- WELCOME TO A LESSON ON LINEAR FUNCTIONS.
THE GOALS OF THIS VIDEO ARE TO DEFINE A LINEAR FUNCTION,
DETERMINE IF A LINEAR FUNCTION IS INCREASING OR DECREASING,
INTERPRET LINEAR FUNCTIONS MODELS,
AND FINALLY DETERMINE LINEAR FUNCTIONS.
LINEAR FUNCTIONS ARE FUNCTIONS
THAT FIT THE FORM F OF X = MX + B
OR F OF X = B + MX.
THESE TWO FORMS ARE EQUIVALENT, BUT NOTICE HOW THIS FORM HERE
SHOULD REMIND US OF SLOPE INTERCEPT FORM
OF A LINEAR EQUATION WITH A FORM Y = MX + B.
IN EITHER CASE B, THE CONSTANT TERM
OF THE LINEAR FUNCTION, IS THE Y-INTERCEPT
AND IT'S ALSO THE INITIAL OR STARTING FUNCTION VALUE
WHEN X = 0.
AND M, THE COEFFICIENT OF X IS THE SLOPE OF THE LINE
AND IT'S ALSO THE CONSTANT RATE OF CHANGE OF THE FUNCTION.
IF M, THE SLOPE OF THE LINEAR FUNCTION,
IS POSITIVE OR >0 THEN WE SAY THE FUNCTION IS INCREASING.
SO FOR EXAMPLE, LOOKING AT THE FUNCTION
F OF X = 1 + 2X NOTICE THE SLOPE IS +2,
WHICH MEANS THE FUNCTION IS INCREASING.
SO IF WE LOOK AT THE GRAPH OF THE LINE, NOTICE AS WE MOVE
FROM LEFT TO RIGHT THE LINE IS GOING UP AND THEREFORE
THE FUNCTION IS INCREASING.
WE CAN ALSO SAY AS X INCREASES SO DOES Y.
AND THEN IF THE SLOPE IS NEGATIVE OR <0
THEN WE SAY THE LINEAR FUNCTION IS DECREASING.
SO FOR EXAMPLE, IF WE LOOK AT THE FUNCTION
F OF X = 5 - X OR IF WE WANT 5 - 1X
NOTICE THE SLOPE IS -1 AND THEREFORE THE FUNCTION
IS DECREASING.
AS WE MOVE FROM LEFT TO RIGHT, NOTICE HOW THE GRAPH GOES DOWN.
WE CAN ALSO SAY AS X INCREASES; Y DECREASES.
NOW LET'S INTERPRET A LINEAR FUNCTION.
THE COST OF A CAB RIDE IS GIVEN BY THE FUNCTION
C OF X = 4 + 2.5X WHERE C OF X IS THE TOTAL COST
AND X IS THE NUMBER OF MILES.
USING THE FORM F OF X = B + MX; INTERPRET THE MEANING OF B,
THE Y-INTERCEPT, AND M THE SLOPE.
SO NOTICE THAT B, THE CONSTANT TERM,
IS +4 AND M, THE COEFFICIENT OF X, IS 2.5.
WELL, REMEMBER B IS THE FUNCTION VALUE
WHEN X = 0, WHICH WE CAN SAY IS THE STARTING
OR INITIAL AMOUNT WHICH MEANS AS SOON
AS THEY GET INTO THE CAB RIDE, EVEN WHEN WE'VE DRIVEN 0 MILES,
THE COST IS STILL $4.00.
SO WE CAN SAY THAT THE STARTING COST
OF THE CAB RIDE IS $4.00.
THIS IS OFTEN CALLED THE FLAG FEE.
NEXT, TO INTERPRET THE SLOPE LETS WRITE
THIS AS A RATIO MEANING 2.5/1.
REMEMBER THE SLOPE IS THE CHANGE IN Y
OR THE CHANGE IN THE FUNCTION VALUE
OVER THE CHANGE IN X.
WELL THE FUNCTION VALUE IS THE COST.
