- WE WANT TO GRAPH THE ABSOLUTE
VALUE FUNCTION F(X) = 1/2 x THE ABSOLUTE VALUE
OF THE QUANTITY X + 2 - 3. THERE ARE A COUPLE WAYS TO GO
ABOUT GRAPHING THIS FUNCTION, BUT IT IS HELPFUL TO KNOW WHAT THE BASIC ABSOLUTE VALUE
FUNCTION LOOKS LIKE, AND WE SHOULD RECOGNIZE IT HERE WHERE WE HAVE G(X) =
THE ABSOLUTE VALUE OF X. NOTICE THE SHAPE IS A V. ALL ABSOLUTE VALUE FUNCTIONS
WILL BE Vs. THEY MAY BE WIDER OR NARROWER
OR OPEN UP OR DOWN, BUT THEY'LL ALWAYS BE
IN THE SHAPE OF A V. EVERY ABSOLUTE VALUE FUNCTION
HAS A VERTEX WHICH IS THE HIGHEST POINT
OR LOWEST POINT ON THE GRAPH. IN THIS CASE,
THIS WOULD BE THE VERTEX, AND THIS IS A KEY POINT
TO FIND WHEN GRAPHING
AN ABSOLUTE VALUE FUNCTION. IF WE CAN FIND THE VERTEX AND THEN JUST ONE POINT
ON THE RIGHT AND ONE POINT ON THE LEFT
OF THE VERTEX, WE CAN MAKE
A NICE ACCURATE GRAPH OF AN ABSOLUTE VALUE FUNCTION. SO GOING BACK TO OUR EXAMPLE, TO FIND THE X COORDINATE
OF THE VERTEX, WE WANT TO SET THE EXPRESSION
INSIDE THE ABSOLUTE VALUE, OR IN THIS CASE X + 2 = 0,
AND SOLVE FOR X. SO IF I SET X + 2 = 0 AND SOLVE
FOR X, NOTICE HOW X = -2. SO IF WE'RE GOING TO COMPLETE
A TABLE OF VALUES TO GRAPH THIS FUNCTION, WE WANT ONE OF THE X VALUES
TO BE -2. AND BECAUSE THE VERTEX
WILL BE AT X = -2, IF WE SELECT ONE X VALUE TO THE
LEFT OF -2 OR LESS THAN -2 AND ONE X VALUE TO THE RIGHT
OF -2 OR GREATER THAN -2, WE CAN MAKE
A NICE ACCURATE GRAPH OF THE ABSOLUTE VALUE FUNCTION. BUT BECAUSE WE HAVE
THIS 1/2 HERE, WE WOULD LIKE X + 2
TO BE A MULTIPLE OF 2. SO FOR A VALUE THAT'S GREATER
THAN -2, I'M NOT GOING TO USE -1 HERE, BECAUSE NOTICE THAT -1 + 2 = +1,
NOT A MULTIPLE OF 2. IT WOULDN'T BE WRONG, BUT THEN WE'D HAVE A FRACTION
FOR THE VALUE OF Y. SO INSTEAD OF USING -1,
LET'S USE X = 0. NOTICE 0 + 2 = 2
WHICH IS A MULTIPLE OF 2. AND NOW FOR A VALUE OF X
THAT'S LESS THAN -2 SO THAT X + 2 IS A MULTIPLE
OF 2, I'M NOT GOING TO USE -3, BECAUSE THAT WOULD BE -3 + 2
OR -1, NOT A MULTIPLE OF 2, SO I'LL USE -4 FOR AN X VALUE
THAT'S LESS THAN -2. IT WOULDN'T BE WRONG TO HAVE
A FRACTION VALUE OF Y, BUT IT WOULD BE MORE DIFFICULT
TO GRAPH. NOW THAT WE'VE FOUND THE
X VALUES THAT WE WANT TO USE, WE'LL EVALUATE THE FUNCTION
AT THESE X VALUES TO FIND THE CORRESPONDING
