Function Transformations Review Guide AP Precalculus Unit 1 • Polynomial & Rational Functions
Topic 1.12: Transformations of Functions
1.12A Translations • 1.12B Dilations
Quick Reference Sheet
The Master Transformation Model Coordinate Rule: \((x, y) \to \left(\frac{x}{b} + h, \; a \cdot y + k\right)\)
\( g(x) = a \cdot f\big(b(x - h)\big) + k \)
Outside (\(a, k\)): Affect \(y\)-values • Acts vertically • Standard operations
Inside (\(b, h\)): Affect \(x\)-values • Acts horizontally • Inverse operations
Vertical Transformations (Outputs)
\(a\) — Vertical Dilation (1.12B) \(y \to a \cdot y\)
• \(|a| > 1\): Stretch vertically by factor \(|a|\)
\(k\) — Vertical Translation (1.12A) \(y \to y + k\)
• \(k > 0\): Shift UP by \(k\) • \(k < 0\): Shift DOWN by \(|k|\)
Horizontal Transformations (Inputs)
\(b\) — Horizontal Dilation (1.12B) \(x \to x / b\)
• \(|b| > 1\): Compress horizontally by factor \(\frac{1}{|b|}\)
\(h\) — Horizontal Translation (1.12A) \(x \to x + h\)
• \(f(x - h)\): Shift RIGHT by \(h\) • \(f(x + h)\): Shift LEFT by \(h\)
Transformation Effects on Graphical & Analytical Attributes
Feature of \(f(x)\)
Transformed \(g(x) = a \cdot f(b(x-h)) + k\)
AP Precalc Key Insight
Domain \([x_1, x_2]\)
\(\left[\frac{x_1}{b}+h, \; \frac{x_2}{b}+h\right]\) (swap if \(b < 0\))
Only affected by inside parameters \(b\) and \(h\).
Range \([y_1, y_2]\)
\([a \cdot y_1 + k, \; a \cdot y_2 + k]\) (swap if \(a < 0\))
Only affected by outside parameters \(a\) and \(k\).
Average Rate of Change (AROC)
\(\text{AROC}_g = (a \cdot b) \cdot \text{AROC}_f\)
Evaluated across the corresponding mapped interval.
Zeros (\(x\)-intercepts)
\(x_{\text{new}} = \frac{x_{\text{old}}}{b} + h\) (if \(k = 0\))
If \(k \neq 0\), zeros change entirely; solve \(g(x) = 0\).
High-Yield AP Precalculus Traps & Test Strategies
1. Factor Inside First: Given \(f(2x + 8)\), write \(f(2(x + 4))\). The horizontal shift is 4 units left, not 8.
2. Table Reading: If \((3, 7)\) is on \(f\) and \(g(x) = 2f(x - 1) + 4\), the point on \(g\) is \((3 + 1, 2(7) + 4) = (4, 18)\).
3. "Dilation Factor" vs "Scale": A horizontal dilation by a factor of 3 means the graph stretches by 3 (\(b = \frac{1}{3}\)). If \(g(x) = f(3x)\), it is compressed by a factor of \(\frac{1}{3}\).
4. Translations Preserve AROC: If \(g(x) = f(x - 3) + 5\), \(\text{AROC}_{[4, 6]}\) on \(g\) equals \(\text{AROC}_{[1, 3]}\) on \(f\).
AP® Precalculus • Topic 1.12: Translations & Dilations of Functions • Quick Review Reference Sheet
Function Transformations Assessment AP Precalculus Unit 1 Assessment • Topic 1.12
Transformations of Functions
Score: _____ / 40
Name:
Date:
Period:
Part A: Multiple Choice (No Calculator) — Select the best answer for each question.
Questions 1–7 (3 pts each)
The point \((-4, 10)\) lies on the graph of a continuous function \(y = f(x)\). If the transformed function \(g\) is defined by \(g(x) = -2f\left(\frac{1}{3}x - 2\right) + 7\), which of the following points is guaranteed to lie on the graph of \(y = g(x)\)?
