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sparrow2

  • one year ago

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

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  1. sparrow2
    • one year ago
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    @ganeshie8

  2. sparrow2
    • one year ago
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    i need prove that it goes right to 1 unit

  3. dan815
    • one year ago
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    |dw:1436470728973:dw|

  4. dan815
    • one year ago
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    |dw:1436470896460:dw|

  5. dan815
    • one year ago
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    can draw the u axis

  6. ganeshie8
    • one year ago
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    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.

  7. sparrow2
    • one year ago
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    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

  8. ganeshie8
    • one year ago
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    Right, but my reference is still the "x" axis, not "x-1" axis.

  9. sparrow2
    • one year ago
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    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?

  10. ganeshie8
    • one year ago
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    Kindof, f(2x) compresses the x axis by a factor of 2, the reference is still x axis

  11. sparrow2
    • one year ago
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    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

  12. sparrow2
    • one year ago
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    in our case argument is a+1-1=a not a+1 so we have to put a not a+1

  13. ganeshie8
    • one year ago
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    the argument to the transformed function is a+1

  14. sparrow2
    • one year ago
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    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

  15. ganeshie8
    • one year ago
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    |dw:1436472503364:dw|

  16. ganeshie8
    • one year ago
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    argument is what ever goes into the function box from outside world output is whatever the function box spits out

  17. ganeshie8
    • one year ago
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    btw it helps to treat f(x) and f(x-1) as two different functions maybe just call g(x) = f(x-1)

  18. sparrow2
    • one year ago
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    |dw:1436472582326:dw|

  19. sparrow2
    • one year ago
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    i think here is main that f is the same (so the rule is the same)

  20. ganeshie8
    • one year ago
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    |dw:1436472673761:dw|

  21. ganeshie8
    • one year ago
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    Think of f(x) and f(x-1) as two separate things

  22. sparrow2
    • one year ago
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    it is main that rules are the same( functions are different because they have different domains and range(mayby)

  23. sparrow2
    • one year ago
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    rules being the same is important

  24. ganeshie8
    • one year ago
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    I see what you're getting at, im still thinking of a better way to convince myself haha

  25. sparrow2
    • one year ago
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    this eats my brain :) just i think it is more theoritocal question

  26. ganeshie8
    • one year ago
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    @dan815

  27. freckles
    • one year ago
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    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?

  28. sparrow2
    • one year ago
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    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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