So what I got was that they want me to reverse everything I think not sure..
R_x-axis:Triangle ABC = T(x,y) to (x, y-3)
The points given for the image of the following are
$$A''(2,5) B''(-4,2) C''(-1,-6) $$
So the first thing i thought i would do was to reverse $$(x,-y)$$
As That's the way you do the reflection across So
The first points i did
$$ A(2,-5)B(-4,-2) C (-1,6)$$
I got this as I thought if the point was (2,5) after the reflection all i had to do was reverse it so i thought the original was (2,-5) so i thought the negative was cancelled when you did the transformation.
than i did the other part of the question as asked
$$A(2,-2) B(-4,5) C(-1,-3) $$ the same way i got the first part of the question, just reversed what the question say so (y-3) than that means i just have to go 3 units up to reverse? But the points i got are wrong
In the terminology of mathematics, the pre-image of a set $X$ under a function $f$ is the set of points $Y$ such that $f(Y)$ is in $X$. In other words, you were given
and are asked to find the set of all inputs which would be turned into the given outputs.
In your particular case, you've got a very simply function - it's surjective (that is, every possible point in 2D space is an output from at least one input) and injective (every point is the output from at most one input). So your function has an inverse, and finding the preimage of a set is a simple matter of finding the inverse of your transformation rule and applying that to your set. The transformation rule $T$ takes points $(x, y)$ and maps them to $(x, y - 3)$. To reverse this, you would indeed map $(x, y)$ to $(x, y + 3)$.