The
Purplemath Forums 
Function Transformations / Translations (page 4 of 4) Sections: Basic rules, Additional rules, Moving the points, Working backwards from the graph The other exercise type is when you are given two graphs, one being the original function and the other being the transformed function, and you're asked to figure out the formula for the transformation.
Comparing the two graphs (and concentrating on the simplest point to keep track of, the vertex), it looks like the original graph was moved up three units (so there is a "+3" outside) and shifted right two units (so there is a "–2" inside). Then the transformation is: f(x – 2) + 3
This one is harder than the previous one, because the graph is mirrored across the yaxis in the transformation. This means there is a "minus" sign on the variable inside the argument. Also, however far a point in the transformation is to the right of the yaxis now, it was that far to the left of it before the mirroring, and vice versa.
I'll keep track of things by picking one point that is off the axes and tracing what it must have done. Then I'll check my work by picking some other point, and making sure that my formula works for that point, too. To start, I'll use the bottom point of the original graph, (2, –1). The point (2,
–1) moves to (1,
–1), and the function
does a flip around the yaxis.
There doesn't appear to be any up/down movement, though. Before the
flip around the axis, (1,
–1) must have been
at (–1, –1).
This means that (2,
–1) was shifted back
three units, so there must be a "+3"
inside the argument. That gives me f(x
+ 3) before the flip
across the axis. The flip requires a "minus" sign on the variable,
so then the transformation must be f(–x
+ 3). Another way to look at this is to flip the graph first, putting a "minus" on the variable. But flipping first moves the graph too far off to the left, taking (5, 2) to (–5, 2), so I have to follow this up with a shift back to the right by three units: f(x) to f(–x) to f(–(x – 3)) = f(–x + 3). Before I hand this in, I'll check to make sure that this moves other points the right way, too. Looking at (3, 1), shifting three left takes the point to (0, 1); the flip around the yaxis leaves the point unchanged. Looking at (5, 2), shifting back three takes the point to (2, 2) and flipping around the yaxis takes it to (–2, 2). So the other points check as well; it appears that I have found the correct transformation. f(–x + 3) In practice, about the only time you'll actually use function transformations (before you get to trigonometry) is with conics; that is, with parabolas and such. For instance, since you know what the basic quadratic x^{2} looks like, you now know that (x + 3)^{2} – 2 is the same parabola shape, but with the whole graph (and, in particular, the point at the origin) having been moved back three units and down two so, in particular, the vertex is now at (–3, –2) in the third quadrant. In a somewhat related way, given that x^{2} + y^{2} = r^{2} is the equation of a circle with radius r centered at the origin, the equation (x + 1)^{2} + (y – 1)^{2} = 4, for instance, would be the equation of the circle with radius 2 and centered at (–1, 1). Of course, a circle equation isn't a function, so this doesn't technically fall under the heading of "function transformations", but you should note the similarity. I think the point that they want you to get is that certain types of equations always have certain kinds of shapes (ax^{2} + bx + c is always a parabola, etc). In practice, though, you won't be using function transformation very much. Make sure you know how to do it for the test (the questions are easy points, once you "get" how they work), but don't worry if you forget all about it later. If you need the subject again (for instance, in trigonometry), they'll cover it again. << Previous Top  1  2  3  4  Return to Index



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