(A) \((-6, -13)\)
(B) \((-6, 27)\)
(C) \(\left(-\frac{10}{3}, -13\right)\)
(D) \((18, -13)\)
A function \(f\) has a domain of \([-6, 10]\) and a range of \([-4, 14]\). The function \(h\) is defined by \(h(x) = 3f(-2x + 4) - 5\). What are the domain and range of \(h\)?
(A) Domain: \([-3, 5]\); Range: \([-17, 37]\)
(B) Domain: \([-5, 3]\); Range: \([-17, 37]\)
(C) Domain: \([-3, 5]\); Range: \([-7, 11]\)
(D) Domain: \([8, -24]\); Range: \([-17, 37]\)
Selected values of a strictly increasing function \(f\) are given in the table below:
\(x\)
-2
0
4
8
\(f(x)\)
6
1
-3
5
The function \(k\) is defined by \(k(x) = \frac{1}{3}f(2x) + 2\). For what value of \(x\) does \(k(x) = 1\)?
(A) \(x = 0\)
(B) \(x = 2\)
(C) \(x = 4\)
(D) \(x = 8\)
AP Precalculus • Topic 1.12 Assessment Page 1 of 4
AP Precalculus • Part A: Multiple Choice (Continued) No Calculator Permitted
The graph of a function \(g\) is obtained from the graph of \(f\) by the following sequence of transformations: a horizontal compression by a factor of \(\frac{1}{2}\), a reflection across the \(y\)-axis, a vertical stretch by a factor of 4, and a vertical translation 6 units downward. Which equation correctly defines \(g(x)\)?
(A) \(g(x) = 4f(-2x) - 6\)
(B) \(g(x) = 4f\left(-\frac{1}{2}x\right) - 6\)
(C) \(g(x) = -4f(2x) - 6\)
(D) \(g(x) = \frac{1}{4}f(-2x) - 6\)
Let \(f\) be a continuous function such that the average rate of change of \(f\) over the interval \([2, 10]\) is \(-4\). The function \(p\) is defined by \(p(x) = 5f\left(\frac{1}{2}x + 1\right)\). What is the average rate of change of \(p\) over the corresponding interval \([2, 18]\)?
(A) \(-40\)
(B) \(-10\)
(C) \(-\frac{8}{5}\)
(D) \(-20\)
The polynomial function \(f\) has exactly three distinct real zeros at \(x = -6\), \(x = 2\), and \(x = 8\). The function \(j\) is defined by \(j(x) = -2f(4x - 12)\). Which of the following is guaranteed to be a zero of \(j\)?
(A) \(x = -36\)
(B) \(x = -1\)
(C) \(x = \frac{7}{2}\)
(D) \(x = 20\)
A marine research probe's depth \(D\) (in meters below sea level) at time \(t\) (in minutes) is given by \(y = D(t)\). An upgraded robotic probe is deployed 15 minutes later, travels at twice the speed (so it reaches any given depth in half the time), and dives three times as deep as the original probe. Additionally, its depth sensor is calibrated starting from a platform 4 meters above the water line (\(+4\)). Which function \(U(t)\) correctly models the upgraded probe's depth at time \(t\)?
(A) \(U(t) = 3D\big(2(t - 15)\big) - 4\)
(B) \(U(t) = 3D\left(\frac{1}{2}(t + 15)\right) + 4\)
(C) \(U(t) = \frac{1}{3}D\big(2(t - 15)\big) - 4\)
(D) \(U(t) = 3D(2t - 15) + 4\)
AP Precalculus • Topic 1.12 Assessment Page 2 of 4
Part B Free-Response & Analytical Modeling
Show all supporting steps and mathematical reasoning
The continuous piecewise linear function \(f\) is defined on the closed domain \([-4, 4]\). The graph consists of line segments connecting the consecutive points: \((-4, -2)\), \((-2, 4)\), \((0, 0)\), and \((4, 4)\). The function \(g\) is defined by \(g(x) = -\frac{1}{2}f(x - 2) + 3\).
