- WELCOME TO A QUADRATIC FUNCTION OVERVIEW.
IN THIS VIDEO WE'LL FIND ALL THE MAJOR COMPONENTS
OF A GIVEN QUADRATIC FUNCTION
AS WELL AS GRAPH THE QUADRATIC FUNCTION.
SO WE'RE GIVEN F OF X = -2X/SQUARED + 5X + 3.
LET'S START BY DETERMINING THE VALUES OF "A", B, AND C
WHERE "A" IS THE LEADING COEFFICIENT
OR THE COEFFICIENT OF X/SQUARED.
B IS THE COEFFICIENT OF X, AND C IS THE CONSTANT TERM.
SO NOTICE THAT "A" IS EQUAL TO -2,
B IS EQUAL TO +5 AND C IS EQUAL TO +3.
SO THE FIRST QUESTION
IS DOES THE GRAPH OF THE QUADRATIC FUNCTION
OR THE PARABOLA OPEN UP OR DOWN.
WE CAN DETERMINE THIS BY DETERMINING THE SIGN OF "A".
IF "A" IS NEGATIVE OR LESS THAN 0
LIKE WE HAVE THE PARABOLA OPENS DOWN.
IF "A" IS POSITIVE OR GREATER THAN 0
THE PARABOLA WOULD OPEN UP.
SO AGAIN WE KNOW "A" = -2 WHICH IS LESS THAN 0
AND THEREFORE THE PARABOLA WILL OPEN DOWN.
THIS ALSO MEANS THE VERTEX WILL BE A HIGH POINT.
NEXT WE'RE ASKED TO FIND THE EQUATION
OF THE AXIS OF SYMMETRY
WHICH IS GIVEN BY X = -B DIVIDED BY 2A.
SO WE'LL HAVE NEGATIVE AND THEN B IS 5 DIVIDED BY 2 x A
WHERE "A" IS -2 SO THIS WOULD BE -5 DIVIDED BY -4
WHICH SIMPLIFIES TO +5/4
OR IN DECIMAL FORM THIS WOULD BE EQUAL TO 1.25.
SO IF THE EQUATION OF THE AXIS OF SYMMETRY
WOULD BE X = 5/4 IN FRACTION FORM OR IF WE WANT X = 1.25.
NOW WE WANT TO FIND THE COORDINATES OF THE VERTEX.
THE VERTEX IS THE ONLY POINT THAT'S ON THE PARABOLA
AS WELL AS ON THE AXIS OF SYMMETRY.
AND THEREFORE, THE X COORDINATE OF THE VERTEX
HAS TO BE -B DIVIDED BY 2A
OR IN OUR CASE 5/4 OR 1.25.
AND NOW TO FIND THE Y COORDINATE OF THE VERTEX
WE'LL HAVE TO EVALUATE THE FUNCTION
AT THE GIVEN VALUE OF X.
TO DO THIS LET'S USE THE FRACTION FORM.
SO WE WANT TO FIND F OF 5/4 SO WE'LL SUBSTITUTE 5/4 FOR X
SO WE'LL HAVE -2
WHICH I'M GOING TO WRITE AS - 2/1 x 5/4/SQUARED
AND THEN + 5 OR 5/1 x 5/4 + 3, WHICH I'LL WRITE AS 3/1.
5/4/SQUARED WOULD BE 25/16 x - 2
THAT WOULD BE -50/16 + THIS WOULD BE 25/4 + 3/1.
NOW -50/16 DOES SIMPLIFY TO -25/8.
NOTICE OUR COMMON DENOMINATOR HERE WOULD BE 8
SO WE'RE GOING TO MULTIPLY THE SECOND FRACTION BY 2/2
AND THE THIRD FRACTION BY 8/8.
SO THIS WOULD GIVE US A COMMON DENOMINATOR OF 8
AND THE NUMERATOR WOULD BE -25 + 50 + 24.
AND THIS WOULD BE 49/8.
SO THIS WOULD BE THE Y COORDINATE IN FRACTION FORM
AND 49 DIVIDED BY 8 IS EQUAL TO 6.125 AS A DECIMAL.
