ACT Math study guide · Chapter 15
Function Transformations and Piecewise Functions
A transformation of f(x) changes its graph in a predictable way. f(x − h) shifts the graph horizontally by h; f(x) + k shifts it vertically by k; −f(x) reflects the graph over the x-axis; f(−x) reflects it over the y-axis; and a·f(x) scales all y-values vertically by a factor of |a|. If a is negative, the graph is also reflected over the x-axis. Anything applied to x inside the function changes the graph horizontally; anything applied to the function's output changes it vertically. For a specific numerical evaluation, simplify the input first, evaluate f, and then apply whatever is outside f. When several horizontal changes combine in the same expression, rewrite that inside expression clearly before deciding the shift or reflection, instead of assuming a fixed order applies to every combination.
A piecewise function uses different rules for different parts of its domain. Evaluating a piecewise function at a specific input means first identifying which condition that input satisfies, then applying only the rule attached to that condition. Using the wrong piece is the most common error on this topic.
Where this appears as a model: a cost that changes structure past a threshold. A shipping cost of $5 for the first pound plus $2 for each additional pound is a piecewise function; a 4-pound package costs 5+2(3)=$11.
▸ Applying a transformation step by step
Worked Example Let f(x) = x² and g(x) = −2f(x + 1) + 3. Find g(1).
Apply the inside shift first, then the outside operations in order.
g(1) = −2f(1+1) + 3 = −2f(2) + 3
= −2(4) + 3
= −8 + 3 = −5
▸ Evaluating a piecewise function
Worked Example A piecewise function is p(x) = 2x + 1 if x < 2, and p(x) = x² − 3 if x ≥ 2. Find p(2) + p(−1).
Check which condition each input satisfies before applying a rule.
p(2): since 2 ≥ 2, use the second piece: 2²−3 = 1
p(−1): since −1 < 2, use the first piece: 2(−1)+1 = −1
p(2) + p(−1) = 1 + (−1) = 0
▸ Practice
Question 1
- AA. −1
- BB. −4
- CC. −5
- DD. 4
Question 2
- FF. −1
- GG. 4
- HH. 0
- JJ. −4
Question 3
- AA. 6
- BB. −6
- CC. 4
- DD. 0
Question 4
- FF. 17
- GG. 8
- HH. 53
- JJ. 35
Question 5
- AA. 8
- BB. 5
- CC. −1
- DD. 2
Question 6
- FF. −11
- GG. 6
- HH. −6
- JJ. 2
Question 7
- AA. 9
- BB. 12
- CC. 3
- DD. 6
Question 8
- FF. 14
- GG. 9
- HH. 5
- JJ. 10
Question 9
- AA. 15
- BB. −17
- CC. 17
- DD. −15
Question 10
- FF. 7
- GG. 5
- HH. 9
- JJ. 3
Question 11
- AA. 36
- BB. 6
- CC. 18
- DD. 9
Question 12
- FF. 25
- GG. −25
- HH. 5
- JJ. −5
Question 13
- AA. 1
- BB. −1
- CC. −7
- DD. 5
Question 14
- FF. 12
- GG. 18
- HH. 36
- JJ. 72
Question 15
- AA. 15
- BB. 9
- CC. 16
- DD. 8
Question 16
- FF. −4
- GG. 19
- HH. 15
- JJ. 10
Question 17
- AA. −9
- BB. 9
- CC. −15
- DD. 15
Question 18
- FF. 16
- GG. 48
- HH. 12
- JJ. 144
Question 19
- AA. −8
- BB. 7
- CC. 2
- DD. 12
Question 20
- FF. 36
- GG. 9
- HH. −36
- JJ. 18
▸ Answers and Explanations
Show the answers and explanations
1: C. g(2) = f(2−2)−5 = f(0)−5 = 0−5 = −5. Choice A evaluates f at the original input (2) instead of first applying the horizontal shift, computing f(2)−5 instead of f(0)−5.
2: G. Since 0 ≤ 0, use the first piece: −(0) + 4 = 4. Choice F applies the second piece, 2x−1, misreading which condition x=0 satisfies.
3: D. f(4) = |4−1| = 3, so g(4) = −(3) + 3 = 0. Choice A drops the negative sign on f(x), computing f(4)+3 instead of −f(4)+3.
