AP® Precalculus review sheet from Aim for Five (aimforfive.com/precalc/units/1/1-12)
Unit 1 · Topic 1.12
1.12 Transformations of Functions
Transformations change a function's graph in predictable ways. Adding shifts it, multiplying stretches or shrinks it, and a negative sign reflects it. Changes outside the function act on outputs (vertically), and changes inside act on inputs (horizontally, and in the opposite way you might expect).
Key terms
- translation
- dilation
- reflection
- additive transformation
- multiplicative transformation
Additive transformations: shifts
g(x) = f(x) + k shifts the graph of f vertically by k units: up if k > 0, down if k < 0.
g(x) = f(x + h) shifts the graph horizontally by −h units. So f(x + 3) moves the graph 3 units left, and f(x − 3) moves it 3 units right. Inside changes undo what they look like, because you need x + 3 to equal the old input, which happens 3 units earlier.
Multiplicative transformations: dilations and reflections
g(x) = a·f(x) with a ≠ 0 is a vertical dilation by a factor of |a|. Every output is multiplied by a. If |a| > 1 the graph stretches away from the x-axis; if |a| < 1 it shrinks toward it. If a < 0, the graph is also reflected over the x-axis.
g(x) = f(bx) with b ≠ 0 is a horizontal dilation by a factor of 1/|b|. So f(2x) squeezes the graph toward the y-axis by a factor of 1/2, and f(x/3) stretches it by a factor of 3. If b < 0, the graph is also reflected over the y-axis.
Combining transformations
A common combined form is g(x) = a·f(b(x − h)) + k. A point (x, y) on f moves to (x/b + h, a·y + k) on g.
- Horizontal: divide the x-coordinate by b, then add h.
- Vertical: multiply the y-coordinate by a, then add k.
Inside order and factoring
If the inside is written as f(bx + c), factor out b first: f(b(x + c/b)). The shift is c/b, not c. A reliable alternative is to solve: set the inside equal to the old input and solve for x.
The original graph is called the preimage and the transformed graph is the image. Transformations can change the domain and range, so recompute them using the same rules.
| Change | Effect on the graph of f |
|---|---|
| f(x) + k | Shift up k (down if k < 0) |
| f(x + h) | Shift left h (right if h < 0) |
| a·f(x) | Vertical dilation by the absolute value of a; reflect over x-axis if a < 0 |
| f(bx) | Horizontal dilation by 1 over the absolute value of b; reflect over y-axis if b < 0 |
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Mapping a point
The point (6, 5) is on the graph of f. Find the corresponding point on the graph of g(x) = −2f(3(x − 1)) + 4.
Show the solutionHide the solution
- Step 1: Horizontal: the inside 3(x − 1) must equal the old input 6. So x − 1 = 2 and x = 3. (Same as 6 ÷ 3 + 1 = 3.)
- Step 2: Vertical: the new output is −2 times the old output plus 4: −2(5) + 4 = −6.
- Step 3: Check: g(3) = −2f(3·2) + 4 = −2f(6) + 4 = −2(5) + 4 = −6.
Answer: (3, −6)
- Example 2
Domain and range of a transformation
f has domain [−3, 6] and range [0, 5]. Find the domain and range of g(x) = −2f(3(x − 1)) + 4.
Show the solutionHide the solution
- Step 1: Domain: the inside 3(x − 1) must be in [−3, 6]. Divide by 3: x − 1 is in [−1, 2]. Add 1: x is in [0, 3].
- Step 2: Range: multiply the outputs [0, 5] by −2 to get values from −10 to 0. The reflection flips the order, so the interval is [−10, 0].
- Step 3: Add 4: [−6, 4].
Answer: Domain [0, 3]; range [−6, 4].
- Example 3
Trap: shifting before factoring
The point (4, 1) is on the graph of f. Find the matching point on g(x) = f(2x + 6).
Show the solutionHide the solution
- Step 1: A common mistake: “compress by 1/2, then shift left 6,” giving 4/2 − 6 = −4.
- Step 2: Factor the inside: 2x + 6 = 2(x + 3). The graph is compressed by a factor of 1/2 and shifted left 3, not 6. That gives 4/2 − 3 = −1.
- Step 3: Or solve directly: set 2x + 6 = 4, so x = −1. The output is f(4) = 1.
Answer: (−1, 1)
Common mistakes
- Shifting the wrong way horizontally. f(x − 4) moves the graph right 4, not left.
- Using the shift from f(bx + c) without factoring out b. The shift is c/b.
- Treating f(2x) as a stretch. Multiplying the input by 2 squeezes the graph horizontally by a factor of 1/2.
- Forgetting that a negative a flips the range interval when you multiply.
On the exam
- Expect to match a transformed function with its graph, or to write the new function from a description like “vertical stretch by 3, then shift down 2.”
- Transformations come back throughout the course: exponential, log, sinusoidal and tangent functions all use the same rules. Learn them once here.
Connected topics
Videos
Check yourself
4 questions on 1.12 Transformations of Functions. Pick an answer to see if you got it, and why.
The point (4, 5) is on the graph of the function f. The function g is defined by g(x) = −2f(x − 3) + 1. Which point must be on the graph of g?
The function f has domain [−2, 6] and range [1, 5]. The function g is defined by g(x) = 3f(2x) − 1. What are the domain and range of g?
The graph of g is obtained from the graph of f by a horizontal dilation by a factor of 3, followed by a reflection over the x-axis. Which of the following could define g?
The function f has exactly two zeros, x = −2 and x = 6. Let g(x) = f(2x + 4). What are the zeros of g?
0 of 4 answered