pre calc

tdc123

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Jan 22, 2011
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question is: the point (2,3) is on the graph of f(x). If you were to perform the following graph shifts on f(x), what point is on the shifted graph? and the graph shift is f(x-3)-7.
This is a email i had with the prof.:
I wrote:
I am working on the group work problem and i could use a little clarification on something. I am working on problem B, 3. and i am a little confused on the notation. is f(x - 3) - 7 the same as f(x)=(x-3)-7? with that logic, i just shifted the point (2,3) 3 units to the right and 7 units down (5,-4). i feel like i am missing something. am i supposed to use the original format and use algebra to find the new x or just do a translation like i had originally thought. Thank You
She responded
No, f(x-3)-7 isn't equivalent to f(x)=(x-3)-7. The way you have written it, it would just be a line. But f(x-3)-7 is actually some unknown function. f(x) may be a line, but it may a parabola, it may be a sine wave (which we will study later), or any other function.

f(x-3)-7 means that you would take whatever function f(x) represents and plug x-3 in as the input and then subtract 7 from whatever comes out of the function. So for example, if f(x)=x^2, then f(x-3)-7 would mean to find (x-3)^2 and then subtract 7 from it. (x-3)^2-7. That would be the new resulting function.

But the real point is that it doesn't matter one bit what the function is. When you subtract 3 from the inputs and then subtract 7 from the output, regardless of what type of graph you are looking at, it will always move the exact same way. That is powerful information since once you know the basic shape of a graph you know all the possible shifts for that same graph. So for example, once I know what y=x^2 looks like, I can figure out what y=x^2+5 would look like.

I am still confused anyone understand?
 
I think you have the right answer. I believe this problem is trying to get you to conceptualize functions. f(x-3)-7 just represents an unknown function that is being shifted right then down. If (2,3) is on f(x) then (5,-4) is on the unknown function of f(x-3)-7. I may be wrong though, I am pretty tired right now. Maybe some one else will chime in and either validate or falsify this.
 
Alkurz, everything you said is correct.

Let me try to clarify the example your teacher was giving you.

If \(\displaystyle f(x) = x^2\), then \(\displaystyle f(x-3)-7 = (x-3)^2-7\)

If we let \(\displaystyle g(x) = f(x-3)-7\), then \(\displaystyle g\) is a new function which has the same exact "shape" as \(\displaystyle f\). They are both parabolas.
\(\displaystyle f\) has its vertex at the point \(\displaystyle (0,0)\). \(\displaystyle g\) is \(\displaystyle f\) shifted 3 units right and 7 units down, and so it has its vertex at the point \(\displaystyle (3,-7)\).
 
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