In my last post, I took a general point (x,y) and through the series of transformations:
A. (x,y)
B. Dilate/reflect by factor a along y-axis
C. Dilate/reflect by factor n along x-axis
D. Translate k units along y-axis
E. Translate h units along x-axis
the point was changed to:
A(x, y) —-> B(x, ay) —-> C(nx, ay) —-> D(nx, ay+k) —->
E(nx+h, ay+k).
Now let’s restrict the original point to be one that satisfies the equation y = x2. How does this equation change so that all of its points are transformed correctly?
Think of the transformation process as changing old points to new ones. So the old points, (x, y), are changed to the new points (nx+h, ay+k) = (x‘, y‘), where the apostrophe ‘ is used to distinguish the new x‘s and the new y‘s from the old ones. So under this transformation, x‘ = nx+h and y‘ = ay+k.
Now the equation y = x2 is the equation that the old points satisfy. To find the equation that the new points satisfy, we need to find what the old variables are in terms of the new ones. Solving for the old x and y in the above equations, we get
\[x’=nx+ h\Longrightarrow x = \frac{1}{n}\left(x’-h\right)\]
\[y’=ay+ k\Longrightarrow y = \frac{1}{a}\left(y’-k\right)\]
If we replace the old variables in the equation with the right side of the above equations, then we will get an equation with the new variables, which can be cleaned up with a little algebra:
\[\frac{1}{a}(y’-k)=\left[\frac{1}{n}(x’-h)\right]^2\Longrightarrow y’=a\left[\frac{1}{n}(x’-h)\right]^2+k\]
Now that we have the new equation, we don’t need the ‘ anymore. So the new equation is:
\[y=a\left[\frac{1}{n}(x-h)\right]^2+k\]Notice that if we were given this new equation and were asked, “what are the transformations that generated this equation from y = x2 ?”, then we can pick off the dilations and the translations. Be aware though, that we must assume that the dilations occurred first. Textbooks will tell you this to keep answers consistent. With this caution, if a = 2, n = 3, h =-5, and k =4, then the new equation would be
\[y=2\left[\frac{1}{3}(x+5)\right]^2+4\]The graph below shows the effect of these transformations on the original curve:
So given a set of transformations, the steps that I did above for the particular equation y = x2, can be done for any equation.
What if the problem is reversed: given the end result, what is the set of transformations that created the new equation from and original one? Using functional notation, notice that:
\[f(x)={x}^{2}\Longrightarrow af\left[\frac{1}{n}(x-h)\right]+k=2{\left[\frac{1}{3}(x+5)\right]}^{2}+4\]If you were given this final result, you can pick off the dilations along both axes (These are assumed to be first) and the translations along both axes. If either the a and/or the n were negative, then there would be reflections as well.