gildartz
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#1
Report Thread starter 8 years ago
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Simple question: How do you transform the graph f(x) = e^x to f(x) = e^kx?
I thought you had to stretch it by factor e^k in the x-direction but it's actually by factor 1/k. Don't understand why this is :confused:
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Hopple
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You're making the independent variable increase k times as fast, so you squish it. You can check it for x=1, the old f(x) will be e, the new will be e^k, so for k>1 it will have shot up faster, for k<1 it will be a slower increase.
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Goldfishy
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For a function f(x), f(kx) will stretch it parallel to the x-axis with scale factor 1/k.

If you can't remember the transformations, try considering the graphs of  y = x^2 (for translations and reflections) and  y = sinx for stretches. Works for me.

(You seem to have got negged for no reason, so here's a free rep back )
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Phil_Waite
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#4
Report 8 years ago
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(Original post by gildartz)
Simple question: How do you transform the graph f(x) = e^x to f(x) = e^kx?
I thought you had to stretch it by factor e^k in the x-direction but it's actually by factor 1/k. Don't understand why this is :confused:
The reason is that for every x value in the new function, the value of kx is k times larger than the value of x in f(x)=e^x. This means that the new x inputs need to be k times smaller to produce the same outputs as before.

x \times  k \times  \frac{1}{k} = x

This results in a stretch of scale factor 1/k parallel to the x axis.
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gildartz
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Report Thread starter 8 years ago
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Ah I see, thanks for the help everyone

(Original post by Goldfishy)
For a function f(x), f(kx) will stretch it parallel to the x-axis with scale factor 1/k.

If you can't remember the transformations, try considering the graphs of  y = x^2 (for translations and reflections) and  y = sinx for stretches. Works for me.

(You seem to have got negged for no reason, so here's a free rep back )
Haha, thanks for the rep, you can have some yourself
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