AND OUR NEXT QUESTION IS WE WANT TO FIND
THE X AND Y INTERCEPTS OF THE FUNCTION.
SO TO FIND THE X INTERCEPTS WE'LL SET Y OR F OF X = 0
AND SOLVE FOR X
AND TO FIND THE Y INTERCEPT WE'LL SET X = 0
AND SOLVE FOR Y.
LET'S START BY FINDING THE X INTERCEPTS.
SO WE SET F OF X OR Y = 0
WE WANT TO SOLVE THE EQUATION 0 = -2X/SQUARED + 5X + 3.
LET'S SEE IF WE CAN FACTOR THIS
SO LET'S START BY FACTORING OUT A NEGATIVE
OR IF WE WANT -1
SO THE LEADING COEFFICIENT WOULD BE POSITIVE.
IF WE FACTOR OUT -1
IT'S GOING TO CHANGE THE SIGN OF EACH TERM.
SO WE'LL HAVE +2X/SQUARED - 5X - 3.
WHICH I THINK WILL FACTOR INTO TWO BINOMIALS
WHERE THE FIRST TERMS MUST COME FROM THE FACTORS
OF 2X/SQUARED WHICH WOULD BE 2X, X.
THE SECOND POSITIONS MUST COME FROM THE FACTORS OF -3
SO THAT THE SUM OR THE INNER PRODUCT AND OUTER PRODUCT
IS EQUAL TO -5X.
SO IF WE PUT A FACTOR OF -3 HERE AND A FACTOR OF +1 HERE
WE'D HAVE -6X + 1X WHICH DOES = -5X.
SO THIS PRODUCT EQUALS 0 WHEN 2X + 1 = 0 OR X - 3 = 0.
SO TO SOLVE FOR X HERE WE WOULD SUBTRACT 1
AND DIVIDE BY 2.
X= -1/2 OR ADDING 3, X = +3.
THESE ARE THE TWO X INTERCEPTS OF OUR FUNCTION.
ONE WOULD BE -1/2, 0 AND THE OTHER WOULD BE 3, 0.
AND NOW, LET'S GO AHEAD AND FIND THE Y INTERCEPT
BY SETTING X = 0 OR FINDING F OF 0.
SO IF WE SUBSTITUTE 0 FOR X WE WOULD JUST HAVE 0 + 0 + 3
SO F OF 0 = 3,
THE Y INTERCEPT IS 3
WHICH WOULD BE THE POINT WITH COORDINATES 0, 3.
NOW LET'S TAKE THE INFORMATION THAT WE JUST FOUND
AND GRAPH THE PARABOLA.
LET'S START BY SKETCHING THE AXIS OF SYMMETRY
WHICH IS A VERTICAL LINE X = 5/4 OR 1.25
WHICH WOULD BE HERE.
THE VERTEX IS ON THIS LINE WITH A Y COORDINATE OF 6.125,
WHICH WOULD BE HERE JUST ABOVE 6.
NEXT, THE X INTERCEPTS ARE -1/2, 0 WHICH WOULD BE HERE
AND 3, 0 WHICH WOULD BE HERE.
NOTICE THESE TWO POINTS ARE SYMMETRICAL ACROSS THE AXIS
AND THEN FINALLY THE Y INTERCEPT IS THE POINT 0, 3,
WHICH WOULD BE THIS POINT HERE
BUT THERE MUST BE A SYMMETRICAL POINT
ON THE OTHER SIDE OF THE AXIS OF SYMMETRY RIGHT HERE.
AND NOW WE CAN MAKE A NICE GRAPH OF THE PARABOLA
THROUGH THESE 5 POINTS.
NOW THAT WE HAVE THE GRAPH OF THE PARABOLA
LET'S FIND THE DOMAIN AND RANGE
AS WELL AS DETERMINE THE INTERVALS
FOR WHICH THE FUNCTION IS INCREASING AND DECREASING
BUT LET'S GO AND DO THIS ON THE NEXT SLIDE.
