Changing a formula in four standard ways moves its graph in four predictable ways. Everything done outside the function acts on the output and behaves as written; everything done inside acts on the input and behaves in the opposite direction, all captured by one point map.
Core ideas
The graph of f is supplied. The task is not to know a future family of curves; it is to send known points to their new coordinates and use that mapping to explain the four familiar moves.
In 2f(x) + 5 the function runs first and the result is doubled and raised by 5. Output changes are vertical, and they do what the symbols suggest: +5 is up, ×2 is taller, a minus sign flips the picture over the horizontal axis.
In f(x − 3) the input is altered before f sees it. The new graph at x = 5 shows the old value at 2, so the whole picture moves right by 3, not left. Every horizontal effect is inverted this way, and it catches everyone the first time.
f(2x + 6) is not a shift by 6. Writing it as f(2(x + 3)) shows a squash by 1/2 and a shift left by 3. Until b is factored out, the number next to it is not the shift.
The material
Why inside and outside behave differently, where a known point lands, and how the four constants follow from that calculation.
A parent function here is simply the supplied reference graph y = f(x). The core examples use a line, |x| and x³, whose arithmetic is already available. The lab also offers root and reciprocal previews, but those families are not part of this lesson's completion standard.
This is the one point in the lesson that repays a proof rather than a rule. Nothing about the graph of f changes; what changes is which input is asked for.
Let g(x) = f(x − 3). Take any point (t, f(t)) on the graph of f. Ask where g takes that same value: g(x) = f(t) as soon as x − 3 = t, which is x = t + 3.
So the point (t, f(t)) on the graph of f corresponds to the point (t + 3, f(t)) on the graph of g. Same height, input three larger. Every point moves three to the right, so the graph does.
The subtraction inside produces an addition to the coordinate because the equation x − 3 = t is solved for x. That single rearrangement is the entire reason horizontal effects are inverted, and it applies equally to the stretch: g(x) = f(2x) takes the value f(t) at x = t/2, so distances from the vertical axis are halved.
□
The two reflections are worth separating. −f(x) negates the output and turns the picture upside down; f(−x) negates the input and turns it left to right. For f(x) = x + 1, they are −x − 1 and −x + 1, visibly different parallel lines. At the parent point (0, 1), the vertical flip lands at (0, −1) while the horizontal flip leaves (0, 1) fixed. One point separates the two readings without requiring a new curve family.
Take g(x) = −2|x − 3| + 5, in the template with parent f(x) = |x|, a = −2, b = 1, h = 3, k = 5. Move each parent point by (x₀, y₀) → (x₀ + 3, −2y₀ + 5). The corner (0, 0) lands at (3, 5); the two points (−1, 1) and (1, 1) land at (2, 3) and (4, 3). Those three points determine the transformed V without any future quadratic technique.
The coordinate map makes the order explicit. The new x-coordinate is found from the inside equation b(x − h) = x₀, giving x = x₀/b + h; the new y-coordinate is found after the parent has run, as ay₀ + k. Mixing those routes — adding k to an x-coordinate or dividing a y-coordinate by b — is not a small slip but a change of which quantity each constant acts on.
Let f(x) = x³ and r(x) = f(2x + 6). The 6 is not the shift. Factoring gives f(2(x + 3)), so b = 2 and h = −3. A parent point with input x₀ lands at x₀/2 − 3: the graph is half as wide and moved left by 3.
The point calculation settles both horizontal constants at once. Reading 6 as a shift left 6 would send (0, 0) to (-6, 0), but direct substitution shows r(−6) = f(−6), not f(0). The factored 3 and the division by 2 are not a mnemonic; they are the solution of 2x + 6 = x₀ for the new input.
If (4, −2) lies on y = f(x), then on y = 3f(x − 5) + 1 it becomes (9, −5): the inside equation x − 5 = 4 gives x = 9, while the outside calculation 3(−2) + 1 gives −5. Checking one mapped point forces each constant to act on the coordinate it actually controls.
Interactive
a f(b(x - h)) + k, with the parent behind it
Method
Identify the parent, factor the input, then read the constants in a fixed order.
Identify the supplied reference graph y = f(x). For assessed work it will be a line, |x| or x³; the lesson is about moving its known points, not recalling an unlearned family shape.
Turn f(2x + 6) into f(2(x + 3)) before reading anything horizontal. Skipping this is the single most common source of a wrong shift, because 6 is sitting there looking like the answer.
a scales the output and flips it when negative; k raises it. These need no reinterpretation, which is why doing the vertical part first builds confidence for the horizontal part.
For a parent point (x₀, y₀), solve b(x − h) = x₀ to get x = x₀/b + h, then compute y = ay₀ + k. Substitute that new x into the transformed formula; the result must be the mapped y.
The first maps three landmarks of |x|; the second factors the input before moving three points of x³.
The first point map handles all four constants without asking the learner to picture four moves at once. Equal parent heights at −1 and 1 become equal image heights at 2 and 4, a built-in check.
The second formula shows why factoring comes before naming h. The raw 6 disappears into 2(x + 3), and each direct input check confirms the half-width and the shift left by 3.
For a faded point map, suppose (2, −1) is on y = f(x) and q(x) = 3f(2(x + 1)) − 4. The supplied inside equation is 2(x + 1) = 2. Solve for the new x, compute the new y, and verify the mapped point by substituting its x into the displayed rule.
Given a supplied graph and y = a f(b(x − h)) + k, you should factor the input, map at least three known points with (x₀/b + h, ay₀ + k), and verify one in the displayed rule. Given f(4) = −2, for example, you should locate the matching point on 3f(2(x + 1)) − 4 without needing to know any new parent-function family.
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