I’m revising graph transformations and I keep tripping over the horizontal ones, especially when there’s more than one thing happening inside the brackets.
Starting from y = f(x), I want to sketch y = -2 f(3(x – 1)) + 4. Here’s how I’m thinking about it, but I’m not confident:
– Inside f: 3(x – 1) should mean a horizontal compression by factor 3 and a shift right by 1. But I’m unsure about the order: do I shift right 1 first, then compress by 3, or compress first, then shift? I’ve heard that inside transformations are “reversed” or that the order matters differently from the outside ones, and that’s where I get stuck.
– Outside f: -2 and +4 seem clearer – reflect in the x-axis and stretch vertically by 2, then shift up 4.
To test myself, I tried mapping a point. If f(2) = -1, then I set 3(x – 1) = 2, so x = 1 + 2/3 = 5/3, and the corresponding y-value would be -2 * (-1) + 4 = 6. So I think (5/3, 6) lands on the new graph. That feels right, but I’m second-guessing because of the horizontal order issue.
Could someone explain the correct order for the horizontal transformations here, and a reliable step-by-step way to map points from y = f(x) to y = -2 f(3(x – 1)) + 4? More generally, for something like y = f(a(x – b)), how do I systematically handle the “inside” part without getting mixed up?
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3 Responses
Great question – the “inside” moves can feel like you’re steering in a mirror! The reliable rule is: to see what happens horizontally, ask “what x would f normally see?” For y = -2 f(3(x – 1)) + 4, set u = 3(x – 1). That’s the input the original f sees. To map points from the base graph, invert this: x = u/3 + 1. So each original point (u, f(u)) becomes (u/3 + 1, -2 f(u) + 4). In other words, horizontally you first divide x by 3 (a compression by factor 3 toward the y-axis), then shift right by 1; vertically you reflect in the x-axis and stretch by 2 (because of -2), then shift up 4. Your test is spot on: since f(2) = -1, the new point is at x = 2/3 + 1 = 5/3 and y = -2(-1) + 4 = 6, so (5/3, 6) lies on the transformed graph.
A small analogy I like: imagine the graph is a rubber picture. The inside 3(x – 1) says “squeeze it 3× narrower” (divide x-coordinates by 3), then “slide the whole picture 1 unit to the right.” Only after that do you apply the outside edits: flip it across the x-axis, stretch it taller by 2, and finally lift it up by 4. The reason the horizontal order feels reversed is that you’re undoing the inside map to figure out which original x produced the same f-value – that’s why x = u/3 + 1 (not 3u − 1).
General recipe you can reuse: for y = A f(a(x − b)) + c, every base point (u, f(u)) goes to (b + u/a, A f(u) + c). So horizontally it’s “scale by 1/a, then shift by +b”; vertically it’s “multiply by A, then add c.” A nice walkthrough with more examples: https://tutorial.math.lamar.edu/classes/alg/graphingtransforms.aspx
A handy way to remember it: “inside = undo,” so for y = -2 f(3(x – 1)) + 4 you first compress horizontally by 3 then shift right 1 (since 3(x – 1) = u ⇒ x = u/3 + 1), then reflect in the x-axis and stretch by 2 (multiply y by -2), and finally shift up 4. Example: if f(2) = -1, then solve 3(x – 1) = 2 to get x = 5/3 and y = -2(-1) + 4 = 6, so (5/3, 6) is on the new graph; in general for y = f(a(x – b)), set a(x – b) = old_x so x_new = old_x/a + b (nice refresher: https://www.khanacademy.org/math/algebra2/functions-and-graphs/transformations-of-functions/v/transformations-of-functions).
Inside stuff runs “backwards”: for y = -2 f(3(x − 1)) + 4, take any point (u, f(u)) and send it to (1 + u/3, -2 f(u) + 4)-in general y = f(a(x − b)) maps (u, f(u)) to (b + u/a, f(u)), i.e., divide x by a first, then shift by b. Example: if f(2) = −1, then (2, −1) → (1 + 2/3, −2(−1) + 4) = (5/3, 6); more details here: https://www.khanacademy.org/math/algebra2/manipulating-functions/transformations-of-functions/a/transformations-of-functions.