Graphing (Precalculus) Homework: Transforming Functions for y=(x-3)^2 and y=x^n

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Homework Statement


Sketch the graph of the function y=(x-3)^2 by transforming the graph of an appropriate function of the form y=x^n



Homework Equations


y=(x-3)^2, y=x^n


The Attempt at a Solution


Precisely speaking, I am just totally confused with what the question is asking. i mean i know how to graph y=(x-3)^2 bit idk what the y=x^n is about. thanks so much!
 
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If you look carefully at the exponents, you might guess that they mean n = 2.
So the question is, through what (geometric) operation(s) can you obtain the graph of (x - 3)3 from that of x2?

Since x2 is a basic graph (you should be able to draw it by heart), given this set of operations you can also draw (x - 3)2.
 
Generally speaking, anything that is done to x before the "basic function" is a change in x and so changes the graph horizontally. Anything that is done after the "basic function" is a change in y and so changes the graph vertically.

Here, the "basic function" is x^2 and (x- 3)^2 is just x^2 with 3 subtracted from x before the square. The graph of y= x^2 itself is a parabola with vertex at (0, 0). x- 3= 0 when x= 3. The graph of y= (x- 3)^2 is the graph of y= x^2[/tex] shifted three places to the right- a parabola with vertex at (3, 0).
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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