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- IN THIS EXAMPLE WE'RE GIVEN A QUADRATIC FUNCTION
IN GENERAL FORM
WE WANT TO WRITE THE EQUATION IN STANDARD FORM.
THE BENEFIT OF STANDARD FORM OR THE FORM WE SEE HERE
IS THAT WE CAN EASILY IDENTIFY THE VERTEX
AND THE EQUATION OF THE AXIS.
THE VERTEX HAS COORDINATES H, K
WHERE H IS HERE AND K IS HERE
AND THE EQUATION OF THE AXIS OF SYMMETRY IS X = H.
THAT'S ALSO TRUE IF "A" IS POSITIVE THE PARABOLA OPENS UP
AND IF "A" IS NEGATIVE THE PARABOLA OPENS DOWN.
SO THE MAIN THING TO RECOGNIZE HERE
IS THAT IF WE WANT TO PUT THIS EQUATION IN THIS FORM
WE HAVE THIS QUANTITY HERE X - H/SQUARED
SO WE'LL HAVE TO CREATE A PERFECT SQUARE TRINOMIAL.
SO TO START THIS PROCESS WE'RE GOING TO ISOLATE
THE X/SQUARED TERM AND THE X TERM.
SO WE'RE GOING TO HAVE Y = 3X/SQUARED - 12X,
NOW BECAUSE WE WANT TO CREATE A PERFECT SQUARE TRINOMIAL
WE'RE GOING TO HAVE TO ADD A CONSTANT HERE.
SO WE'LL LEAVE SPACE FOR THAT AND THEN WE'LL HAVE + 11
AND THEN WHATEVER CONSTANT WE ADD HERE
WE'LL HAVE TO SUBTRACT HERE TO MAINTAIN EQUALITY.
BUT BEFORE WE CREATE A PERFECT SQUARE TRINOMIAL
WE DO WANT TO HAVE A LEADING COEFFICIENT OF 1 HERE
SO WHAT WE'RE GOING TO DO IS FACTOR A 3
BUT WE'LL FACTOR A 3 ONLY OUT OF THE FIRST TWO TERMS.
SO WE'LL HAVE Y = 3 x THE QUANTITY X/SQUARED - 4X.
WE'LL STILL HAVE + SOME CONSTANT
AND THEN WE'LL HAVE + 11 - A CONSTANT.
NOW REMEMBER TO MAKE THIS A PERFECT SQUARE TRINOMIAL
WE'LL TAKE 1/2 OF B AND SQUARE IT
WHERE B IS THE COEFFICIENT OF THE X TERM.
AGAIN, WE TAKE 1/2 OF B OR DIVIDE B BY 2
AND THEN SQUARE IT
TO MAKE THIS A PERFECT SQUARE TRINOMIAL.
SO IN THIS CASE, SINCE B IS -4,
-4 DIVIDED BY 2 IS -2/SQUARED WHICH WOULD BE +4.
SO WE'RE GOING TO ADD 4 HERE
AND HERE'S WHERE WE HAVE TO BE CAREFUL.
IF WE ADD 4 HERE BECAUSE OF THIS 3 OUT HERE
WE'RE ACTUALLY ADDING 12.
REMEMBER IF WE DISTRIBUTE THIS 3, 3 x 4 IS EQUAL TO 12.
SO UNDO ADDING 12 WE ACTUALLY HAVE TO SUBTRACT 12 HERE.
SO IF WE HAVE A CONSTANT OUT HERE WE NEED TO MAKE SURE
THAT WE DISTRIBUTE SO THAT WE KNOW WHAT TO SUBTRACT.
OKAY, SO NOW WE'LL FACTOR THIS TRINOMIAL
AND THEN COMBINE OUR TERMS HERE.
SO WE'LL HAVE Y = 3, WE'LL GO AHEAD AND FACTOR THIS
INTO TWO BINOMIALS TO BEGIN WITH
AND THEN + 11 - 12 IS GOING TO BE - 1.