(a) Complete the coordinate transformation table:
Map each given vertex of \(f(x)\) to its image on \(g(x)\):
Point on \(f(x)\)
Point on \(g(x)\)
\((-4, -2)\)
\((-2, 4)\)
\((0, 0)\)
\((4, 4)\)
State domain & range of \(g(x)\):
(b) Sketch and clearly label \(g(x)\):
y x O
Scale: 1 grid mark = 1 unit
Selected values of a continuous function \(h\) are given in the table:
\(x\)
-2
2
6
\(h(x)\)
14
2
-6
(a) Calculate the average rate of change of \(h\) over the interval \([-2, 6]\).
(b) The function \(w\) is defined by \(w(x) = -4h(2x) + 7\). Determine the mapped interval \([x_1, x_2]\) corresponding to \([-2, 6]\) on \(h\), and compute the average rate of change of \(w\) over that interval. Explain your calculation.
AP Precalculus • Topic 1.12 Assessment Page 3 of 4
AP Precalculus • Part B: Free-Response (Continued) Question 10 (10 points total)
Two precalculus students, Maya and Leo, are discussing the function \(r(x) = f(3x + 12)\), which is obtained from the graph of \(y = f(x)\).
• Maya asserts: "To obtain the graph of \(r\), we must first compress the graph horizontally by a factor of \(\frac{1}{3}\), and then translate the graph 12 units to the left."
• Leo asserts: "To obtain the graph of \(r\), we must first translate the graph 12 units to the left, and then compress the graph horizontally by a factor of \(\frac{1}{3}\)."
(a) Determine which student, if either, is correct. Explain your reasoning algebraically by factoring the argument of \(f\) and clearly stating the correct sequence of transformations under both interpretations.
(b) Suppose \(f(x)\) has an \(x\)-intercept at \((6, 0)\) and a \(y\)-intercept at \((0, -9)\). Find the exact coordinates of the corresponding \(x\)-intercept on the graph of \(r(x)\). Show the algebraic step used to find this value.
(c) Does the \(y\)-intercept of \(r(x)\) necessarily have a \(y\)-coordinate of \(-9\)? Explain why or why not, and express the value of \(r(0)\) in terms of \(f\).
AP Precalculus • Topic 1.12 Assessment Page 4 of 4 • End of Assessment
Function Transformations Answer Key Teacher Solutions & Rubric AP Precalculus • Topic 1.12
Assessment Answer Key & Scoring Guide
Complete Explanations
Quick Answer Matrix
Q1A
Q2A
Q3B
Q4A
Q5B
Q6C
Q7A
Q8FRQ
Q9FRQ
Q10FRQ
1. Coordinate Mapping • Correct: (A) 1.12A / 1.12B
Solution: The point \((-4, 10)\) satisfies \(f(-4) = 10\). To evaluate \(g(x) = -2f\left(\frac{1}{3}x - 2\right) + 7\), set the argument equal to \(-4\): \[ \frac{1}{3}x - 2 = -4 \implies \frac{1}{3}x = -2 \implies x = -6 \] Substitute \(x = -6\): \(g(-6) = -2f(-4) + 7 = -2(10) + 7 = -20 + 7 = -13\). The point is \((-6, -13)\).
Distractor Analysis: (B) incorrectly computed \(-2(10) + 7 = 27\) by failing to apply the negative sign. (C) evaluated \(\frac{1}{3}(-4) - 2\) directly without inverting operations. (D) multiplied \(-4\) by 3 and added 2 incorrectly.
2. Domain & Range Transformation • Correct: (A) 1.12A / 1.12B
Solution: Domain is bounded by the original domain \([-6, 10]\): \[ -6 \le -2x + 4 \le 10 \implies -10 \le -2x \le 6 \implies 5 \ge x \ge -3 \implies x \in [-3, 5] \] Range of \(f\) is \([-4, 14]\). The output is scaled by \(3\) and shifted down by \(5\): \[ y_{\min} = 3(-4) - 5 = -17, \quad y_{\max} = 3(14) - 5 = 42 - 5 = 37 \implies y \in [-17, 37] \]
Distractor Analysis: (B) forgot to flip inequality signs when dividing by \(-2\). (C) miscalculated range outputs. (D) incorrectly altered domain using outside parameters.
