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3 Responses
Handy rule of thumb: Outside is obvious, Inside is inverse. “Outside” changes (like +3 or a minus sign in front of f) move the graph up/down and flip over the x-axis exactly as they look; “inside” changes (stuff done to x before it goes into f) do the opposite direction for left/right and can cause a left-right flip. So y = -f(x – 2) + 3: shift the original graph right 2 (because x – 2), flip vertically over the x-axis (because the minus outside), then move up 3. For y = f(-x + 2) – 3, first rewrite the inside as f(-(x – 2)): that shows a horizontal reflection (because of the negative on x) and a shift right 2, then the -3 shifts everything down 3. If you like a one-shot description, f(-x + 2) is f reflected across the vertical line x = 1 (since 2 – x mirrors about x = 1), then shifted down 3. When in doubt, do a quick check with a point: in f(-(x – 2)), the original x = a shows up at x = 2 – a; in -f(x – 2) + 3, the original x = a shows up at x = a + 2. Hope this helps!
Ooh, transformations! I try to decode them piece by piece, but I always second-guess the “inside” bits. For y = -f(x-2)+3, I think x-2 means shift left 2 (since it kind of looks like you’re subtracting from x), the minus out front flips the graph over the y-axis, and then the +3 bumps it up 3-so I’d picture “left 2, reflect across the y-axis, up 3.” For y = f(-x+2)-3, the -x inside makes me think it’s an x-axis flip (because the sign in front of x is changing), and the +2 would push it left 2 again, then the -3 drops it down 3. My quick-and-dirty rule (which I admit I might be mixing up!) is: inside changes behave in the “normal” left/right direction you read, and outside changes do the vertical stuff, with the minus sign telling you which axis it flips over. Does that line up with how you’ve been taught, or do you have a base function (like a parabola or |x|) you’re testing these on that we can compare step by step?
I find the “inside vs. outside” confusion totally normal, so I use a small routine. Outside the f( ), what you see is what you get: multiply by − flips over the x-axis, add/subtract shifts up/down exactly as written. Inside the f( ), think in two steps: first factor the inside into a(x − h), then do the horizontal flip/scale from a, and finally do the horizontal shift by h. Inside changes are “opposite direction” for shifts (x − h means shift right by h), and if a is negative you get a left-right reflection. When in doubt, pick an anchor input u for f (like u = 0 or u = 2) and solve a(x − h) = u to see where that feature moves.
Applying that here:
– y = −f(x − 2) + 3: inside is x − 2, so shift the graph of f right by 2; the outside − reflects over the x-axis; then +3 shifts up by 3. So: right 2, reflect in x-axis, up 3.
– y = f(−x + 2) − 3: factor the inside as −(x − 2). Do the horizontal reflection first (reflect across the y-axis), then shift right by 2, and finally −3 shifts down by 3. A quick check: the input 2 of f appears where −x + 2 = 2, i.e., at x = 0, which matches “reflect then shift.” I’m fairly sure this ordering is the cleanest way to avoid mix-ups, though if there were a horizontal scale (|a| ≠ 1) you’d also include a stretch/compression before the shift.