NOTICE HOW I HAVE A NICE COMPUTERIZED GRAPH
OF THE PARABOLA HERE.
REMEMBER THE DOMAIN REPRESENTS ALL THE POSSIBLE X VALUES.
GRAPHICALLY THE AXIS MOVES FROM LEFT TO RIGHT
SO WE'RE ALSO ASKING
HOW FAR LEFT AND RIGHT DOES THIS GRAPH GO.
NOTICE HOW IT DOESN'T HAVE ANY BREAKS
AND WHILE IT IS MOVING DOWN VERY QUICKLY
IT'S ALSO MOVING LEFT AND RIGHT FOREVER
AND THEREFORE THE DOMAIN WOULD BE ALL REAL NUMBERS
WHICH USING INTERVAL NOTATION
THE INTERVAL FROM NEGATIVE INFINITY TO POSITIVE INFINITY.
THIS WILL ALWAYS BE THE DOMAIN FOR EVERY QUADRATIC FUNCTION.
AND THEN, THE RANGE WILL BE THE SET
OF ALL POSSIBLE Y VALUES.
WELL NOTICE THAT THE Y AXIS MOVES UP AND DOWN
SO WE'RE REALLY TRYING TO DETERMINE
HOW THIS GRAPH BEHAVES MOVING UP AND DOWN.
WELL NOTICE HOW IT DOES GO DOWN FOREVER
BUT IT DOES NOT GO UP FOREVER.
THE HIGHEST POINT WOULD BE HERE
WHICH WOULD BE THE Y COORDINATE OF THE VERTEX
WHICH WE KNOW IS 49/8 OR 6.125.
WHICH MEANS THE RANGE WOULD BE DOWN FOREVER,
MEANING FROM NEGATIVE INFINITY UP TO A HIGH VALUE OF 6.125
OR 49/8.
IT INCLUDES THIS Y VALUE SO IT'S CLOSED ON 49/8.
SO WE USE A SQUARE BRACKET USING INTERVAL NOTATION.
OF COURSE WE COULD ALSO SAY Y IS LESS THAN OR EQUAL TO 49/8
TO REPRESENT THE SAME INTERVAL.
IN OUR LAST QUESTION
WE'RE ASKED TO DETERMINE THE INTERVALS
FOR WHICH THE FUNCTION IS INCREASING AND DECREASING.
SO AGAIN LOOKING AT THE GRAPH
AS THEY MOVE FROM LEFT TO RIGHT ALONG THE GRAPH
IF WE'RE MOVING UPHILL THE FUNCTION IS INCREASING
SO IT'S INCREASING ON THIS INTERVAL
ALL THE WAY TO THE VERTEX
AND THEN NOTICE TO THE RIGHT OF THE VERTEX
THE FUNCTION IS GOING DOWNHILL AND THEREFORE IT'S DECREASING.
WHEN WE USE X VALUES TO DESCRIBE THE INTERVALS
AND BECAUSE AGAIN THE VERTEX HERE
HAS AN X COORDINATE OF 5/4 AND A Y COORDINATE OF 49/8.
TO THE LEFT OF THE VERTEX WOULD BE THE INTERVAL
FROM NEGATIVE INFINITY TO 5/4.
IT'S OPEN AT 5/4 OR THE VERTEX,
IT'S NOT INCREASING OR DECREASING
THAT WOULD BE A HIGH POINT OR A MAXIMUM.
THEN THE FUNCTION IS DECREASING ON THE INTERVAL
FROM 5/4 TO INFINITY
MEANING TO THE RIGHT OF THE VERTEX.
SO AGAIN, THE FUNCTION IS INCREASING
TO THE LEFT OF THE VERTEX
BECAUSE THE GRAPH IS MOVING UPHILL
AND THEN IT'S DECREASING TO THE RIGHT OF THE VERTEX
BECAUSE THE GRAPH IS GOING DOWNHILL.
OKAY. THAT'S GOING TO DO IT FOR THIS OVERVIEW.
I HOPE YOU FOUND THIS HELPFUL.