transform y=f(x) to y=f(x-1)

- sparrow2

transform y=f(x) to y=f(x-1)

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- schrodinger

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- sparrow2

@ganeshie8

- sparrow2

i need prove that it goes right to 1 unit

- dan815

|dw:1436470728973:dw|

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## More answers

- dan815

|dw:1436470896460:dw|

- dan815

can draw the u axis

- ganeshie8

Say \((a,b)\) is a point on \(f(x)\) before the transformation, then we have
\[f(a) = b\tag{1}\]
Next, plugin \(x=\color{red}{a+1}\) into the transformed function \(f(x-1)\),
\[f(\color{red}{a+1}-1) = f(a) = b \tag{2}\]
That means \((a,b)\) has moved to \((a+1, b)\) under the transformation \(f(x-1)\).
After the transformation, the \(x\) value is \(a+1\) which is exactly \(1\) unit to the right of\(a\).
We can conclude the transformation shifts \(x\) coordinates of all the points to right by \(1\) unit.

- sparrow2

i understand that but here is what i don't undertand: an argument of the transformed function is x-1, so when you plug for x=a+1, argument is a+1-1=a not a+1. @ganeshie8

- ganeshie8

Right, but my reference is still the "x" axis, not "x-1" axis.

- sparrow2

x axis mean that we put arguments there(not the meaning of x, like in y=f(2x) i put 2x meaning not x to x axis) right?

- ganeshie8

Kindof, f(2x) compresses the x axis by a factor of 2, the reference is still x axis

- sparrow2

i think it doesn't matter what we call x or not.. main is that we put arguments there.f(2z) i put the meaning of 2z to x axis too

- sparrow2

in our case argument is a+1-1=a not a+1 so we have to put a not a+1

- ganeshie8

the argument to the transformed function is a+1

- sparrow2

it is f(x-1) you plug x=a+1(this is not the argument) argument is x-1=a+1-1=a i see this like that

- ganeshie8

|dw:1436472503364:dw|

- ganeshie8

argument is what ever goes into the function box from outside world
output is whatever the function box spits out

- ganeshie8

btw it helps to treat f(x) and f(x-1) as two different functions
maybe just call g(x) = f(x-1)

- sparrow2

|dw:1436472582326:dw|

- sparrow2

i think here is main that f is the same (so the rule is the same)

- ganeshie8

|dw:1436472673761:dw|

- ganeshie8

Think of f(x) and f(x-1) as two separate things

- sparrow2

it is main that rules are the same( functions are different because they have different domains and range(mayby)

- sparrow2

rules being the same is important

- ganeshie8

I see what you're getting at, im still thinking of a better way to convince myself haha

- sparrow2

this eats my brain :) just i think it is more theoritocal question

- ganeshie8

@dan815

- freckles

Example:
\[f(x)=x^2 \\ f(x-1)=(x-1)^2 \\ \text{ Let } g(x)=(x-1)^2 \\ g(x+1)=(x+1-1)^2=x^2 =f(x) \\ \text{ the argument for } g \text{ was } x+1 \text{ on that last line }\]
I'm not sure this answers your question or not.
I'm still struggling with what you were talking about the other day.
I'm not trying to be slow but what is it exactly you are looking clarification on?

- sparrow2

i'm still confused..it will be good if you show the transformation from f(x) to f(x-1) ,note that in f(x-1) an argument is x-1 not x,

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