shifting functions vertically and horizontally

Question

dannyrod2000 · Answer

attachment

dannyrod2000 · Answer

@navk    @skullpatrol   @zzr0ck3r

e.mccormick · Answer

Shifts in x use postive values to go left and negative to go right.  So $x^2$ shifted right one is $(x-1)^2$.  The y values use positive to go up and negative to go down.  So $x^2$ shifted up one is $x^2+1$.

dannyrod2000 · Answer

@jtryon

e.mccormick · Answer

So all you need to do is count the number up/down and left/right and put it in the proper place with the proper sign.

$y=(x\mp h)^2\pm k$ where h is the horizontal shift and k it the vertical.

dannyrod2000 · Answer

thanks

e.mccormick · Answer

So, did you come up with a solution?

dannyrod2000 · Answer

(x+4)^2+1?

e.mccormick · Answer

The $(x+4)^2$ oaer is good.  That brings it left 4.  But up 1?

e.mccormick · Answer

part... who knows where that other word came from. LOL

dannyrod2000 · Answer

so is it correct?

e.mccormick · Answer

No, you need to go down, and more than 1.

dannyrod2000 · Answer

?

e.mccormick · Answer

Well, in what you have a pic of, does it go up or down when you go from $f(x)$ to $g(x)$?

dannyrod2000 · Answer

it goes up one time

e.mccormick · Answer

Umm, that is not what is in the picture you posted.

dannyrod2000 · Answer

oops i put the wrong picture i will put the correct

dannyrod2000 · Answer

attachment

e.mccormick · Answer

Ah, that one goes up one.  But it only goes left 2, not 4.

dannyrod2000 · Answer

how do you go left? do you start from the side of the line?

e.mccormick · Answer

The sign tells you if it is up or down and left or right.

$(x+L)^2$ left
$(x-R)^2$ right
$x^2+U$ up
$x^2-D$ down

They can be mixed with things like:

$(x-R)^2+U$ right and up

dannyrod2000 · Answer

o ok so it would be (x+2)^2+1?

e.mccormick · Answer

Yes.

e.mccormick · Answer

And the great thing is, those rules for shifting work on more than just a parabola.  The same basics work for a lot of graphs.