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What's the transformation of y=(x-3)^2+4?

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4 Answers

Hi, Ana.

The number inside the parentheses with the x tells us the horizontal shift; a -3 means the graph moves to the right 3.

The constant tells us the vertical movement of the graph; the +4 tells us that the graph moves up 4.

Your vertex is now at (3,4).

Not appicable to this case, but a coefficient multiplied by the (x-3) would tell us the vertical stretch of the graph.

Hope this helps!


One note - the 'vertical stretch' coefficient should be applied outside the parentheses.  Putting a coeffient inside the parentheses will also create a vertical stretch, but it will also have other [possibly undesirable] effects such as additional horizontal translation and horizontal stretching/compression (it depends on how you look at it).

You're right - I wanted to mention the information about the coefficient's effect on the graph, but it was not worded well. I'll clarify that.

Hi Ana, 

To visualize how the graph moves, rewrite y = (x - 3)^2 + 4 so that it is easier to compare with y = x^2. 

old: y = x^2

new: (y' - 4) = (x' - 3)^2

Now you can see that the transformation changed y to (y' - 4) and x to (x' - 3).  

y = y' - 4 ----> y' = y + 4.  This means that the new y' is the old y shifted up 4.

x = x' - 3 ----> x' = x + 3.  This means that the new x' is the old x shifted up 3.

A very general rule:

If g(x) = f(x-h) + k, then the graph of g(x) is produced by shifting the graph of f(x) by h units right and k units up.

For example, if f(x) = x2then your equation is just y = f(x-3) + 4. So your function is just a shift of y = x2 by 3 units right and 4 units up.

It actually depends on the original function.

If the original function is y = x2, then y = (x-3)^2 moves y = x2 to the right by 3 units, and y = (x-3)^2 + 4 moves y = (x-3)^2 up by 4 units.


Extention: Redo the problem for y=(2x-3)^2+4.