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Translations help

HI
Capture.PNG

I get that it is a stretch in x direction but i don't understand how you work out the scale factor?
Any help?
Many thanks
Reply 1
Original post by pygmypossum
HI
Capture.PNG

I get that it is a stretch in x direction but i don't understand how you work out the scale factor?
Any help?
Many thanks


If you call the first one f(x)f(x) then you should spot the second one is f(2x)f(2x). Can you see why this is true?

The transformation f(ax)f(ax) is a stretch scale factor 1a\frac{1}{a} parallel to the x-axis.
Reply 2
Original post by Zacken
If you call the first one f(x)f(x) then you should spot the second one is f(2x)f(2x). Can you see why this is true?

The transformation f(ax)f(ax) is a stretch scale factor 1a\frac{1}{a} parallel to the x-axis.


No sorry I don't understand how its (2x)??
Reply 3
Original post by pygmypossum
No sorry I don't understand how its (2x)??


If you have f(x)=x3+1f(x) = \sqrt{x^3 + 1} then if you replace all the x's with 2x you get f(2x)=(2x)3+1=8x3+1f(2x) = \sqrt{(2x)^3 +1} = \sqrt{8x^3 + 1}, which is precisely the second function you have.
Reply 4
Original post by Zacken
If you have f(x)=x3+1f(x) = \sqrt{x^3 + 1} then if you replace all the x's with 2x you get f(2x)=(2x)3+1=8x3+1f(2x) = \sqrt{(2x)^3 +1} = \sqrt{8x^3 + 1}, which is precisely the second function you have.


Ah that makes sense! So is the answer 1/2?
Reply 5
Original post by pygmypossum
Ah that makes sense! So is the answer 1/2?


Yeah.
Reply 6
Original post by Zacken
Yeah.


Thank you!
Reply 7
Original post by pygmypossum
Thank you!


No problem. :-)

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