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

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- Jan 18th 2013, 12:02 PMelmidgeTransformations of quadratic equations
Can you please explain the transformations involved when transforming y = x^2 to y = 4(3x -6)^2 + 1

- Jan 18th 2013, 01:29 PMBariothRe: Transformations of quadratic equations
I dont understand what your question is sorry

- Jan 18th 2013, 01:42 PMSorobanRe: Transformations of quadratic equations
Hello, elmidge!

Quote:

$\displaystyle \text{Can you please explain the transformations involved}$

$\displaystyle \text{when transforming }\,y \:=\: x^2\,\text{ to }\,y\:=\:4(3x -6)^2 + 1$

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

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

$\displaystyle \text{We have: }\:y \:=\:(x\,{\color{red}-\,2})^2$

. . $\displaystyle \text{The graph is moved 2 units to the right.}$

$\displaystyle \text{Then: }\:y \:=\:{\color{red}36}(x-2)^2$

. . The graph is stretched vertically by a factor of 36.

$\displaystyle \text{Finally: }\:y \:=\:36(x-2)^2\, {\color{red}+ 1}$

. . The graph is moved one unit upward.

- Jan 18th 2013, 02:14 PMelmidgeRe: Transformations of quadratic equations
Thank you can you also show me the transalations involved with transforming y = x^2 to y = 9x^2 -6x

- Jan 18th 2013, 03:47 PMProve ItRe: Transformations of quadratic equations
- Jan 18th 2013, 04:03 PMFuryanRe: Transformations of quadratic equations
The vertex of $\displaystyle y = x^2$ is at $\displaystyle (0,0)$

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

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

Can you do that?