Learning Objectives
Describe vertical and horizontal shifts, stretches, and reflections.
Map a point on an original graph to its transformed position.
Use transformations to determine a domain and range.
Prerequisites: function notation, coordinate points, and domain/range. Treat a transformation as a change in where an input-output pair appears, rather than memorizing a list of directions without explanation.
1. Changes Outside f Affect Outputs
For y = af(x) + k, multiply each original output by a and then add k. If a > 1, the graph stretches vertically by factor a. If 0 < a < 1, it compresses vertically. If a < 0, there is also reflection across the x-axis. Adding k moves the result up when k > 0 and down when k < 0.
Worked Example: Transform an Output
The point (2, 5) lies on y = f(x). Where does it go on y = −2f(x) + 3?
Its input stays 2. Its output becomes −2(5) + 3 = −7. The new point is (2, −7).
Multiply before adding. Applying the shift first would produce a different graph.
2. Changes Inside f Affect Inputs
For y = f(x − h), the new input x must equal u + h to reproduce the old input u. Thus the graph shifts right by h when h > 0. For y = f(bx), an old input u appears at x = u/b, provided b ≠ 0. The horizontal scale factor is therefore 1/|b|, with reflection across the y-axis if b < 0.
New function | Where (u,v) moves |
|---|---|
f(x) + k | (u, v + k) |
af(x) | (u, av) |
f(x − h) | (u + h, v) |
f(bx), b ≠ 0 | (u/b, v) |
3. Combine the Changes with a Point Rule
For g(x) = af(b(x − h)) + k, with b ≠ 0, a point (u, v) on f maps to (h + u/b, av + k). Solve b(x − h) = u for the new x-coordinate, then transform the output. This avoids ambiguity about the order of horizontal changes.
Worked Example: Read the Inner Expression Carefully
Let g(x) = 2f(3x − 6) + 1. Factor the inner expression as 3(x − 2), so b = 3 and h = 2.
The point (6, −1) on f maps to (2 + 6/3, 2(−1) + 1) = (4, −1).
Check directly: at x = 4, the inner input is 3(4) − 6 = 6, and the output is 2f(6) + 1 = −1.
4. Track Domain and Range
Worked Example: A Transformed Square Root
Find the domain and range of g(x) = −2√(x − 3) + 4. The original square-root input must be nonnegative: x − 3 ≥ 0.
Thus the domain is [3, ∞). The square root's outputs are nonnegative, so multiplication by −2 gives nonpositive outputs, and adding 4 gives values at most 4.
The range is (−∞, 4]. The endpoint (0, 0) of the parent graph becomes (3, 4), matching both interval endpoints.
Common Mistakes
Factor an inner linear expression before reading its horizontal shift.
A horizontal scale uses the reciprocal of the inside multiplier.
Negative vertical multipliers reverse the order of range endpoints; express the final interval from smaller to larger.
5. Practice
1. Where does (4, 3) move under g(x) = f(x − 2) − 5?
2. Where does (6, 2) move under g(x) = −f(2x)?
3. Find the vertex and range of g(x) = 3(x + 1)² − 4 on all real x.
4. Find the domain of g(x) = √(2x + 6).
Worked Solutions
1. The point moves right 2 and down 5, becoming (6, −2).
2. The input coordinate is divided by 2 and the output is negated. The new point is (3, −2).
3. The parent vertex (0, 0) moves to (−1, −4). The positive multiplier preserves upward opening, so the range is [−4, ∞).
4. Require 2x + 6 ≥ 0, giving x ≥ −3. The domain is [−3, ∞).
6. Summary
- Outside operations act directly on outputs.
- Find new input coordinates by solving the inside expression.
- Use a known point to check the transformation.
- Recalculate domain and range rather than copying the parent's intervals.