SO 2.5 REPRESENTS THE CHANGE IN COST
WHICH WOULD BE 2.5 DOLLARS OR $2.50 PER THE CHANGE IN X
WOULD BE THE CHANGE IN MILES SO THIS WOULD BE
$2.50 PER 1 MILE.
SO THE SLOPE OF THE LINEAR FUNCTION
TELLS US THE COST PER MILE.
SO THE LINEAR FUNCTION TELLS US THERE'S A STARTING COST
OF $4.00 TO GET INTO THE CAB AND THEN ONCE WE START DRIVING
THE COST IS AN EXTRA $2.50 PER MILE.
FOR PART B WE WANT TO FIND THE COST FOR A 5 MILE CAB RIDE;
WHICH MEANS WE WANT TO FIND C OF 5.
FOR C OF 5 WE'LL SUBSTITUTE 5 FOR X SO WE'LL HAVE
4 + 2.5 x 5, WHICH WOULD BE 4 + 2.5 x 5
IS = TO 12.5.
SO C OF 5 = 16.5 AND THEREFORE THE COST OF A 5 MILE CAB RIDE
IS $16.50.
NOW LET'S TAKE A LOOK AT ANOTHER EXAMPLE.
HERE WE'RE TOLD F OF X IS LINEAR AND F OF 2 = -3 AND F OF 4 = 7.
WE WANT TO DETERMINE THE LINEAR FUNCTION.
SO TO DO THIS WE'LL HAVE TO DETERMINE
THE SLOPE OF THE LINEAR FUNCTION AS WELL AS THE Y-INTERCEPT.
LET'S START BY DETERMINING THE SLOPE.
NOW, THERE ARE A COUPLE WAYS OF DOING THIS.
WE COULD USE THE SLOPE FORMULA M = Y SUB 2 - 1 SUB 1 DIVIDED
BY X SUB 2 - X SUB 1, BUT REMEMBER Y VALUES
ARE FUNCTIONS VALUES SO WE CAN WRITE THIS FORMULA
USING FUNCTION NOTATION AS F OF X SUB 2 - F OF X
SUB 1 DIVIDED BY X SUB 2 - X SUB 1.
SO AGAIN IF F OF 2 = -3 THAT MEANS THE LINEAR FUNCTION
CONTAINS THE POINT 2, -3 AND IF F OF 4 = 7
THE LINEAR FUNCTION CONTAINS THE POINT 4, 7.
LET'S GO AHEAD AND FIND THE SLOPE USING
OUR FUNCTION NOTATION.
SO THE SLOPE M WILL BE = TO THE CHANGE
IN THE FUNCTION VALUE DIVIDED BY THE CHANGE OF X.
SO THE CHANGE IN THE FUNCTION VALUES
WOULD BE F OF 4 - F OF 2 DIVIDED BY THE CHANGE IN X
WHICH WOULD BE 4 - 2.
WELL, F OF 4 IS 7 - F OF 2 IS -3 DIVIDED BY 4 - 2,
THAT'S 2.
WELL, 7 - -3 IS THE SAME AS 7 + 3 OR 10,
10 DIVIDED BY 2 = 5.
SO NOW WE KNOW THAT OUR LINEAR FUNCTION
MUST BE IN THE FORM OF F OF X = 5X + B.
AND NOW WE CAN USE ONE OF THESE LINEAR FUNCTIONS
TO DETERMINE B OR THE Y-INTERCEPT.
SO FOR EXAMPLE, IF WE KNOW F OF 2 = -3;
TO FIND F OF 2 WE WOULD SUBSTITUTE 2 FOR X.
SO WE'D HAVE 5 x 2 + B, BUT THIS MUST = -3.
SO HERE WE'LL HAVE 10 + B = -3.
SOLVE FOR B, WE WOULD SUBTRACT 10 ON BOTH SIDES B = -13.
SO NOW WE KNOW THE VALUED OF B, WE HAVE OUR LINEAR FUNCTION
F OF X = 5X.
