- IN THIS EXAMPLE, WE WANT TO DETERMINE
THE EQUATION OF THE GRAPH USING THE SINE FUNCTION.
BUT BEFORE WE DO THAT, LET'S REVIEW
SOME OF THE COMPONENTS OF THE BASIC SINE FUNCTION.
SO WE'LL FOCUS ON THE GRAPH FROM HERE TO HERE,
WHICH IS ONE PERIOD OF THE BASIC SINE FUNCTION.
THE PERIOD IS FROM 0 TO 2PI.
THE FUNCTION, IT STARTS AT THE MIDLINE,
GOES UP TO A MAXIMUM, DOWN TO THE MIDLINE,
DOWN TO A MINIMUM, AND BACK UP TO THE MIDLINE.
THE MAXIMUM IS +1 AND THE MINIMUM IS -1.
WE USE THESE FEATURES TO HELP US DETERMINE THE TRANSFORMATION
THAT OCCURS ON THE GRAPH THAT'S GIVEN.
SO THE FIRST THING WE SHOULD DO IS TAKE A LOOK
AT THE GRAPH AND HIGHLIGHT ONE PERIOD OF THE GRAPH
SO WE CAN FOCUS ON THAT TO COME UP WITH THE EQUATION.
AND DEPENDING ON WHICH PIECE WE HIGHLIGHT,
IT WILL AFFECT THE EQUATION THAT WE COME UP WITH.
LET'S GO AHEAD AND FOCUS ON THIS GRAPH
STARTING AT X = 0.
AND NOTICE HOW IT STARTS AT -1, DECREASES, INCREASES
TO A MAXIMUM AND THEN DECREASES BACK TO THE MIDLINE.
SO WE'RE GOING TO FOCUS ON THIS GRAPH FROM 0 TO 6.
SO, RIGHT AWAY IF ON THE INTERVAL FROM 0 TO 6,
THIS FUNCTION COMPLETES ONE CYCLE
AND THEN STARTS TO REPEAT ITSELF,
THAT TELLS US THAT THE PERIOD IS GOING TO BE EQUAL TO 6 UNITS.
REMEMBER, B IS THE VALUE THAT AFFECTS THE PERIOD.
AND THE FORMULA FOR THE PERIOD IS 2PI DIVIDED BY B,
SO 2PI DIVIDED BY B MUST EQUAL 6 IN THIS CASE.
SO IF WE PUT THIS OVER ONE AND CROSS MULTIPLY,
WE'RE GOING TO HAVE 6B IS EQUAL TO 2PI.
SO WE DIVIDE BOTH SIDES BY 6.
WE HAVE B = PI OVER 3.
NOW, LET'S GO BACK TO OUR GRAPH AND DETERMINE THE MIDLINE.
THE MIDLINE IS GOING TO BE AT Y = -1.
YOU COULD THINK OF IT AS THE HORIZONTAL LINE
THAT KIND OF SPLITS THE WAVE IN HALF.
NOW, BECAUSE THE MIDLINE IS AT Y = -1,
THAT TELLS US THAT OUR GRAPH HAS BEEN SHIFTED DOWN ONE UNIT.
AND THE VERTICAL SHIFT IS CONTROLLED BY THE VALUE OF C.
SO, BECAUSE IT SHIFTED DOWN ONE UNIT, C MUST BE EQUAL TO -1.
SO, SO FAR WE KNOW THAT B = PI OVER 3 AND C = -1.
NOW, LET'S TALK ABOUT THE AMPLITUDE.
THE AMPLITUDE IS AFFECTED BY THE VALUE OF "A."
AND THE AMPLITUDE IS A DISTANCE FROM THE MIDLINE TO A MAXIMUM
OR THE DISTANCE FROM THE MIDLINE TO A MINIMUM.
SO NOTICE THAT FROM -1 TO 2, THAT WOULD BE THREE UNITS,
THAT DISTANCE HERE,
AS WELL AS THE DISTANCE FROM THE MIDLINE TO THE MINIMUM.
SO OUR AMPLITUDE IS EQUAL TO 3, BUT THERE'S ONE MORE THING
TO NOTICE ABOUT THIS GRAPH.
NOTICE HOW THE BASIC SINE FUNCTION
STARTS AT THE MIDLINE, THEN GOES UP, DOWN, DOWN, UP,
WHERE THIS FUNCTION STARTS AT THE MIDLINE
AND GOES DOWN, UP, UP, DOWN.
SO THIS HAS BEEN REFLECTED ACROSS THAT MIDLINE,
WHICH TELLS US THAT EVEN THOUGH THE AMPLITUDE IS 3,
"A" IS GOING TO BE EQUAL TO -3.
REMEMBER, IF "A" IS LESS THAN 0, IT'S A REFLECTION
ACROSS THE MIDLINE.
THE LAST THING TO THINK ABOUT IS D, WHICH AFFECTS
THE PHASE SHIFT OR A SHIFT LEFT OR RIGHT.
BUT SINCE WE'RE USING THE SINE FUNCTION,
AND AT 0 IT'S STARTING AT THE MIDLINE,
WHICH IS THE SAME FOR THE BASIC SINE FUNCTION,
THIS GRAPH DOES NOT HAVE A PHASE SHIFT.
SO WE HAVE ALL THE INFORMATION WE NEED TO WRITE THE EQUATION
OF THIS GRAPH.
WE JUST NEED TO FILL IN THE VALUES FOR A, B, AND C.
SO WE'RE GOING TO HAVE Y = -3 x SINE OF THE QUANTITY.
REMEMBER, B WAS PI OVER 3, SO WE'LL HAVE PI OVER 3 x X,
AND THEN PLUS -1 OR JUST MINUS 1.
THIS WILL BE THE EQUATION OF THE GIVEN GRAPH.
AND, AGAIN, REMEMBER, IF WE WOULD HAVE FOCUSED
ON A DIFFERENT PIECE OF THE GRAPH,
IT WOULD AFFECT OUR EQUATION, BUT THE RESULT
WOULD BE THE SAME GRAPH.
AND I'LL LEAVE YOU WITH A SUMMARY OF THE NOTES
THAT WE USE TO DETERMINE THE VALUES OF A, B, C, AND D.