3. Tabular Transformation & Inversion • Correct: (B) 1.12A / 1.12B
Solution: We want \(k(x) = \frac{1}{3}f(2x) + 2 = 1\). Isolate \(f(2x)\): \[ \frac{1}{3}f(2x) = 1 - 2 = -1 \implies f(2x) = -3 \] From the table of \(f(x)\), \(f(x) = -3\) when \(x = 4\). Therefore, \(2x = 4 \implies x = 2\).
Distractor Analysis: (A) confused \(k(x)=1\) with the table value \(f(0)=1\). (C) found the input to \(f\) (\(2x=4\)) but forgot to solve for \(x\). (D) multiplied by 2 instead of dividing.
4. Formulating Transformation Equations • Correct: (A) 1.12B Dilation Form
Solution: Horizontal compression by factor \(\frac{1}{2}\) requires \(b = 2\). Reflection across the \(y\)-axis negates the input: \(b = -2\). Vertical stretch by 4 gives \(a = 4\). Vertical shift down 6 gives \(k = -6\). Putting it together: \(g(x) = 4f(-2x) - 6\).
Distractor Analysis: (B) used \(-\frac{1}{2}x\) (horizontal stretch by 2). (C) reflected across the \(x\)-axis instead of the \(y\)-axis (\(-4\)). (D) vertically compressed by \(\frac{1}{4}\).
AP Precalculus • Topic 1.12 Solutions Guide Page 1 of 3
Solutions Guide • Questions 5–8 Part A (cont.) & Part B Scoring
5. Average Rate of Change of Transformed Functions • Correct: (B) 1.12B Rate Rule
Solution: The transformed function is \(p(x) = 5f\left(\frac{1}{2}x + 1\right)\). Over \([2, 18]\): \[ \text{At } x = 2: \frac{1}{2}(2) + 1 = 2 \implies p(2) = 5f(2); \quad \text{At } x = 18: \frac{1}{2}(18) + 1 = 10 \implies p(18) = 5f(10) \] \[ \text{AROC}_p = \frac{p(18) - p(2)}{18 - 2} = \frac{5[f(10) - f(2)]}{16} = \frac{5}{2} \cdot \left(\frac{f(10) - f(2)}{8}\right) = \frac{5}{2} \cdot (-4) = -10 \] AP Rule: \(\text{AROC}_{\text{new}} = (a \cdot b) \cdot \text{AROC}_{\text{old}} = \left(5 \cdot \frac{1}{2}\right) \cdot (-4) = -10\).
6. Zeros Under Horizontal Transformation • Correct: (C) 1.12A / 1.12B
Solution: \(j(x) = -2f(4x - 12) = 0 \iff f(4x - 12) = 0\). The zeros of \(f\) are \(u = -6, 2, 8\). Set \(4x - 12 = 2 \implies 4x = 14 \implies x = \frac{7}{2}\). (The other zeros of \(j\) are \(4x - 12 = -6 \implies x = \frac{3}{2}\) and \(4x - 12 = 8 \implies x = 5\).)
7. Contextual Dilation & Translation Model • Correct: (A) 1.12 Applied
Solution: Deployment 15 minutes later is a horizontal shift right: replace \(t\) with \(t - 15\). Twice the speed means time elapsed is compressed by \(\frac{1}{2}\), which requires multiplying inside by \(2\): \(2(t - 15)\). Diving 3 times as deep scales depth by 3: \(3D(\dots)\). Platform 4 meters above sea level shifts position downward relative to depth: \(- 4\). Thus \(U(t) = 3D\big(2(t - 15)\big) - 4\).
8. Free-Response Scoring Rubric • Piecewise Function (6 points) 6 Points
Part (a) Coordinate Mapping (3 pts):
Rule: \((x, y) \to \left(x + 2, \; -\frac{1}{2}y + 3\right)\)
Original Vertex
Mapped Image
\((-4, -2)\)
\((-2, 4)\)
\((-2, 4)\)
\((0, 1)\)
\((0, 0)\)
\((2, 3)\)
\((4, 4)\)
\((6, 1)\)
Domain: \([-2, 6]\) • Range: \([1, 4]\)
Part (b) Graph & Point Breakdown (3 pts):
• 1 pt: Correctly maps all 4 vertex coordinates in table.
