Mysterious Property When Translating Graphs
Hey
I am wondering why adding to the x-value causes the graph to shift right instead of left. I know that algebriaclly:
(x+d) ; started with addition
x+d=0
x=-d ; negative means shift left
(x-d) ; started with subtraction
x-d=0
x=d ; positive means shift right
Can anyone elaborate why this property exist!?
Sam
If your basic function is y= f(x), you need to distinguish between "changes to x" (before applying the function f) and "changes to y" (after applying the function f). Obviously, any changes to y affect the graph vertically: y= f(x)+ a moves the graph up by a, y= f(x)- a moves the graph down. Multiplying, y= af(x) stretches (if a> 1) or squeezes (if a< 1) the graph vertically.Originally Posted by ArcherSam
Changes to x (before applying the function f) are horizontal changes. But when you talk about graphing y= f(x), as the notation indicates, you are talking about values of y, not x. Changes before applying f, that is, changes to x, are more subtle: Suppose. Then
will be equal to 0 when
, that is, when x= -a, not a. More generally, if
, then
when
or when
.
@HallsofIvy
This is what I have interpreted from your reply:
WRONG
y = f(x) = (x-3)^2
y = f(0) = (0-3)^2 = 9 ; (0,9)
y = f(1) = (1-3)^2 = 4 ; (1,4)
y = f(2) = (2-3)^2 = 1 ; (2,1)
y = f(3) = (3-3)^2 = 0 ; (3,0)
CORRECT
y = f(x) = (x-3)^2
y = f(0) = (0-3)^2 = 9 ; (3,9)
y = f(1) = (1-3)^2 = 4 ; (4,4)
y = f(2) = (2-3)^2 = 1 ; (5,1)
y = f(3) = (3-3)^2 = 0 ; (6,0)
y = f(x) = (x-a)^b where (x-a) = x or (x-a) = f(x-a) ; and this is possible because of functional notation.
What I am failing to understand is how the notation is allowed to be equal to the part of the equation that modifies the x-value.How can I take the input(before) and make it equal to the change(during) to still equal the input(after)?
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