With f(x) = x − 2, at x = −2 the graph of |f(x)| reads 4 and the graph of f(|x|) reads 0. Two bars in two different places, and two completely different functions.
The dashed line is the original in every view. Each transformation is read off it, which is the whole method: you are never given a formula to manipulate.
Both start from the same graph and neither needs any algebra. The difference is which axis the folding happens about.
| Graph | What to do to the original | Result |
|---|---|---|
| y = |f(x)| | Anything below the x-axis is reflected up. | Never negative. |
| y = f(|x|) | Delete the left half and replace it with a mirror image of the right half. | Always even. |
Say them in that form and they stop being confusable. Outside the function acts on the output, so it folds about the x-axis. Inside the function acts on the input, so it folds about the y-axis. That is the same inside-and-outside rule as 2.11, applied to a modulus.
For f(x) = x − 2:
At x = −2 the first is 4 and the second is 0. At x = −3 they are 5 and 1. They are equal only for x ≥ 2, where the original was already positive and neither fold did anything.
Count the roots as a check. |x − 2| = 0 at one place, x = 2. |x| − 2 = 0 at two, x = 2 and x = −2, because the right half was copied over. Different numbers of roots is the fastest proof that they are different functions.
y = [f(x)]². Square the heights. So:
Here [f(x)]² = (x − 2)², which is 4 at x = 0 and 1 at both x = 1 and x = 3: the symmetry about x = 2 is forced by the squaring.
y = 1/f(x). Invert the heights. So:
1/f is 0.5 at x = 4 and −0.5 at x = 0, closing on the x-axis in both directions.
f(2x + 1) is the one that catches people. The original f(x) = x − 2 has its root at x = 2. Where is the root of f(2x + 1)?
Set the input equal to the original root: 2x + 1 = 2, so 2x = 1 and x = 0.5.
Not 1. The common answer is 1, from subtracting the b and stopping. You subtract the b and then divide by the a, in that order, because the a was applied last going in and so comes off first coming out.
As a transformation: a translation 1 to the left followed by a horizontal squash by 2, or equivalently squash first then translate by half. Either way, f(2x + 1) = 2x − 1 here, which you can check is zero at 0.5 and −1 at x = 0.
Store f once and define the rest in terms of it. Then the transformed graph really is built from the original rather than retyped, which is the point the sub-topic is making, and a typing error cannot make the two disagree.
When you may use it. Analysis Paper 1 is non-calculator, and these questions are nearly always Paper 1: you are given a sketch of f with no formula at all and asked to sketch a transformation of it. The calculator is for building the habit, not for the exam.
The mark people lose. Drawing y = f(|x|) by reflecting the part below the axis. That is the other graph. The two instructions sound alike and do entirely different things, and a sketch drawn with the wrong one is wrong everywhere on the left. The habit: say out loud which axis you are folding about before you draw a single line. Outside means the x-axis; inside means the y-axis.
Throughout: f(x) = x − 2.
1. Find |f(−2)|, which is the outside version.
2. Find f(|−2|), which is the inside version.
3. Find 1/f(x) at x = 4.
4. At what x is f(2x + 1) equal to zero?
5. How many roots does y = f(|x|) have here, and why?
1 markThe correct part of the original kept unchanged.
1 markThe correct fold, reflection or asymptote.
1 markKey points marked: roots, vertices, asymptotes.
These are sketch marks, so label the features. A V with its point at (2, 0) marked and labelled earns more than a neater V with nothing on it, and an asymptote has to be dashed and labelled with its equation.
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