4: J. g(3) = f(2·3) = f(6) = 36−1 = 35. Choice F evaluates f at the original input (3) instead of first doubling it, computing f(3) instead of f(6).
5: B. Since 3 ≥ 3, use the second piece: 2(3)−1 = 5. Choice A applies the first piece, x+5, misreading which condition x=3 satisfies.
6: H. g(−3) = −f(−3+1)−2 = −f(−2)−2 = −(4)−2 = −6. Choice F computes f(x)+1 instead of f(x+1), misreading which operation happens inside the function versus outside.
7: B. g(5)=f(5−2)+3=f(3)+3=9+3=12. Choice A reports f(3) alone, forgetting to add 3 afterward.
8: F. p(1)=1+4=5 (first piece, x<3); p(5)=2(5)−1=9 (second piece, x≥3); 5+9=14. Choice G reports p(5) alone instead of completing the requested sum p(1) + p(5).
9: D. g(4)=−f(4)+1=−16+1=−15. Choice A drops the negative sign on f(x), computing f(4)+1 instead of −f(4)+1.
10: G. Since 3≥3, use the second piece: 2(3)−1=5. Choice F applies the first piece, x+4, misreading which condition x=3 satisfies.
11: A. g(2)=f(3·2)=f(6)=36. Choice B reports the tripled input (3×2=6) instead of squaring it.
12: F. g(−5)=f(−(−5))=f(5)=25. Choice G keeps the negative sign that squaring would have removed, computing −f(x) instead of f(−x).
13: B. p(−2)=3(−2)=−6 (first piece, x≤0); p(3)=3²−4=5 (second piece, x>0); −6+5=−1. Choice A adds the two piece values with a sign error, computing p(−2) as +6 instead of −6.
14: J. g(6)=2f(6)=2(36)=72. Choice F reports 2 times the original input instead of 2 times the squared value.
15: A. g(1)=f(1+3)−1=f(4)−1=16−1=15. Choice B evaluates f at the original input (1) instead of first adding 3, computing f(1)−1 instead of f(4)−1.
16: H. p(−2)=−2−2=−4 (first piece, x<1); p(4)=4²+3=19 (second piece, x≥1); −4+19=15. Choice F reports p(−2) alone instead of completing the requested sum p(−2) + p(4).
17: A. g(5)=−f(5−2)=−f(3)=−9. Choice B drops the negative sign, computing f(x−2) instead of −f(x−2).
18: G. g(4)=3f(4)=3(16)=48. Choice F reports f(4) alone, forgetting to multiply by 3 afterward.
19: C. p(−4)=2(−4)+1=−7 (first piece, x<−2); p(0)=0²−1=−1 (second piece, −2≤x<3); p(5)=3(5)−5=10 (third piece, x≥3); −7+(−1)+10=2. Choice A sums only the first two piece values, forgetting to include p(5) in the total.
20: F. g(3)=f(−2·3)=f(−6)=36. Choice G evaluates f at the original input (3) instead of first applying −2x, computing f(3) instead of f(−6).
▸ Reading transformations from a description
A transformation question can also run in reverse: instead of transforming a given equation, you're told what happened to a graph and asked to build the resulting equation, or to match a description to the correct graph among several options.
Question 1
- AA. y=(x+4)²−1
- BB. y=−(x+4)²−1
- CC. y=−(x−4)²−1
- DD. y=(x−4)²+1
Question 2
- FF. y=|x+3|+2
- GG. y=|x−3|−2
- HH. y=|x−3|+2
- JJ. y=|x+3|−2
Question 3
- AA. y=f(−x)+5
- BB. y=−f(x)+5
- CC. y=f(x−5)
- DD. y=f(−x−5)
▸ Answers and Explanations
Show the answers and explanations
1: B. A left shift of 4 replaces x with x+4; a reflection over the x-axis negates the whole function; a down shift of 1 subtracts 1 at the end: y=−(x+4)²−1. Choice A includes the left shift and down shift but omits the reflection over the x-axis.
2: H. A right shift of 3 replaces x with x−3; an up shift of 2 adds 2 at the end: y=|x−3|+2. Choice F applies the shift in the wrong horizontal direction, using x+3 instead of x−3.
3: A. Reflecting over the y-axis replaces x with −x inside the function; shifting up 5 adds 5 outside the function: y=f(−x)+5. Choice B reflects over the x-axis (negating the whole function) instead of over the y-axis (negating just the input).