL SEE VERY QUICKLY THAT IT DOESN'T MATCH, SO X PLUS 1 TIMES X(SQUARED) PLUS 2.

AND IF I GO TO THE TABLE, I'D BETTER SEE ALL FOUR OF THOSE POINTS, AND RIGHT AWAY I CAN SEE -2,0 ISN'T THERE, IT'S -2,-6, SO I KNOW THAT A IS NOT THE RIGHT CHOICE.

WHAT IT WILL LOOK LIKE IS, IF YOU DO CHOOSE THE RIGHT ONE, OR YOU DO TEST THE RIGHT ONE, I'VE GOT X MINUS 1 TIMES X(SQUARED) MINUS 4, AND I GO INTO MY TABLE, I AM GOING TO SEE ALL OF THOSE SPECIFIC VALUES THAT I'VE LISTED ON MY GRAPH.

I'VE GOT -2,0, I'VE GOT -1,6, I'VE GOT 1,0, AND I'VE GOT 2,0.

SO IN THIS CASE, CHOICE C WOULD BE THE BEST CASE.

IF YOU WANTED TO SOLVE THIS ALGEBRAICALLY, YOU WOULD MOST LIKELY LOOK AT YOUR THREE 0S AND THEN USE THE ZERO PRODUCT LAW IN CHOICE C TO SOLVE, MEANING YOU WOULD SAY X MINUS 1 EQUALS 0 AND X(SQUARED) MINUS 4 EQUALS 0 TO GET YOUR THREE ZEROS AND MATCH THEM TO THE GRAPH.

UNDOUBTEDLY MY FAVORITE ONE TO SHOW MY STUDENTS IS TESTING INEQUALITIES - I'M ACTUALLY GOING TO SKIP THAT ONE AND GO BACK TO IT JUST BECAUSE I'VE ALREADY STARTED TALKING ABOUT TESTING INEQUALITIES.

SO NUMBER 6 HERE IS A LINEAR INEQUALITY, I HAVE 3/2B PLUS 5 IS LESS THAN 17, AND RIGHT AWAY MY STUDENTS WOULD GET NERVOUS BECAUSE THEY SEE A FRACTION THERE, BUT YOU COULD SOLVE THIS ALGEBRAICALLY BY JUST SUBTRACTING 5 AND MULTIPLYING BY THE RECIPROCAL OF 2/3 TO GET THE SOLUTION B IS LESS THAN 8.

HOWEVER, YOU CAN ABSOLUTELY USE THE CALCULATOR TO PRODUCE A NUMBER LINE TO MATCH TO YOUR ANSWER CHOICES, RIGHT, SO IN Y1 WHAT YOU COULD DO IS JUST INPUT THIS ENTIRE INEQUALITY.

ALPHA Y EQUALS, ENTER, FOR MY FRACTION, SO I'VE GOT 3/2X (INSTEAD OF B), PLUS 5, AND I WANT TO FIND THE INEQUALITY SYMBOL LESS THAN IN MY CALCULATOR, WHICH IS SECOND MATH, ALL MY TEST INEQUALITIES I FIND LESS THAN 17.

WHAT YOUR CALCULATOR WILL DO IF YOU HIT GRAPH IS ACTUALLY PRODUCE A NUMBER LINE, SO I CAN KINDA SEE HERE I'M GONNA HAVE TO ZOOM IN A LITTLE BIT, LET ME SEE IF I CAN ZOOM IN FOR YOU.

IF I SCROLL OVER YOU CAN SEE THAT MY NUMBER LINE STOPS AT A VALUE OF 8, AND I WOULD JUST MATCH THAT TO MY SOLUTION SET HERE, B IS LESS THAN 8.

THIS WOULD BE LIKE THE NUMBER LINE WITH A POINT RIGHT AT 8 GOING TO THE LEFT, JUST LIKE YOU WOULD LEARN RIGHT AWAY IN ALGEBRA WHEN YOU'RE TALKING ABOUT INEQUALITIES.

EVALUATING FUNCTIONS IS A REALLY GOOD WAY TO - YOU CAN USE A CALCULATOR TO EVALUATE FUNCTIONS AS WELL, SORRY FOR BACKTRACKING TO NUMBER 5, BUT, USUALLY WHEN I'VE GOT A FUNCTION, I TELL MY FRESHMEN ESPECIALLY TO PUT IT RIGHT INTO THE CALCULATOR, THERE'S A LOT OF TIMES YOU CAN LEARN A LOT ABOUT THE FUNCTION JUST BY GRAPHING IT OR BY GOING TO THE TABLE AND SEEING WHAT YOU GOT.

SO YOU CAN EVALUATE FUNCTIONS USING THE CALCULATOR BY JUST INPUTTING A FUNCTION INTO Y1 AND THEN LOOKING AT THE TABLE.

SO I'VE GOT 2 TIMES 3(TO THE X POWER) PLUS 1, AND IF I GO INTO THE TABLE, IN THIS CASE I'VE GOT AN EXPONENTIAL FUNCTION, SO MY VALUES ARE INCREASING EXPONENTIALLY, BUT I JUST WANNA LOOK AND SEE, WELL, I'VE GOT THE POINT 2,19 AND I CAN KIND OF TELL I'VE GOT AN X VALUE OF 2 AND A Y VALUE OF 19, I CAN MATCH UP MY FUNCTION THAT WAY.