SINCE B IS NEGATIVE WE COULD WRITE + -13 OR JUST - 13.
NOW LETS SEE IF WE CAN FIND LINEAR FUNCTION
GIVEN AN APPLICATION PROBLEM.
YOU HAVE $350 IN THE BANK AND CAN SAVE $90 PER MONTH.
WRITE A LINEAR FUNCTION THAT WOULD MODEL
YOUR TOTAL SAVINGS.
LET'S FIRST DEFINE OUR VARIABLES.
LETS LET X = THE NUMBER OF MONTHS AND LETS LET S OF X,
OUR FUNCTION, BE = TO THE TOTAL SAVINGS.
SO OUR FUNCTION, S OF X, MUST = THE INITIAL
OR STARTING AMOUNT; WHICH WOULD BE
$350 + THE CONSTANT RATE OF CHANGE x X.
AND YOU'RE SAVING $90 PER MONTH SO IT WOULD BE + 90 x X.
NOTICE, EVERY TIME X INCREASES BY 1
WE'D ADD ANOTHER $90 TO THE TOTAL SAVINGS.
NOW FOR THE NEXT QUESTION: HOW MUCH WILL YOU HAVE SAVED
IN 6 MONTHS?
SO NOW WE WANT TO FIND S OF 6.
SO WE'LL HAVE THE STARTING AMOUNT, $350 + $90 x 6,
THIS AMOUNT HERE IS THE AMOUNT SAVED OVER THE 6 MONTHS.
SO THIS WOULD BE $350 + THIS WOULD BE $540.
SO THIS TOTAL WOULD BE $890.
SO AFTER 6 MONTHS YOU HAVE SAVED $890.
AND NOW FOR THE LAST QUESTION: HOW LONG WILL IT TAKE
TO SAVE AT LEAST $1,000?
SO REMEMBER THE TOTAL SAVINGS IS S OF X.
SO NOW WE'LL SUBSTITUTE 1,000 FOR S OF X AND SOLVE FOR X.
LETS GO AHEAD AND DO THIS ON THE NEXT SLIDE.
SO STARTING WITH OUR FUNCTION
WE'LL NOW SUBSTITUTE 1,000 FOR S OF X AND SOLVE FOR X.
NOW I KNOW THIS DOES SAY AT LEAST $1,000 SO WE COULD
PUT AN EQUALITY SYMBOL IN HERE, BUT LETS GO AHEAD AND LEAVE IT
AS AN EQUATION AND THEN WE'LL ADJUST OUR ANSWER
ONCE WE FINISH.
SO TO SOLVE FOR X WE WOULD SUBTRACT 350
ON BOTH SIDES THAT WOULD GIVE US 650 = 90X AND WE'D DIVIDE
BOTH SIDES BY 90.
SO WE HAVE X = THESE BOTH HAVE A COMMON FACTOR OF 10,
THIS WOULD BE 65 DIVIDED BY 9.
IF WE CONVERT THIS TO A DECIMAL IT'S APPROXIMATELY 7.22,
BUT NOW REMEMBER WE'RE ASSUMING WE'RE MAKING A $90 DEPOSIT
EACH MONTH.
SO NOTICE HOW THIS IS TELLING US IT TAKES MORE THAN 7 MONTHS
TO SAVE $1,000 WHICH MEANS WE HAVE TO ROUND THIS UP
TO THE NEXT MONTH BECAUSE WE'RE MAKING A DEPOSIT
OF $90 EACH MONTH.
SO IT'LL ACTUALLY TAKE 8 MONTHS TO SAVE AT LEAST $1,000.
AFTER 8 MONTHS WE'LL HAVE MORE THAN $1,000,
BUT AFTER 7 MONTHS WE'LL HAVE LESS THAN $1,000.
OKAY, THAT'S GOING TO DO IT FOR THIS LESSON.
I HOPE YOU FOUND THIS HELPFUL.