- 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.