YOU CAN EVALUATE ANY FUNCTION THAT WAY AS LONG AS YOU KNOW THAT X IS ALWAYS THE VALUE INSIDE OF THOSE PARENTHESES.

YOU CAN ABSOLUTELY JUST TEST SOLUTIONS AS WELL, ESPECIALLY EQUATIONS WITH ONE VARIABLE, LIKE NUMBER 7 HERE WHICH SAYS, "WHAT IS THE SOLUTION TO THE EQUATION 3/5 TIMES THE QUANTITY OF X PLUS 4/3, WHICH IS EQUAL TO 1.04.

SO AGAIN THIS IS AN EXAMPLE OF A QUESTION THAT WOULD PROBABLY ELICIT SOME SHUDDERS FROM PEOPLE TAKING THE EXAM, YOU'VE GOT FRACTIONS AND DECIMALS RIGHT IN THE SAME PROBLEM, BUT YOU CAN ABSOLUTELY USE THE CALCULATOR TO TEST YOUR ANSWERS AND SEE WHICH ONE MATCHES.

SO WHAT I MEAN BY THAT, IS YOU JUST WANT TO REPLACE X OR SUBSTITUTE VALUES IN FOR X IN YOUR CALCULATOR AND JUST SEE IF IT MATCHES THAT 1.04.

SO IF I GO AND I TEST 3/5 TIMES 0.4 PLUS 4/3 AND I HIT ENTER, MY VALUE BETTER BE 1.04, AND THAT'S HOW I KNOW I GOT THE RIGHT ANSWER.

RIGHT, IF YOU SUBSTITUTE ONE OF THOSE OTHER ONES IN, YOU'RE NOT GOING TO GET 1.04, AND THAT'S HOW YOU CAN TELL IT'S NOT A CORRECT SOLUTION.

SAME THING WITH NUMBER 8, IT'S JUST A TWO-SIDED WHERE I'VE GOT X ON BOTH SIDES, BUT YOU CAN ABSOLUTELY ON THE CALCULATOR SUBSTITUTE X INTO THE LEFT SIDE, AND THEN SUBSTITUTE X INTO THE RIGHT SIDE AND SEE IF THEY MATCH.

SO LET'S SEE, -2 TIMES 1 MINUS 4, X IN THIS CASE IS 2 THAT I'M TESTING, SO THAT LEFT SIDE WHEN I PLUG IT INTO WAS 14, I'D BETTER HAVE THE RIGHT SIDE BE 14 TO SEE IF X IS THE RIGHT SOLUTION.

SO 3 TIME 2 PLUS 8, I'VE GOT 14 ON THE RIGHT, WHEN I PLUG IN X IS EQUAL TO 2, SO THAT'S AN EASY WAY TO SOLVE PROBLEMS LIKE THAT.

NOW CERTAINLY YOU COULD DO THE ALGEBRA OUT FOR ALL THESE PROBLEMS, THE CALCULATOR IS EITHER JUST A REALLY HELPFUL CHECK OR IF YOU'RE TRULY STUMPED YOU CAN USE THE CALCULATOR TO SOLVE PROBLEMS THAT LOOK LIKE THIS.

THANK YOU SO MUCH.

>> IN CONCLUSION, WE'D LIKE TO SAY THANK YOU FOR WATCHING OUR VIDEO.

WE WANT TO WISH YOU GOOD LUCK ON THE EXAM, IF YOU ARE WATCHING THIS VIDEO AND WORKING ALONG WITH IT, THEN YOU ARE DEFINITELY PREPARING YOURSELF AND YOU WILL DO WELL.

MAKE SURE THAT YOU GET A GOOD NIGHT'S SLEEP THE NIGHT BEFORE, EAT A GOOD BREAKFAST, HAVE YOUR CALCULATOR CHARGED AND READY TO GO, AND GOOD LUCK ON YOUR TEST!

>> GOOD LUCK, THANK YOU ALL SO MUCH.

>> THANK YOU!

[MUSIC] ♪ ♪ ♪ ♪ >> IF YOU'RE LOOKING FOR ADDITIONAL REGENTS EXAM INFO, THEN JUMP ONTO YOUR COMPUTER AND LOG ONTO REGENTSREVIEWNY DOT NET.

IT'S THE OFFICIAL WEBSITE FOR THE SERIES AND IT'S LOADED WITH TEST PREP RESOURCES THAT ARE GUARANTEED TO CURE THOSE REGENTS EXAM BLUES.

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AND DON'T FORGET; ALL OF THE PROGRAMS IN THE SERIES WILL BE AVAILABLE FOR STREAMING ON THE SITE ONCE THE TELEVISION BROADCAST SCHEDULES ARE COMPLETE.

SO WHAT ARE YOU WAITING FOR?

INCREASE YOUR CHANCES FOR A BETTER SCORE ON A TEST DAY.

REGENTS REVIEW 2.0: THE NEXT GENERATION OF REGENTS EXAM PREP.

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