How to do this question in regards of matrices and transformation? Feb 3, 2018

$\frac{y ' - 4}{2} = - 2 {\left(\frac{x ' + 1}{4}\right)}^{3} + 6 \left(\frac{x ' + 1}{4}\right)$

or simplified

$y = - {x}^{3} / 16 - \frac{3 {x}^{2}}{16} + \frac{45 x}{16} + \frac{111}{16}$

Explanation:

From the matrices, the function is going to be dilated by a factor of 4 from the y-axis, 2 from the x-axis, translated 1 unit to the left and 4 units up.

Do the matrix multiplication first

$A X = \left(\begin{matrix}4 & 0 \\ 0 & 2\end{matrix}\right) \left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}4 x \\ 2 y\end{matrix}\right)$

$A X + B = \left(\begin{matrix}4 x \\ 2 y\end{matrix}\right) + \left(\begin{matrix}- 1 \\ 4\end{matrix}\right) = \left(\begin{matrix}4 x - 1 \\ 2 y + 4\end{matrix}\right)$

Put it together to get

$X ' = A X + B = \left(\begin{matrix}4 x - 1 \\ 2 y + 4\end{matrix}\right)$

$X ' = \left(\begin{matrix}x ' \\ y '\end{matrix}\right)$

$\left(\begin{matrix}x ' \\ y '\end{matrix}\right) = \left(\begin{matrix}4 x - 1 \\ 2 y + 4\end{matrix}\right)$

Now that we have expressions for x' and y', rearrange them to solve for x and y.

$x ' = 4 x - 1 \Rightarrow x = \frac{x ' + 1}{4}$

$y ' = 2 y + 4 \Rightarrow y = \frac{y ' - 4}{2}$

Finally, substitute the expressions for x and y into the original function to get the image function

$y = - 2 {x}^{3} + 6 x$

$\Rightarrow \frac{y ' - 4}{2} = - 2 {\left(\frac{x ' + 1}{4}\right)}^{3} + 6 \left(\frac{x ' + 1}{4}\right)$