Transformations of Graphs
Describe and apply the effect of f(x) + a, f(x + a), -f(x) and af(x) on a graph, and give the coordinates of a marked point after the change.
What a learner can do afterwards
- Explain the direction of horizontal shifts such as f(x + 3), translate allowed-input domains, and identify what translations preserve or undo
- Describe shifts, reflections and positive scale factors; track marked points, including turning points; compare transformed outputs and recover transformation parameters
- Match transformed sketches or stated graph features to equations, and identify equivalent transformed equations
1 · Read
In the last stop, swapping a point's coordinates reflected an inverse graph in y = x. Here you track points through translations, scaling and reflections in the coordinate axes. A point on y = f(x) has coordinates (u, v), where v = f(u). Read the new rule to find its image. These moves apply to many different graph shapes.
For f(x) + a, (u, v) becomes (u, v + a): positive a moves up; negative a moves down. For af(x), it becomes (u, av). Distances from the x-axis scale by |a|, the magnitude of a. Above 1 stretches; between 0 and 1 compresses; 1 leaves distances unchanged. Negative a also reflects in the x-axis, reversing highest and lowest points. An output of 0 stays 0. At a = 0, all defined outputs become 0; excluded inputs stay excluded.
For f(x + a), solve x + a = u to keep the old output v. The image is (u - a, v). Positive a moves left; negative a moves right; zero leaves points unchanged. Thus f(x - a) moves right by a when a is positive. Horizontal translations preserve y-values, highest and lowest values, and horizontal distances. Every allowed input moves by the same amount. Equal opposite translations cancel: subtract a from x, then add a back.
Suppose f has its lowest point at (2, -5). Under f(x) + 2, it becomes (2, -3) and remains the lowest point. Under -f(x), it becomes (2, 5) and is the highest point. Translations preserve the graph's shape. Reflections reverse its orientation, while keeping distances unchanged.
A vertical reflection changes y's sign and leaves x unchanged.
Suppose f passes through (4, -4) and has its highest point at (1, -1). Under -2f(x), these become (4, 8) and (1, 2). The highest point becomes the lowest. Distances from the x-axis double, and the graph reflects in that axis. Adding a constant afterwards shifts every image by the same amount, so the new lowest point remains the lowest.
The scale factor is 2. The negative sign supplies the reflection.
Combine moves by tracking each coordinate. Under Af(x - h) + k, (u, v) becomes (u + h, Av + k). Multiply y before adding k. A turning point at (-3, 4) becomes (-3, 9) under 2f(x) + 1. But 2(f(x) + 1) gives y = 2(4 + 1) = 10. Expanding those brackets gives 2f(x) + 2, so the two rules differ.
Read the brackets to decide the order of the output changes.
For f(ax), with a nonzero, solve ax = u. The image of (u, v) is (u/a, v). Distances from the y-axis scale by 1/|a|. Negative a also reflects in the y-axis. For example, (6, 9) becomes (2, 9) under f(3x). Under f(-x), it becomes (-6, 9). At a = 0, the new rule is the constant f(0) if that output exists; otherwise it has no allowed inputs.
Horizontal scaling divides x-coordinates by the input multiplier.
Recover a point by undoing coordinate changes when scaling is nonzero. Under 3f(x - 2) - 4, image (1, 11) came from (-1, 5): x = 1 - 2 and y = (11 + 4)/3. Zero output scaling loses the original y. To find A in Af(x), solve v' = Av; divide by v when v != 0. If (2, 6) becomes (2, 20) under 3f(x) + k, solve 18 + k = 20 to get k = 2.
Undo output changes in reverse order.
Substitute the rule for f. If f(x) = x^2, then Af(x - h) + k = A(x - h)^2 + k. When A != 0, its turning point is (h, k). One unit from x = h, its y-value is k + A, which helps identify A. When A = 0, the graph is the constant y = k and has no turning point. For f(x) = x^3, substitution gives A(x - h)^3 + k. Track points by the coordinate rule even when the graph has no turning point.
Use the turning point and another point to distinguish equations.
Absolute value, |x|, is the distance of x from zero. It gives a V-shaped graph with vertex (0, 0). Transform its points using the same coordinate rules. Also, |cv| = |c||v|. For example, 2|x + 1| = |2x + 2|. These two equations describe the same graph, even though one puts the multiplier outside and the other puts it inside.
Different-looking equations can describe the same transformed graph.
A domain is the set of allowed inputs. A horizontal translation moves both ends of a restricted domain. If f is defined for -2 <= x <= 6, then f(x + 1) needs -2 <= x + 1 <= 6. Subtract 1 throughout: -3 <= x <= 5. The outputs stay the same. A vertical translation leaves the domain unchanged.
Allowed inputs move with the points in a horizontal translation.
Compare two vertical translations at the same allowed input. If g(x) = f(x) + r and h(x) = f(x) + s, then g(x) - h(x) = r - s: the common f(x) cancels. For example, f(x) + 8 is always 7 units above f(x) + 1. You do not need to know the original output.
Subtract corresponding outputs to compare translated graphs.
Track each coordinate. Added outputs shift y; added inputs shift x in the opposite direction. Scaling uses positive magnitudes, with reflection separate. Read brackets before combining moves.
The lesson, practice questions and flashcards were written for this dataset.
The extra questions and reading were written for this dataset.
More to read
Track the coordinates
Write an original point as (u, v), with v = f(u). You can track its image without knowing a formula for the whole graph.
For g(x) = Af(x - h) + k, the image is (u + h, Av + k). The horizontal change adds h to the point's x-coordinate. The output change multiplies v by A, then adds k. Brackets can change that order: A(f(x) + k) multiplies the added k as well.
Geometric scale factors are positive. A negative output multiplier has two effects: its magnitude scales vertical distances, and its sign reflects the graph in the x-axis. A magnitude of 1 leaves those distances unchanged. A multiplier of 0 sends every defined output to 0, but does not make an excluded input valid.
For g(x) = f(Ax), with A nonzero, the image is (u/A, v). Solve Ax = u to find the new input. The horizontal scale factor is 1/|A|, and a negative A adds reflection in the y-axis.
To sketch, track several points and useful features, such as a turning point. A horizontal translation preserves the set of output values. A negative output multiplier swaps highest and lowest values.
You can recover an original point from its image by undoing invertible coordinate changes in reverse order. For the image (p, q) under Af(x - h) + k, this gives u = p - h and v = (q - k)/A when A != 0. If A = 0, every original output maps to k. The image alone cannot tell you the original output v.
2 · Watch
En anglais3 · Play
Add 3 outside the bracket, then inside it, and watch the curve go two different ways.
Take it off screen
Where it sits
Where this leads
Jobs that lean on this skill. Follow one to see everything it is built on.
36 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.