Y VALUES. SO WHEN X IS -2,
WE WANT TO FIND F(-2) WHICH WOULD BE 1/2
x THE ABSOLUTE VALUE OF -2 + 2 AND THEN - 3. WELL, THIS IS GOING TO BE 1/2
x THE ABSOLUTE VALUE OF 0 - 3. WELL, THE ABSOLUTE VALUE OF 0
IS 0 x 1/2 IS 0, SO WE JUST HAVE -3 HERE. SO WHEN X IS -2, Y IS -3, SO THIS WOULD BE THE VERTEX
OF THE ABSOLUTE VALUE FUNCTION WHICH WOULD BE HERE. NOW WE HAVE X = 0. SO TO FIND THE CORRESPONDING
Y VALUE, WE WANT TO FIND F(0) WHICH IS GOING TO BE 1/2 x
THE ABSOLUTE VALUE OF 0 + 2 - 3, SO THIS WILL BE 1/2 x, THE ABSOLUTE VALUE OF +2 IS 2
WHICH I'LL WRITE AS 2/1 - 3. NOTICE HERE THIS SIMPLIFIES
NICELY GIVING US JUST 1 - 3 WHICH IS -2. SO THE (0,-2)
OR THIS POINT HERE, WOULD BE A POINT
ON THE ABSOLUTE VALUE FUNCTION. AND THEN FINALLY,
WE HAVE X = - 4. SO TO FIND THE CORRESPONDING
Y VALUE, WE WANT TO FIND F(-4). SO WE'LL HAVE 1/2 x THE ABSOLUTE
VALUE OF -4 + 2 - 3, SO WE HAVE 1/2 x, THIS IS THE ABSOLUTE VALUE OF -2
WHICH IS STILL +2 OR 2/1 - 3. NOTICE HOW THE FUNCTION VALUE
WHEN X IS -4 IS THE SAME AS THE FUNCTION
VALUE WHEN X IS 0. WE HAVE 1 - 3 OR -2. SO (-4,-2) OR THIS POINT HERE
IS ON THE FUNCTION. NOW WE CAN MAKE
A NICE ACCURATE GRAPH OF OUR ABSOLUTE VALUE FUNCTION. SINCE THIS IS THE VERTEX, THE RIGHT SIDE OF THE GRAPH
WOULD LOOK LIKE THIS, AND THE LEFT SIDE WOULD LOOK
LIKE THIS. NOW THERE'S ALSO A WAY TO GRAPH
THIS ABSOLUTE VALUE FUNCTION BY USING TRANSFORMATIONS OF THE
BASIC SQUARE ROOT FUNCTION. IF WE COMPARE THE BASIC ABSOLUTE
VALUE FUNCTION GRAPHED HERE IN RED TO OUR ABSOLUTE VALUE FUNCTION
HERE, FOR THE FUNCTION F(X) = 1/2 x THE ABSOLUTE VALUE
OF THE QUANTITY X + 2 - 3, IT WOULD TAKE THE POINTS ON THE
BASIC ABSOLUTE VALUE FUNCTION, AND THEN BECAUSE WE HAVE X + 2
INSIDE THE ABSOLUTE VALUE, IT WOULD SHIFT THE POINTS LEFT
2 UNITS. AND THEN BECAUSE OF THIS
1/2 HERE, WE WOULD MULTIPLY
ALL OF THE Y COORDINATES OF THE POINTS ON THE ABSOLUTE
VALUE FUNCTION BY 1/2, HORIZONTALLY COMPRESSING
THE FUNCTION, AND THEN THE - 3 HERE SHIFTS
ALL THE POINTS DOWN 3 UNITS. SO WE COULD SAY F(X) TAKES
THE ABSOLUTE VALUE FUNCTION, SHIFTS IT LEFT 2 UNITS, HORIZONTALLY COMPRESSES IT
BY 1/2, AND THEN SHIFTS IT DOWN 3 UNITS. I HOPE YOU FOUND THIS HELPFUL.