Two changes in a row
Each of the earlier lessons changes a graph once: a translation, a stretch parallel to the y-axis, or a stretch parallel to the x-axis. An equation such as y = 2f(x) + 1 makes two changes, and the question is which comes first.
Following one point answers it. With , the point (4, 2) is on y = f(x). On y = 2f(x) + 1 the height at x = 4 is 2 × 2 + 1 = 5, so the point ends at (4, 5).
Outside the bracket: in order
Outside the bracket the changes act on the height, and they happen in the order of the arithmetic. 2f(x) + 1 multiplies the height by 2 and then adds 1: a stretch of scale factor 2 parallel to the y-axis, then a translation 1 up.
In the other order the result is different. Translating 1 up gives f(x) + 1, and stretching that gives 2(f(x) + 1) = 2f(x) + 2: the point (4, 2) would end at (4, 6), not (4, 5).
Inside the bracket: backwards
Inside the bracket the changes act on x, and they work backwards. The new graph y = f(2x + 4) has the old height 2 wherever 2x + 4 equals the old input 4.
Solving 2x + 4 = 4 gives x = 0, so (4, 2) moves to (0, 2). In the same way the end point (0, 0) of moves to where 2x + 4 = 0, at x = −2, which is where the domain of begins.
Describing the inside change
Replacing x by x + 4 translates the graph 4 to the left, giving f(x + 4). Then replacing x by 2x is a stretch of scale factor ½ parallel to the x-axis, giving f(2x + 4).
The other order also works, with a different translation: stretch first to get f(2x), then translate 2 to the left to get f(2(x + 2)) = f(2x + 4). Writing the bracket as 2(x + 2) shows that version directly.
The general rule
On y = af(bx + c) + d, a point (p, q) of y = f(x) moves to aq + d). The height goes through ×a then +d, forwards; the x goes through −c then ÷b, which is 2x + 4 = p solved for x.
A negative factor is a reflection: −f(x) reflects the graph in the x-axis, and f(−x) reflects it in the y-axis.
The usual slips
Adding before multiplying. In 2f(x) + 1 the 2 acts first; 2(f(x) + 1) is a different function.
Applying an outside factor to x. A number outside the bracket changes heights only.
Running the inside rule forwards. Putting the old x into 2x + 4 gives the wrong point; solve 2x + 4 = old x instead.
Undoing the + but not the ×. For the point (10, q) on y = f(x), solving 2x + 4 = 10 gives x = 3; subtracting 4 alone gives 6.