The last two easy transformations involve flipping functions upside down (flipping them around the x-axis), and mirroring them in the y-axis.
The first, flipping upside down, is found by taking the negative of the original function; that is, the rule for this transformation is –f (x).
To see how this works, take a look at the graph of h(x) = x2 + 2x – 3. The graph of the original function looks like this:
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To imagine this graph flipping upside-down, imagine that the graph is drawn on a sheet of clear plastic that has been placed over a drawing of just the y-axis, and that the x-axis is a skewer stuck through the sheet. To flip the graph, turn the skewer 180°. (Pictures here.) Then the new graph, being the graph of –h(x), looks like this:
Flipping a function upside-down always works this way: you slap a "minus" on the whole thing. The "flipping upside-down" thing is, slightly more technically, a "mirroring" of the original graph in the x-axis. If you think of taking a mirror and resting it vertically on the x-axis, you'd see (a portion of) the original graph upside-down in the mirror. When they talk about "mirroring" or "reflecting" in or about an axis, this is the mental picture they have in mind.
To keep straight what this transformation does, remember that f (x) is the exact same thing as y. So, by putting a "minus" on everything, you're changing all the positive (above-axis) y-values to negative (below-axis) y-values, and vice versa. (Any points on the x-axis stay right where they are. It's only off-axis points that move.)
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The previous reflection was a reflection in the x-axis. This leaves us with the transformation for doing a reflection in the y-axis.
For this transformation, I'll switch to a cubic function, being g(x) = x3 + x2 – 3x – 1.
Here's the graph of the original function:
If I put –x in for x in the original function, I get:
g(–x) = (–x)3 + (–x)2 – 3(–x) – 1
= –x3 + x2 – (–3x) – 1
= –x3 + x2 + 3x – 1
This new function graphs like this:
This transformation rotated the original graph around the y-axis. Any points on the y-axis stay on the y-axis; it's the points off the axis that switch sides. This is always true: g(–x) is the mirror image of g(x); plugging in the "minus" of the argument gives you a graph that is the original reflected in the y-axis.
To keep straight what this transformation does, remember that you're swapping the x-values. Whatever you'd gotten for x-values on the positive (or right-hand) side of the graph, you're now getting for x-values on the negative (or left-hand) side of the graph, and vice versa. Since the inputs switched sides, so also does the graph.
They've given me Graph A:
...and Graph B:
Comparing Graphs A and B with the original graph, I can see that Graph A is the upside-down version of the original graph. It's been reflected across the x-axis. This means that it's the "minus" of the original function; it's the graph of –f (x).
Graph B has its left and right sides swapped from the original graph; it's been reflected across the y-axis. That means that this is the "minus" of the function's argument; it's the graph of f (–x).
Graph A represents –f (x)
Graph B represents f (–x)
Well, "appropriately" is a little vague; I'll just be sure the label everything very clearly.
I need to find the simplified functional statements for each of the reflections. One of the reflections involves putting a "minus" on the function; the other involves putting a "minus" on the argument of the function. So I'll do each of these.
First up, I'll put a "minus" on the argument of the function:
f (–x) = (–x)2 + (–x) – 3
= x2 – x – 3
Now for the other reflection:
–f (x) = –(x2 + x – 3)
= –x2 – x + 3
Putting a "minus" on the argument reflects the graph in the y-axis. Putting a "minus" on the whole function reflects the graph in the x-axis. So my (clearly labelled) answer is:
reflection in the y-axis:
f (–x) = x2 – x – 3
reflection in the x-axis:
–f (x) = –x2 – x + 3
Many textbooks don't get any further than this. If these are all the rules you need, then write 'em down and make sure you've done enough practice to be able to keep them straight on the next test: