Question

How to do this Q.13 question regarding matrices and transformations ?

Answer 1

`color(blue)(y=-4/27x^3-16/9x^2-29/9x-94/27)`

Explanation:

All points on the line `y=2x^3+6x^2+2` will be transformed by the matrix

`bb(A)=((-3,0),(0,-2))`

and then translated by the matrix `bb(B)=((-1),(2))`

Any point on the line will have coordinates of the form. `(k,-2k^3+6k^2+2)`

Thus:

`((x'),(y'))=((-3,0),(0,-2))((k),(-2k^3+6k^2+2))+((-1),(2))`

`color(white)(888888)=((-3k+0),(0-2(-2k^3+6k^2+2)))=((-3k),(4k^3-12k^2-4))+((-1),(2))`

`=((-3k-1),(4k^3-12k^2-2))`

So:

`x'=-3k-1`

`y'=4k^3-12k^2-2`

Eliminating `bbk`

`x=-3k-1=>k=(x+1)/-3`

`y=4((x+1)/-3)^3-12((x+1)/-3)^2-2`

`=-4/27x^3-16/9x^2-29/9x-94/27`

So the image of `y=-2x^3+6x^2+2` under the transformation followed by the translation is:

`color(blue)(y=-4/27x^3-16/9x^2-29/9x-94/27)`

We can test this to verify it is the image.

Generate an `bbx` and corresponding `bby` from original equation. Put theses through the transformation followed by the translation.

For: `x=2` , `y=10`

`((-3,0),(0,-2))((2),(10))+((-1),(2))=((-7),(-18))`

Using the unexpanded form:

`y=4((x+1)/-3)^3-12((x+1)/-3)^2-2`

Plugging in `-7`

`y=4(((-7)+1)/-3)^3-12(((-7)+1)/-3)^2-2=-18`

So the transformed point does lie on the curve.

In previous transformations of straight lines we have had an alternate method of finding the image equation, .i.e. by generating two points from the transformation and using these to find the equation of the image line. It is possible to find the equation of this problem by transforming points, but it would take four points and the simultaneous solving of four equations, not really practical in this situation.

Important

Do not make the mistake of including the translation before the transformation takes place.