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Function transformations

by WannabeFeynman
Tags: function, transformations
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WannabeFeynman
#1
Jan30-14, 08:02 PM
P: 55
Hello all, I need some help to clear my doubts.

Why does a horizontal translation (f(x + c)) move to the left if c is positive?

Can someone graphically explain what effect a stretch and compression (vertical and horizontal) has on the original parent function?

Similar to the first question, why does f(ax) actually stretch by a factor of 1/a instead of a?

Thanks, I might have more questions later.
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#2
Jan31-14, 07:28 AM
P: 540
Is you graph y1=f(x) and y2=f(x+C) and then compare them, whatever "y1" you see at any "x1" will be seen to the left (assuming C>0) at x2="x1-C" for y2 because y2 = f(x2+C) = f(x1-C+C) = f(x1) = y1.
Similarly:
f(Ax) will horizontally thin the graph by a factor of "A" (or stretch it by 1/A)
Af(x) will vertically stretch the graph by a factor of "A".
Seydlitz
#3
Jan31-14, 11:49 PM
Seydlitz's Avatar
P: 253
Take a look at this simple example, suppose ##f(x)=x##, a simple straight line through the origin. If ##x=0##, then ##f(x)=0## as well. But if you take ##f(x+a)##, the origin will be at ##-a##. If you draw the line, the origin will move to the left because the origin is now in negative part of the x-axis.

Mark44
#4
Feb1-14, 01:51 PM
Mentor
P: 21,214
Function transformations

Quote Quote by Seydlitz View Post
Take a look at this simple example, suppose ##f(x)=x##, a simple straight line through the origin. If ##x=0##, then ##f(x)=0## as well. But if you take ##f(x+a)##, the origin will be at ##-a##. If you draw the line, the origin will move to the left because the origin is now in negative part of the x-axis.
The origin doesn't move around, but the x-intercept does.


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