# Mysterious Property When Translating Graphs

• September 28th 2011, 10:26 AM
ArcherSam
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=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
• September 28th 2011, 11:18 AM
HallsofIvy
Re: Mysterious Property When Translating Graphs
Quote:

Originally Posted by ArcherSam
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=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.

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 $y_0= f(0)$. Then $f(x+a)$ will be equal to 0 when $x+ a= 0$, that is, when x= -a, not a. More generally, if $y_1= f(x_1)$, then $f(x+a)= y_1$ when $x+ a= x_1$ or when $x= x_1- a$.
• September 28th 2011, 12:51 PM
ArcherSam
Re: Mysterious Property When Translating Graphs
@HallsofIvy

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)?