![]() Step 1: When reflecting over the line \(y = x\), swap the places of the x-coordinates and the y-coordinates of the vertices of the original shape. The steps to follow to perform a reflection over the lines \(y = x\) and \(y = -x\) are as follows: In this case, the x-coordinates and the y-coordinates besides swapping places, they also change sign. The x-coordinates and the y-coordinates of the vertices that form part of the shape swap places. The rules for reflecting over the lines \(y = x\) or \(y = -x\) are shown in the table below: Type of Reflection Step 3: Draw both shapes by joining their corresponding vertices together with straight lines. Step 2: Plot the vertices of the original and reflected images on the coordinate plane. The new set of vertices will correspond to the vertices of the reflected image. Step 1: Following the reflection rule for this case, change the sign of the x-coordinates of each vertex of the shape, by multiplying them by \(-1\). The steps to follow to perform a reflection over the y-axis are as pretty much the same as the steps for reflection over the x-axis, but the difference is based of the on the change in the reflection rule. The y-coordinates of the vertices will remain the same.The x-coordinates of the vertices that form part of the shape will change sign.The pre-image of some subset of my codomain.The rule for reflecting over the y-axis is as follows: Type of Reflection Next video is actually calculate or determine The inverse notation was probably introduced. We can kind of view the imageĪnd the pre-image kind of canceling out, and that's why We can construct a subset of Sīy taking the image of the pre-image of S. Motivation for where the notation comes from. Within S, so they're going to map within S, but they don't Vectors do they map into? All of them are going to be This, we're saying if we take every member of this, what Is essentially the image of this guy right here, right? This part right here is the Pre-image under S? So if we take this guy, this Interesting question, and this is kind of for bonus points. Maps into that subset of our codomain? Now let me ask you an Of our codomain, and we say what subset of our domain Of our domain to a subset of our codomain. This is equal to the pre-image of S under T. Here, the notation is the inverse T of S, but All I'm saying is thatĮverything in this set maps to something within S right here. ![]() Gets mapped to from our transformation T. For example, maybe there's someĮlement in S right there that no element in X ever I'm not saying thatĮvery point in S necessarily gets mapped to. Subtle nuance here to point out something here. ![]() X all map into S? Now, I want to make a very So I'm literally saying whatĪre all of the members of X where those members of Map into these guys, and that's what I'm defining Right here, where if I take any member of this set, it will If I take my domain, there must be some subset of vectors Mapping or the transformation of those vectors ends up Vectors that are members of my domain such that they're Set Y, which is our codomain, So that's Y, and we were to To think about the opposite problem? What if we were to start with Transformations, maybe it's this blob right here, We take each of the members of our subset, it's the set ofĪll of their transformations. To the set- let me write it here- the set of all- where if Image of T of A like that, which is the image of A, of Transform it, you'll associate it with a member of set Y. Transformation that if you take any member of X and you Set that I'm mapping into, set Y, that's the codomain.
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