• 1 pt: Correct domain \([-2, 6]\) and range \([1, 4]\).
• 1 pt: Correctly plots and connects vertices with linear segments on \([-2, 6]\).
Common Error: Students often reflect over the \(y\)-axis instead of the \(x\)-axis, or add 2 to the \(y\)-values instead of \(x\)-values.
AP Precalculus • Topic 1.12 Solutions Guide Page 2 of 3
Solutions Guide • Free-Response Questions 9 & 10 Scoring Rubrics & Sample Work
9. Free-Response Scoring Rubric • Tabular Rates of Change (5 points) 5 Points
Part (a) Average Rate of Change of \(h\) (2 pts): \[ \text{AROC} = \frac{h(6) - h(-2)}{6 - (-2)} = \frac{-6 - 14}{8} = \frac{-20}{8} = -\frac{5}{2} = -2.5 \] • 1 pt for difference quotient setup; 1 pt for final correct value (\(-\frac{5}{2}\)).
Part (b) Transformed Interval & AROC of \(w\) (3 pts):
Mapped interval: set \(2x = -2 \implies x_1 = -1\); set \(2x = 6 \implies x_2 = 3\). Thus interval is \([-1, 3]\). Evaluate endpoints: \(w(3) = -4h(6) + 7 = -4(-6) + 7 = 31\); \(w(-1) = -4h(-2) + 7 = -4(14) + 7 = -49\). \[ \text{AROC}_w = \frac{w(3) - w(-1)}{3 - (-1)} = \frac{31 - (-49)}{4} = \frac{80}{4} = 20 \] Alternatively, \(\text{AROC}_w = (a \cdot b) \cdot \text{AROC}_h = (-4 \cdot 2) \cdot \left(-\frac{5}{2}\right) = 20\).
• 1 pt for identifying interval \([-1, 3]\); 1 pt for endpoint evaluation or scaling formula; 1 pt for correct rate of change (\(20\)).
10. Free-Response Scoring Rubric • Justification & Synthesis (8 points) 8 Points
Part (a) Student Claims Analysis (4 pts):
Conclusion: Leo is correct (under standard order of operations on the input variable \(x\)).Algebraic Justification: In \(r(x) = f(3x + 12)\), if we start with \(f(x)\), a horizontal translation 12 units left replaces \(x\) with \(x + 12\), giving \(f(x + 12)\). Then, a horizontal compression by factor \(\frac{1}{3}\) replaces \(x\) with \(3x\), producing \(f(3x + 12)\). (Leo's order).Note: Maya would only be correct if the shift were 4 units left: \(f(3x + 12) = f(3(x + 4))\).
• 1 pt for selecting Leo; 2 pts for algebraic demonstration of compositions; 1 pt for referencing factored form \(3(x+4)\).
Part (b) Corresponding \(x\)-intercept (2 pts):
An \(x\)-intercept of \(f\) at \((6, 0)\) means \(f(6) = 0\). For \(r(x) = f(3x + 12) = 0\), set \(3x + 12 = 6 \implies 3x = -6 \implies x = -2\). The coordinates are (-2, 0).
• 1 pt for equation \(3x + 12 = 6\); 1 pt for coordinates \((-2, 0)\).
Part (c) \(y\)-intercept Analysis (2 pts):
Answer: No. The \(y\)-intercept of \(r(x)\) is found at \(x = 0\): \(r(0) = f(3(0) + 12) = f(12)\). The value of \(r(0)\) depends on the function's output at \(x = 12\), not at \(x = 0\). The original \(y\)-intercept \((0, -9)\) has shifted horizontally to \(x = -4\) (since \(3(-4) + 12 = 0\)).
• 1 pt for stating \(r(0) = f(12)\); 1 pt for complete explanation of horizontal displacement.
AP Precalculus Scoring Insight: On the AP Exam, students frequently lose points on justification questions by providing vague verbal descriptions instead of formal function notation like \(f(3x + 12) = f(3(x+4))\). Insist that students explicitly show input substitution when justifying order of transformations.
AP Precalculus • Topic 1.12 Solutions Guide Page 3 of 3 • End of Scoring Guide
