- IN THIS EXAMPLE, WE'RE GIVEN
A QUADRATIC FUNCTION
IN GENERAL FORM
AS WE SEE HERE.
WE WANT TO WRITE THE EQUATION
IN STANDARD FORM,
WHICH WOULD BE THIS FORM HERE.
REMEMBER IF WE HAVE
A QUADRATIC FUNCTION
IN STANDARD FORM
IF "A" IS GREATER THAN ZERO
THE PARABOLA OPENS UP.
IF "A" IS LESS THAN ZERO
THE PARABOLA OPENS DOWN.
THE VERTEX HAS COORDINATES
H, K,
AND THE EQUATION OF THE AXIS
OF SYMMETRY IS X = H.
SO TO PUT THIS EQUATION
IN STANDARD FORM
NOTICE HOW WE HAVE
THIS QUANTITY X - H/SQUARED
WHICH WOULD BE
A PERFECT SQUARED TRINOMIAL.
SO WE'RE GOING TO START
BY ISOLATING THE X/SQUARED
AND X TERM
AND THEN EQUATE
A PERFECT SQUARE TRINOMIAL.
SO WE'LL WRITE THIS
AS Y = X/SQUARED + 2X,
NOW BECAUSE WE'RE GOING
TO CREATE
A PERFECT SQUARE TRINOMIAL
WE KNOW WE'LL HAVE TO ADD
A CONSTANT HERE
SO I'LL PUT + SOME CONSTANT
AND THEN WE HAVE OUR - 4
AND THEN TO MAINTAIN EQUALITY
SINCE WE'RE ADDING
A CONSTANT HERE
WE'LL HAVE TO SUBTRACT
THAT SAME CONSTANT HERE.
WE'RE ADDING AND SUBTRACTING
HERE TO MAINTAIN EQUALITY
BECAUSE WE'RE ON THE SAME SIDE
OF THE EQUAL SIGN.
NOW TO MAKE THIS
A PERFECT SQUARE TRINOMIAL
WE'RE GOING TO TAKE 1/2 OF B
OR THE COEFFICIENT
OF THE X TERM AND SQUARE IT.
SO IF WE TAKE 1/2 OF B
AND SQUARE IT
WE WOULD HAVE 2 DIVIDED BY 2
THAT'S 1
AND 1/SQUARED IS EQUAL TO 1.
SO THAT MEANS WE'RE GOING
TO ADD 1 HERE
THAT WILL MAKE THIS
A PERFECT SQUARE TRINOMIAL
AND THEN TO UNDO THE + 1
WE'LL SUBTRACT 1 HERE.
NOW WE'RE GOING TO FACTOR
THIS TRINOMIAL
AND THEN COMBINE
THE TERMS HERE.
SO WE'LL HAVE Y EQUALS,
AGAIN THIS WILL FACTOR
IN TO 2 BINOMIAL FACTORS
AND THEN WE'LL HAVE - 5.
SO THE FACTORS OF X/SQUARED
GO HERE AND HERE
AND THE FACTORS OF +1 THAT ADD
TO +2 WOULD BE +1 AND +1.
SO WE HAVE + 1 HERE
AND + 1 HERE.
NOTICE HOW WE HAVE
A PERFECT SQUARE TRINOMIAL.
SO WE CAN WRITE THIS
AS Y = THE QUANTITY
X + 1/SQUARED - 5.
SO FROM HERE
WE SHOULD RECOGNIZE
THAT "A" IS GOING TO BE
EQUAL TO +1.
SO THE PARABOLA OPENS UP
SOMETHING LIKE THIS.
NEXT, THE VERTEX WILL HAVE
AN X COORDINATE OF -1
AND A Y COORDINATE OF -5.
REMEMBER WE HAVE TO BE CAREFUL
ABOUT THESE SIGNS.
IT HAS TO BE IN THE FORM
OF X - H
SO THEREFORE THE X COORDINATE
IS -1
AND THEN THE Y COORDINATE IS K
AND BECAUSE WE HAVE - 5
THE Y COORDINATE IS -5.
AND THEN BECAUSE THE VERTEX
HAS AN X COORDINATE OF -1
WE KNOW THE EQUATION
OF THE AXIS OF SYMMETRY
WOULD BE X =  -1.
NOW LET'S GO AHEAD
AND GRAPH THE PARABOLA.
TO DO THIS WE'LL GO AHEAD
AND SKETCH THE AXIS
OF SYMMETRY AT X = -1
WHICH WOULD BE
THIS VERTICAL LINE HERE.
THE VERTEX OF COURSE
IS ON THE AXIS OF SYMMETRY.
-1, -5 IS HERE.
NOW WE KNOW
THIS PARABOLA OPENS UP
BUT LET'S GO AHEAD AND FIND
SOME ADDITIONAL POINTS
TO MAKE A MORE ACCURATE
SKETCH.
REMEMBER IF WE WANTED
TO FIND THE Y INTERCEPT
WE WOULD HAVE TO SET X = 0.
IF WE DO THIS IN THE ORIGINAL
FORM OF THE EQUATION
NOTICE THAT THIS WOULD BE 0,
THIS WOULD BE 0,
AND Y WOULD BE EQUAL TO -4.
SO THE Y INTERCEPT IS THE
POINT WITH COORDINATE 0, -4,
WHICH WOULD BE THIS POINT HERE
AND THEN BECAUSE WE HAVE
THE AXIS OF SYMMETRY SKETCHED
WE KNOW THERE'S GOING
TO BE ANOTHER POINT ONE UNIT
TO THE LEFT OF THE AXIS.
SO THERE MUST ALSO BE A POINT
HERE
DUE TO THE SYMMETRY
ACROSS THE AXIS OF SYMMETRY.
NOW LET'S GO AHEAD AND FIND
TWO MORE ADDITIONAL POINTS.
LET'S EVALUATE THE FUNCTION
AT X = 2.
IF WE SUBSTITUTE 2
INTO THIS EQUATION HERE
WE'D HAVE 2 + 1 THAT'S 3,
3/SQUARED IS 9.
9 - 5 = 4.
SINCE F OF 2 IS EQUAL TO 4
WE KNOW THE GRAPH MUST
ALSO CONTAIN THE POINT 2, 4.
SO HERE'S THE POINT 2, 4.
NOTICE HOW IT'S 1, 2, 3,
AND THAT IT'S TO THE RIGHT
OF THE AXIS OF SYMMETRY
SO THERE MUST BE ANOTHER POINT
THREE UNITS TO THE LEFT
OF THE AXIS OF SYMMETRY
SO THERE MUST ALSO BE A POINT
HERE.
SO NOTICE AS LONG AS WE HAVE
THE AXIS OF SYMMETRY SKETCHED,
FOR EVERY POINT THAT WE FIND
ON THE PARABOLA
WE CAN ACTUALLY PLOT
TWO DO TO THE SYMMETRY
ACROSS THE AXIS OF SYMMETRY.
SO OUR PARABOLA WOULD LOOK
SOMETHING LIKE THIS.
AND OF COURSE WE CAN ALWAYS
VERIFY THIS
BY GRAPHING THIS WITH SOFTWARE
WHICH I'VE DONE HERE,
WHICH DOES MATCH THE GRAPH
WE DID BY HAND.
I HOPE YOU FOUND THIS HELPFUL.
