sparrow2
  • sparrow2
transform y=f(x) to y=f(x-1)
Mathematics
schrodinger
  • schrodinger
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sparrow2
  • sparrow2
sparrow2
  • sparrow2
i need prove that it goes right to 1 unit
dan815
  • dan815
|dw:1436470728973:dw|

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dan815
  • dan815
|dw:1436470896460:dw|
dan815
  • dan815
can draw the u axis
ganeshie8
  • 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
  • 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
  • ganeshie8
Right, but my reference is still the "x" axis, not "x-1" axis.
sparrow2
  • 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
  • ganeshie8
Kindof, f(2x) compresses the x axis by a factor of 2, the reference is still x axis
sparrow2
  • 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
  • sparrow2
in our case argument is a+1-1=a not a+1 so we have to put a not a+1
ganeshie8
  • ganeshie8
the argument to the transformed function is a+1
sparrow2
  • 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
  • ganeshie8
|dw:1436472503364:dw|
ganeshie8
  • ganeshie8
argument is what ever goes into the function box from outside world output is whatever the function box spits out
ganeshie8
  • 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
  • sparrow2
|dw:1436472582326:dw|
sparrow2
  • sparrow2
i think here is main that f is the same (so the rule is the same)
ganeshie8
  • ganeshie8
|dw:1436472673761:dw|
ganeshie8
  • ganeshie8
Think of f(x) and f(x-1) as two separate things
sparrow2
  • sparrow2
it is main that rules are the same( functions are different because they have different domains and range(mayby)
sparrow2
  • sparrow2
rules being the same is important
ganeshie8
  • ganeshie8
I see what you're getting at, im still thinking of a better way to convince myself haha
sparrow2
  • sparrow2
this eats my brain :) just i think it is more theoritocal question
ganeshie8
  • ganeshie8
freckles
  • 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
  • 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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