SO THE FACTORS OF X/SQUARED WOULD BE X AND X.
THE FACTORS OF +4 THAT ADD TO -4 WOULD BE -2, AND -2.
SO WE HAVE - 2 HERE AND - 2 HERE.
NOTICE HOW WE DO HAVE A PERFECT SQUARE TRINOMIAL HERE.
SO WE FINALLY HAVE THE STANDARD FORM OF THE EQUATION.
IT WOULD BE 3 x THE QUANTITY X - 2/SQUARED - 1.
AND NOW, IN THIS FORM WE CAN RECOGNIZE FIRST THAT
SINCE "A" IS EQUAL TO 3 THE PARABOLA OPENS UP.
WE CAN ALSO IDENTIFY THE COORDINATES OF THE VERTEX
BUT WE DO NEED TO BE CAREFUL
ABOUT THE SIGNS OF THE COORDINATES.
HERE WE HAVE THE QUANTITY X - 2/SQUARED.
SO THE X COORDINATE IS +2.
AND THEN WE HAVE - 1 HERE ON THE END
SO THE Y COORDINATE IS -1.
AND IF THIS IS OUR VERTEX
WE KNOW THE EQUATION OF THE AXIS MUST BE X = 2.
SO LET'S GO AHEAD AND SKETCH THE EQUATION OF THE AXIS
WHICH IS THE VERTICAL LINE X = 2 OR THIS VERTICAL LINE HERE.
WE KNOW THE VERTEX IS ON THE AXIS
AND IT HAS COORDINATES 2, -1
WHICH WOULD BE THIS POINT HERE.
NOW WE KNOW THE PARABOLA OPENS UP
BUT WE SHOULD FIND AT LEAST TWO MORE POINTS
TO MAKE A MORE ACCURATE SKETCH.
SO LET'S EVALUATE THE FUNCTION OF LET'S SAY X = 3.
F OF 3 WOULD BE EQUAL TO 3 x THE QUANTITY 3 - 2/SQUARED.
WELL 3 - 2 IS 1 SO WE HAVE 1/SQUARED - 1.
BUT 1/SQUARED IS 1 x 3 IS 3 - 1 IS EQUAL TO 2.
SO IF F OF 3 IS EQUAL TO 2
WE KNOW THAT THE PARABOLA MUST CONTAIN THE POINT
WITH AN X COORDINATE OF 3 AND A Y COORDINATE OF 2
WHICH WOULD BE THIS POINT HERE.
NOTICE HOW THIS POINT IS ONE UNIT
TO THE RIGHT OF THE AXIS OF SYMMETRY
SO THERE MUST ALSO BE ON POINT ONE UNIT
TO THE LEFT OF THE AXIS OF SYMMETRY,
WHICH WOULD BE THIS POINT HERE.
AND THEN IF WE DID WANT TO FIND THE Y INTERCEPT
WE WOULD HAVE TO SET X = 0 AND SOLVE FOR Y.
LOOKING AT THE VISIONAL FORM OF THE EQUATION
IF X WAS 0 THIS WOULD BE 0, THIS WOULD BE 0,
AND Y WOULD BE = +11.
SO THE Y INTERCEPT IS THE POINT 0, 11,
BUT OF COURSE THAT WOULD BE OFF OUR SCREEN.
SO WE CAN SEE THIS PARABOLA IS VERY NARROW
OR IT'S BEEN STRETCHED VERTICALLY BY A FACTOR OF 3.
SO THE GRAPH OF THE PARABOLA
IS GOING TO LOOK SOMETHING LIKE THIS.
AND OF COURSE WE CAN VERIFY THIS
BY GRAPHING THIS ON SOFTWARE WHICH I'VE DONE HERE
AND THIS GRAPH DOES MATCH THE GRAPH THAT WE DID BY HAND.
OKAY. I HOPE YOU FOUND THIS HELPFUL.