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    This is what I think, but can anyone help here? I may be completely wrong:

    A polynomial has the expression of ax^n+bx^{n-1}+...+c
    So if it was translated by a vector of [p/q] it'd be:

    y-q = a(x-p)^n+b(x-p)^{n-1}+...+c
    y = a(x-p)^n+b(x-p)^{n-1}+...+(c+q)

    So if you had two polynomials such as:

    1. y = (x-3)^3 + 2(x-3)^2+4(x-3)+7
    2. y = (x-5)^3 + 5(x-5)^2+7(x-5)+9

    This means that the vector translation of 1 is [3/7] and 2 is [5/9] so would the translation vector be (from 1 ---> 2) [2/2]?

    (I know that the vector columns are wrong in many ways but e.g [3/7] = [p/q])
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    (Original post by Sulfur)
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    Only if graph 2 is  y = (x-5)^3 + 2(x-5)^2+4(x-5)+9 .

    Method does not work since they are translations of different polynomials.
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    (Original post by lazy_fish)
    Only if graph 2 is  y = (x-5)^3 + 2(x-5)^2+4(x-5)+9 .

    Method does not work since they are translations of different polynomials.
    Thanks! I was wondering about that with polynomials of different degrees etc.
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    (Original post by Sulfur)

    EDIT: bx is to the power of n-1 for all of this post, don't know why it isn't appearing like that)
    If you have more than a single character then you need to use { }.

    bx^{n-1}
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    (Original post by BabyMaths)
    If you have more than a single character then you need to use { }.

    bx^{n-1}
    Thank you!
 
 
 
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