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This is the exact problem from the worksheet. Now I understand that it is giving the parent function for $f(x)$. The only formula I know of for transformations is $y=f(x)\rightarrow y=af(bx+c)+d$ where:
$a$=vertical compression or expansion
$b$=horizontal compression or expansion
$c$=left or right(+=left, -=right)
$d$=up or down (+=up, -=down)
This is all the information I have for transformations. I have done a lot of them.

Edit: I solved the first two and now need a clearer explanation on $c$ and $d$. No answers please.

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    so set $a=1,b=-1,c=0,d=0$. You'll get a mirror effect on the vertical axis for (b) since for $x=-2$ you will get $f(2)=0$.2012-08-07
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    hint for (d) : is there a way (choice of $x$) to see the left part of the picture? (your (c) in the other comment is right)2012-08-07
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    What do you mean by "is there a way of seeing the left part of the picture"? Do you mean left of the $x$-axis?2012-08-07
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    the part $x<0$ of your initial picture2012-08-07
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    Okay, so I still do not get your question.2012-08-07
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    $f(|x|)$ will always be invoked with a nonnegative value so that only $x\ge 0$ part of your picture will be used for the final graph ! (remember you don't want the answer!)2012-08-07
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    That makes sense. So the values less than $0$ are not included in the graph? Meaning it will only be the right $(x\ge0)$?2012-08-07
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    the right part will be conserved but for the left part $x=-2$ for example we will compute $f(|-2|)$ getting...2012-08-07
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    It would be $f(2)$. Any negative value becomes positive.2012-08-07
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    yes a little like in (b) (for the left part)2012-08-07
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    So is the graph like this: left part of normal problem but in the II quadrant and the right part stays the same?2012-08-07
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    yes the right is idem and the left part is the mirror of the right part.2012-08-07
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    Great, so how does $c$ differ from $d$?2012-08-07
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    (d) acts on the result and not on the parameter. For example $f(-2)$ will not become $-1$ but...2012-08-07
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    It would become $2$? I'm a little confused.2012-08-07
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    sorry I wanted to say "(c) acts on the result..." (we are speaking of (c) now!). In (c) you don't change the parameter but the result : we want $|f(-2)|=1$.2012-08-07
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    But how would you get $1$ from $|f(-2)|$ without changing the parameter?2012-08-07
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    you compute $f(-2)$ get (say) $-1$ and only then take the absolute value $\to 1$.2012-08-07
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    So, would I do that for each value? ($-1,0,1,2\cdots$)2012-08-07
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    yes every $y$ of the graph will become $|y|$ (nothing will remain on the bottom ($y<0$).2012-08-07
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    So does the graph look like this: the right side is idempotent and the left side the right side but reflected off of the $y$-axis?2012-08-07
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    in practice yes (for (c) the mirror effect is around the $y=0$ for $x<0$)). For other graphs the right part could be changed and the left part not mirrored...2012-08-07
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    Great. Thanks for helping and being patient with me.2012-08-07

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