- THIS VIDEO WILL REVIEW THE QUADRATIC FUNCTION.
MORE SPECIFICALLY IT WILL REVIEW
HOW TO GRAPH A QUADRATIC FUNCTION IN GENERAL FORM.
LET'S TAKE A LOOK AT SOME OF THE PROPERTIES
OF A QUADRATIC FUNCTION IN GENERAL FORM.
THE GRAPH OF A QUADRATIC FUNCTION,
F OF X = "A"X SQUARED + BX + C IS CALLED A PARABOLA.
IT IS ALWAYS A CUP SHAPED OR U-SHAPED CURVE.
THE GRAPH OF A QUADRATIC FUNCTION,
F OF X = "A"X SQUARED + BX + C IS CALLED A PARABOLA.
IT IS ALWAYS A CUP SHAPED OR A U-SHAPED CURVE.
IT OPENS UPWARD IF "A" IS GREATER THAN ZERO
OR IT OPENS DOWNWARD IF "A" IS LESS THAN ZERO.
THE VERTICAL LINE X = -B/2"A" IS THE LINE OF SYMMETRY.
IT HAS A TURNING POINT OR A VERTEX AT A POINT
WHERE THE X COORDINATE IS -B/2"A"
AND THE Y COORDINATE IS F OF -B/2"A".
LET'S TAKE A LOOK AT THE GRAPH OF THESE IMPORTANT FEATURES.
FIRST OFF, THE PARABOLA IS IN BLUE.
NOTICE HOW IT OPENS UPWARD,
THEREFORE, WE CAN CONCLUDE IN THIS CASE
THAT "A" WOULD BE GREATER THAN ZERO.
THE LINE OF SYMMETRY IS IN RED.
NOTICE HOW--THE REASON IT'S CALLED THE LINE OF SYMMETRY
IS IF YOU WERE TO FOLD THE PARABOLA ACROSS THAT LINE,
IT WOULD MATCH UP PERFECTLY WITH THE OTHER HALF.
THE VERTEX IS THE ONLY POINT THAT IS ON THE GRAPH,
AS WELL AS ON THE LINE OF SYMMETRY.
IT WILL EITHER BE A HIGH POINT OR A LOW POINT ON THE GRAPH,
AND IN THIS CASE SINCE IT OPENS UPWARD,
THE VERTEX IS A LOW POINT.
TWO OTHER IMPORTANT COMPONENTS,
THE X INTERCEPTS AND THE Y INTERCEPTS,
AS WE KNOW FROM OUR PREVIOUS STUDIES,
X INTERCEPTS ARE WHERE THE GRAPHS CROSS THE X AXIS.
SOME PARABOLAS MAY NOT HAVE X INTERCEPTS,
AND OF COURSE, THE Y INTERCEPT
IS WHERE THE GRAPH CROSSES THE Y AXIS.
LET'S EXPLORE THE GRAPH OF A QUADRATIC IN GENERAL FORM.
I'M GOING TO GO TO THIS WEBSITE.
NOW THIS EQUATION IS IN GENERAL FORM
AND WHAT THESE SLIDERS ALLOW YOU TO DO
IS ADJUST THE VALUES OF "A", B AND C.
WE'RE GOING TO LOOK AT WHAT HAPPENS
WHEN WE CHANGE THE VALUE OF "A".
NOTICE HOW AS "A" INCREASES,
THE GREEN PARABOLA GETS NARROWER AND NARROWER,
BUT IT DOES STILL CONTINUE TO OPEN UPWARD.
AND AS SOON AS I SLIDE THIS
SO THAT THE VALUE OF "A" IS LESS THAN ONE,
IT BECOMES WIDER
AND THEN WHEN IT REACHES A NEGATIVE VALUE,
IT DOES OPEN DOWNWARD.
SO ALL OF THE VALUES OF "A" THAT ARE NEGATIVE,
THE PARABOLA WILL OPEN DOWNWARD.
I'LL LEAVE THIS FOR YOU TO PLAY WITH ON YOUR OWN TIME,
BUT LET'S TAKE A QUICK LOOK AT WHAT THE VALUE OF C DOES.
