Thread: Finding the image of the straight line with an equation under a transformation.

1. Finding the image of the straight line with an equation under a transformation.

A transformation is described through the equation TX+B=X^1
where T=⎡4 0⎤ and B= ⎡-1⎤
⎣0 2⎦ ⎣ 4⎦
Find the image of the curve with equation y=-2x^3+6x under the transformation.

I know how to work out these types of questions when the equation is T(X+B)=X^1, but I'm just not to sure how much differently I will need to do the question.

This is what I was doing:
T^-1*T(X+B)=T^-1*X^1
X+B=T^-1*X^1
and X=T^-1*X^1-B
Therefore you then sub in your matrixes. Once you get an equation for x and y, you then sub them into the straight line equation etc.
Does anyone know if that is the right method?

2. I think you mean that T is the matrix
$\begin{bmatrix}4 & 0 \\ 0 & 2\end{bmatrix}$
B is the column matrix
$\begin{bmatrix}-1 \\ 4\end{bmatrix}$
and X is the column matrix
$\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}x \\ -2x^3+ 6x\end{bmatrix}$
but I am not sure what "X^1" is supposed to indicate. Is it X', the new value?

Then
$TX+ B= \begin{bmatrix}4 & 0 \\ 0 & 2\end{bmatrix}\begin{bmatrix} x \\ -2x^3+ 6x}\end{bmatrix}+ \begin{bmatrix}-1 \\ 4\end{bmatrix}$
$= \begin{bmatrix}4x - 1 \\-4x^3+ 12x+ 4\end{bmatrix}$

Now, to write that in the form "y= f(x)", let u= 4x- 1 so that $x= \frac{u+1}{4}$ and $-4x^3+ 12x+ 1= -\frac{(u+1)^3}{16}+ 3(u+1)+ 1$
and the column matrix is
$\begin{bmatrix} u \\ -\frac{(u+1)^3}{16}+ 3(u+ 1)+ 1\end{bmatrix}$

and so $y= f(x)= (-\frac{x+1)^3}{16}+ 3(x+ 1)+ 1= -\frac{1}{16}x^3- \frac{3}{16}x^2+ \frac{45}{16}x+ \frac{15}{16}$.

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how to find eqiation of image line

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