Transformations of quadratic equations

**elmidge** · January 18th 2013, 12:02 PM

Can you please explain the transformations involved when transforming y = x^2 to y = 4(3x -6)^2 + 1

**Barioth** · January 18th 2013, 01:29 PM

I dont understand what your question is sorry

**Soroban** · January 18th 2013, 01:42 PM

Hello, elmidge!

$\text{Can you please explain the transformations involved}$
$\text{when transforming }\,y \:=\: x^2\,\text{ to }\,y\:=\:4(3x -6)^2 + 1$

$\text{We are going from }\,y \:=\:x^2\,\text{ to }\,y \:=\:36(x-2)^2 + 1$

$\text{I assume you know the graph of: }\:y \,=\,x^2.$

$\text{We have: }\:y \:=\:(x\,{\color{red}-\,2})^2$
. . $\text{The graph is moved 2 units to the right.}$

$\text{Then: }\:y \:=\:{\color{red}36}(x-2)^2$
. . The graph is stretched vertically by a factor of 36.

$\text{Finally: }\:y \:=\:36(x-2)^2\, {\color{red}+ 1}$
. . The graph is moved one unit upward.

**elmidge** · January 18th 2013, 02:14 PM

Thank you can you also show me the transalations involved with transforming y = x^2 to y = 9x^2 -6x

**Prove It** · January 18th 2013, 03:47 PM

Originally Posted by elmidge

Thank you can you also show me the transalations involved with transforming y = x^2 to y = 9x^2 -6x

Start by Completing the Square, then the transformations will be easier to see.

**Furyan** · January 18th 2013, 04:03 PM

Originally Posted by elmidge

Thank you can you also show me the transalations involved with transforming y = x^2 to y = 9x^2 -6x

The vertex of $y = x^2$ is at $(0,0)$

The vertex of $y = (x + p)^2 + q$ is at $(- p, q)$ . A translation p units in the negative x direction and q units in the positive y.

If you complete the square for $y = 9x^2 - 6x$ , you will have it in the above, vertex, form and will therefore know the transformations involved.

Can you do that?

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