Understanding Vector Transformation with <1,-1> Translation - Explained

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Applying the vector <-1,-1> to translate f(x) to h(x) involves adding this vector to the function's output values. Specifically, if f(x) is a cubic function, the goal is to find a new function g(x) that represents the graph of f(x) translated down and to the left by one unit. The translation requires adjusting the input of f(x) to account for the shift, leading to the formulation g(x) = f(x + 1) - 1. This results in g(x) being expressed in the standard polynomial form ax^3 + bx^2 + cx + d after expansion. The discussion concludes with the successful derivation of the translated function.
kvzrock
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what does it mean by applying vector <-1,-1> to translate f(x) to h(x)?
 
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"Translation" in general means to add a vector. Since you haven't told us what "f(x)" and "h(x)" represent I can't be more specific but if they are vector valued functions, then I would suspect you are to add the vector <-1,-1> to f(x).
 
f(x)=x^3-3x^2
h(x) should be in the form of ax^3+bx^2+cx+d

edit: problem solved. Thanks.
 
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I suspect that you are intended to find a function whose graph looks exactly line y= f(x) but is translated (one translates geometric things- points and sets of points like graphs- not functions) by <-1,-1>. In particular, that means that, since f(0)= 0, we want g(-1)= -1. First, we want x= -1, in g, to "act like" x=0 in f. That is
g(x)= f(u) for some u so that when x= -1, u= 0: okay the simplest possible thing is u= x+1. If we write g(x)= f(x+1)= (x+1)3- 3(x+1)2, we have g(-1)= f(0) but we are not quite done: g(-1)= f(0)= 0 and we want g(-1)= f(0)-1. Fine: just subtract 1 from what we just got:

g(x)= (x+1)3- 3(x+1)2- 1.

In order to get it in the form "ax3+ b2+ cx+ d", you will need to multiply it out.
 
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