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Transformation of graphs , y= f(2x)

What is the reasoning behind the x being stretch by a factor of 1/2 rather than a factor of 2 ?, since it is 2x , which is rather counter intuitive.
(edited 6 years ago)
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Original post by Bilbao
What is the reasoning behind the x being stretch by a factor of 1/2 rather than a factor of 2 ?, since it is 2x , which is rather counter intuitive.

An example to get you thinking:

f(x) = 2x

f(2x) = g(x) = 2(2x)

f(1) = 2
g(0.5) = 2

f(2) = 4
g(1) = 4

So to get to the same output in g as you do in f, you need to halve the input.
I struggled with this for a bit but recently got it

Say you have the function y = f(x). It goes through the point (2, 4). You also have y = f(2x). Where does the point (2, 4) move to on the new graph? Well you know if you put in 2 into the function you will receive 4 as a value of y. So now you can "solve" for the new X coordinate: 2x = 2, so x = 2/2 = 1. From this it's very intuitive to see why you stretch by 1/2. Think of it as solving an equation.
Original post by Bilbao
What is the reasoning behind the x being stretch by a factor of 1/2 rather than a factor of 2 ?, since it is 2x , which is rather counter intuitive.


It is counter intuitive, yes, but then again there are plenty of other things in maths which are counter intuitive.

Take a function y=f(x)y=f(x) which takes on a certain yy value when x=Xx=X. Then introduce a transformation y=f(2x)y=f(2x). We want to retain the exact same yy value as the previous one, so we need to set 2x=X2x=X which hence means that x=X2x=\frac{X}{2}. Hence the new input into your transformed function is HALF the original one, and this applies to all xx hence we say f(x)f(2x)f(x) \mapsto f(2x) is a stretch by a factor 1/2 parallel to the x-axis.

Same thing works when you transform y=f(x)y=f(x) to 2y=f(x)2y=f(x). You half every yy value, but more often you would rewrite the second eq. by saying that y=12f(x)y=\frac{1}{2}f(x) and hence it is much more obvious to say that yy halves, and so y2yy \mapsto 2y is a transformation of a stretch factor 1/2 parallel to the y-axis.
(edited 6 years ago)

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