AS I INCREASE THE VALUE OF C,
NOTICE HOW IT'S A VERTICAL SHIFT UPWARD.
AS I DECREASE IT, IT'S A VERTICAL SHIFT DOWNWARD.
AND THE OTHER CONNECTION IS THAT THE VALUE OF C
IS ACTUALLY WHERE THE GRAPH CROSSES THE Y AXIS
OR THE Y INTERCEPT.
LET'S GO BACK TO OUR PRESENTATION.
LET'S TAKE A LOOK AT OUR OWN EXAMPLES.
GRAPH F OF X = X SQUARED - 4X - 5.
FIND THE EQUATION OF THE AXIS OF SYMMETRY
OR LINE OF SYMMETRY,
THE VERTEX AND THE X INTERCEPTS.
OKAY, SO REMEMBER TO FIND THE LINE OF SYMMETRY,
WE HAVE AN EQUATION FOR THAT AND THAT WOULD BE X = -B/2"A".
THE FIRST THING I LIKE TO DO
IS IDENTIFY THE VALUES OF "A", B AND C.
THOSE ARE THE COEFFICIENTS.
SO "A" WOULD BE EQUAL TO 1, B WOULD BE EQUAL TO -4,
C IS EQUAL TO -5.
SO DOING OUR SUBSTITUTIONS INTO OUR FORMULA
FOR THE AXIS OF SYMMETRY,
WE WOULD HAVE -4 ALL OVER 2 x 1,
WHICH WILL GIVE US X = 2 FOR OUR LINE OF SYMMETRY.
REMEMBER WHEN WE GO TO FIND THE VERTEX,
THE VALUE OF X FOR THE LINE OF SYMMETRY
ALSO GIVES US THE X COORDINATE OF THE VERTEX.
IN ORDER TO FIND THE Y COORDINATE,
WE HAVE TO SUBSTITUTE 2 INTO THE ORIGINAL FUNCTION,
F OF 2 IS EQUAL TO -9,
THEREFORE, OUR VERTEX IS 2, -9.
RECALL TO FIND THE X INTERCEPTS OF ANY FUNCTION
YOU HAVE TO SET Y EQUAL TO ZERO AND SOLVE FOR X.
SO WE WOULD HAVE 0 = X SQUARED - 4X - 5.
LUCKILY THIS IS FACTORABLE.
THE SOLUTIONS TO THIS QUADRATIC
ARE X = 5 AND X = -1.
THESE ARE ALSO OUR X INTERCEPTS.
SO WE CAN WRITE OUR X INTERCEPTS AS THE TWO POINTS,
5, 0 AND -1, 0.
LET'S GO AHEAD AND PUT ALL THESE PIECES TOGETHER
IN A GRAPH.
HERE IT IS IN RED WE HAVE OUR LINE OF SYMMETRY,
X = 2, X INTERCEPT OF POSITIVE 5, X INTERCEPT OF -1,
OUR VERTEX WITH THE COORDINATES 2, -9.
NOTICE HOW WITH THESE THREE POINTS WE COULD ACTUALLY MAKE
A NICE GRAPH JUST BY HAND
WITHOUT THE USE OF ANY TECHNOLOGY.
LET'S TAKE A LOOK AT ONE MORE EXAMPLE.
GIVEN F OF X = -2X SQUARED + 10X - 7,
FIND THE EQUATION OF THE AXIS OF SYMMETRY,
AGAIN, OR LINE OF SYMMETRY AND THE VERTEX.
THEN GRAPH WITH THE HELP OF THE GRAPHING CALCULATOR.
GOING BACK TO OUR EQUATION FOR THE AXIS OF SYMMETRY,
THE FIRST THING WE NEED TO DO
IS IDENTIFY THE VALES OF "A" AND B.
OUR "A" VALUE WOULD BE A -2, B 10 AND WE DON'T NEED C,
BUT C WOULD BE -7.
