How do you solve for g(x) using given points on the graph of f(x)?

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To solve for g(x) using the graph of f(x), it is essential to understand the transformation defined by g(x) = 2f(x-1). For example, to calculate g(0), substitute x with 0, resulting in g(0) = 2f(-1), where the value of f(-1) can be obtained from the graph. This transformation indicates that the graph of g(x) is a rightward shift of f(x) by one unit, followed by a vertical stretch by a factor of 2. Consequently, the point (-1,3) on f(x) transforms to (0,6) on g(x). Understanding these relationships allows for the calculation of g(1), g(2), and g(3) similarly.
Peter G.
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Hi, :smile:

The sketch (attachment) shows part of the graph of y = f (x) which passes through the points:

A (-1,3) B (0,2), C (1,0), D(2,1) and E (3,5)

A second function is defined by g (x) = 2f (x-1)

a) Calculate g (0), g(1), g(2) and g (3)

I am confused about what the question is asking... Could anyone maybe give me a clue/guide with one of the points so I can try and do the rest on my own?

Thanks,
Peter G.
 

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Think about how g(x) is related to f(x).

eg. g(0) = 2f(-1)

You could use the graph to find f(-1)
 


so, when g (0), we sub x for 0, giving us: 2f(0-1) = 2f(-1)

Then, I can go to the graph, find the value for y when x = -1 and multiply by 2?
 


Yep. You could also think about it in terms of transformations; the graph of g(x) is just the graph of f(x) that has been shifted to the right by 1 unit and then stretched parallel to the y-axis by a factor of 2. This means that the point with coordinate (-1,3) becomes (0,6).
 


Yeah, that was what I thought about first but I am a bit sleepy and for some reason couldn't do it :-p

Thanks a lot,
Peter G.
 
The working out suggests first equating ## \sqrt{i} = x + iy ## and suggests that squaring and equating real and imaginary parts of both sides results in ## \sqrt{i} = \pm (1+i)/ \sqrt{2} ## Squaring both sides results in: $$ i = (x + iy)^2 $$ $$ i = x^2 + 2ixy -y^2 $$ equating real parts gives $$ x^2 - y^2 = 0 $$ $$ (x+y)(x-y) = 0 $$ $$ x = \pm y $$ equating imaginary parts gives: $$ i = 2ixy $$ $$ 2xy = 1 $$ I'm not really sure how to proceed from here.
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