OKAY, LET'S DO THE SUBSTITUTION
AND SEE WHAT WE GET.
-B WOULD BE -10 OVER 2 TIMES OUR "A" VALUE OF -2.
IT LOOKS LIKE WE'D HAVE 10/4 OR 5/2.
LINE OF SYMMETRY IS X = 5/2 OR IF WE WANT, 2.5.
OF COURSE, THE VERTEX AGAIN---
WE KNOW THE X COORDINATE OF THE VERTEX HAS TO BE 5/2.
IN ORDER TO FIND THE Y COORDINATE OF THE VERTEX,
WE HAVE TO DO THE SUBSTITUTION INTO THE ORIGINAL EQUATION
WITH X EQUAL TO 5/2
AND YOU CAN GO AHEAD AND DO THAT BUT I CAME UP WITH 11/2.
AND NOTICE HOW THIS PROBLEM DOES NOT ASK US TO FIND
THE X INTERCEPTS OF THIS PARABOLA,
AND THE REASON IT DOESN'T ACTUALLY
IS BECAUSE IF YOU SET Y EQUAL TO ZERO,
THE QUADRATIC EQUATION IS NOT FACTORABLE.
SO LET'S GO AHEAD AND GET OUR GRAPHING CALCULATORS OUT
AND USE SOME TECHNOLOGY.
IF YOU HIT Y EQUALS--I'VE ALREADY TYPED IN THE FUNCTION.
NOW IF I HIT GRAPH, THE FIRST THING I WANT YOU TO NOTICE
IS IT DOES OPEN DOWNWARD,
AND THE REASON IT OPENS DOWNWARD, AGAIN,
IS BECAUSE "A" IS NEGATIVE.
HOWEVER, IT'S VERY DIFFICULT TO GET ADDITIONAL POINTS
ON THE GRAPH FROM THIS SCREEN,
SO IT'S OFTEN MORE HELPFUL IF YOU HIT SECOND GRAPH
AND PICK SOME POINTS FROM THE T-TABLE.
FOR EXAMPLE, I COULD PICK THE POINTS 1, 1, 2, 5
AND MAYBE 0, -7.
OF COURSE, THERE WERE A BUNCH OF OTHER POINTS
THAT WE COULD PICK.
I JUST DECIDED TO PICK THESE THREE.
LET'S SEE IF WE CAN GRAPH THIS
AND PUT ALL OF THESE KEY COMPONENTS TOGETHER.
THE FIRST THING WE NOTICE HERE
IS OUR AXIS OF SYMMETRY OF LINE OF SYMMETRY,
AND WE SAID THAT WAS EQUAL TO X = 2.5 OR 5/2.
HERE IS THE VERTEX WHERE WE FOUND THE X COORDINATE
TO BE 5/2, OF COURSE,
AND THE Y COORDINATE TO BE 11/2.
THERE IT IS.
AND WE FOUND A FEW OTHERPOINTS ON THE GRAPH.
I BELIEVE WE FOUND THE POINT 1, 1, 0, -7
AND I THINK WE FOUND 2, 5.
AND NOTICE HOW WITH THE AXIS OF SYMMETRY,
IF YOU HAVE THESE POINTS ON THE LEFT SIDE OF THE AXIS,
YOU COULD EASILY FIND THEIR MIRROR IMAGES ON THE RIGHT,
MEANING IF THERE IS A POINT HERE 2, 5,
THERE HAS TO BE ANOTHER POINT THAT'S A MIRROR IMAGE
OF THAT ON THE OTHER SIDE OF THE AXIS OF SYMMETRY.
SAME THING WITH THIS POINT,
THERE HAS TO BE ANOTHER POINT SOMEWHERE OVER HERE.
THAT HELPS YOU FINDING ADDITIONAL POINTS,
AND THAT'S WHY IT'S SO IMPORTANT
TO FIND THE AXIS OF SYMMETRY.
I HOPE THAT HELPS YOU REVIEW
HOW TO GRAPH QUADRATIC FUNCTIONS IN GENERAL FORM.
HAVE A GREAT DAY.