Linear Transformations

\(y\)-int \(b\): -2

Transformed Equation:

\(g(x) = x - 2\)

Teacher Answer Key • Linear Transformations Page 1 of 4

Teacher Resource • Solutions

Dual Transformations • Answer Key

Model answers for simultaneous slope and shift transformations.

Problems 7 – 12

Algebra 1 • Solutions

Problem 7 Flatter + Left 4 / Up 2

x y -4 -2 2 4 2 -2

Transformations:

Flatter (compress 1/2) • Shift up 2

Also: shifted left 4

Slope \(m\): 1/2

\(y\)-int \(b\): 2

Transformed Equation:

\(g(x) = \frac{1}{2}x + 2\)

Problem 8 Reflected + Shift Up 3

x y -4 -2 2 4 3 -3

Transformations:

Reflected across \(x\)-axis • Shift UP 3

Negative slope with \(b = 3\)

Slope \(m\): -1

\(y\)-int \(b\): 3

Transformed Equation:

\(g(x) = -x + 3\)

Problem 9 Reflected + Steeper

x y -4 -2 2 4 4 -4

Transformations:

Reflected across \(x\)-axis • Stretch 2

Negative & steeper (\(m = -2\))

Slope \(m\): -2

\(y\)-int \(b\): 0

Transformed Equation:

\(g(x) = -2x\)

Problem 10 Steeper + Shift Down 3

x y -4 -2 2 4 3 -3

Transformations:

Vertical stretch 2 • Shift DOWN 3

Steeper line with \(y\)-int \(-3\)

Slope \(m\): 2

\(y\)-int \(b\): -3

Transformed Equation:

\(g(x) = 2x - 3\)

Problem 11 Negative + Flatter + Down

x y -4 -2 2 4 2 -2

Transformations:

Reflected • Compress 1/2 • Down 2

Negative, flatter slope (\(m = -1/2\))

Slope \(m\): -1/2

\(y\)-int \(b\): -2

Transformed Equation:

\(g(x) = -\frac{1}{2}x - 2\)

Problem 12 Negative + Steeper + Up

x y -4 -2 2 4 4 -4

Transformations:

Reflected • Stretch 2 • Shift UP 4

Negative, steeper line (\(m = -2\))

Slope \(m\): -2

\(y\)-int \(b\): 4

Transformed Equation:

\(g(x) = -2x + 4\)

Teacher Answer Key • Linear Transformations Page 2 of 4

Teacher Resource • Solutions

Multi-Step Transformations • Answer Key

Model checklist choices, descriptive breakdowns, and final equations.

Problems 13 – 16

Multi-Step Transformations

Problem 13 3 Transformations

Solutions

x y -4 -2 2 4 2 -2

Transformation Analysis:

Slope: □ Positive ✓ Negative

Steepness: □ Steeper ✓ Flatter

Shift: ✓ Up □ Down by 2

Describe Transformations:

Reflected across \(x\)-axis, vertically compressed by 1/2 (flatter), shifted UP 2 units.

Slope \(m\): -1/2

\(y\)-intercept \(b\): 2

Equation:

\(g(x) = -\frac{1}{2}x + 2\)

Problem 14 3 Transformations

Solutions

x y -4 -2 2 4 3 -3

Transformation Analysis:

Slope: □ Positive ✓ Negative

Steepness: ✓ Steeper □ Flatter

Shift: □ Up ✓ Down by 2

Describe Transformations:

Reflected across \(x\)-axis, vertically stretched by factor of 3 (steeper), shifted DOWN 2.

Slope \(m\): -3

\(y\)-intercept \(b\): -2

Equation:

\(g(x) = -3x - 2\)

Problem 15 2-3 Transformations

Solutions

x y -4 -2 2 4 3 -3

Transformation Analysis:

Slope: ✓ Positive □ Negative

Steepness: ✓ Steeper □ Flatter

Shift: ✓ Up □ Down by 5

Describe Transformations:

Vertically stretched by factor of 2 (steeper), shifted UP 5 (or left 3 and down 1: \(2(x+3)-1\)).

Slope \(m\): 2

\(y\)-intercept \(b\): 5

Equation:

\(g(x) = 2x + 5\)

Problem 16 2 Transformations

Solutions

x y -3 3 2 -2

Transformation Analysis:

Slope: ✓ Positive □ Negative

Steepness: □ Steeper ✓ Flatter

Shift: □ Up ✓ Down by 2

Describe Transformations:

Vertically compressed by 1/3 (flatter), shifted DOWN 2 units.

Slope \(m\): 1/3

\(y\)-intercept \(b\): -2

Equation:

\(g(x) = \frac{1}{3}x - 2\)

Teacher Answer Key • Linear Transformations Page 3 of 4

Teacher Resource • Solutions

Advanced Transformations • Answer Key

Model solutions and master parameters for problems 17 – 20.

Problems 17 – 20

Advanced Transformations

Problem 17 3 Transformations

Solutions

x y -4 -2 2 4 1 -1

Transformations:

Reflected across \(x\)-axis • Compress 1/4 (flatter) • Shift UP 1

\(m\): -1/4

\(b\): 1

Equation:

\(g(x) = -\frac{1}{4}x + 1\)

Problem 18 2 Transformations

Solutions

x y -4 -2 2 4 3 -3

Transformations:

Vertically stretched by factor of 4 (steeper) • Shift DOWN 3

\(m\): 4

\(b\): -3

Equation:

\(g(x) = 4x - 3\)

Problem 19 3 Transformations

Solutions

x y -4 -2 2 4 1 -1

Transformations:

Reflected across \(x\)-axis • Stretch 2 (steeper) • Shift DOWN 1

\(m\): -2

\(b\): -1

Equation:

\(g(x) = -2x - 1\)

Problem 20 3 Transformations

Solutions

x y -4 -2 2 4 1 -1

Transformations:

Reflected across \(x\)-axis • Compress 1/3 (flatter) • Shift DOWN 1

\(m\): -1/3

\(b\): -1

Equation:

\(g(x) = -\frac{1}{3}x - 1\)

Teacher Grading Key Summary:

All 20 equations verified • Integer grid intercept points checked Total Points: 20 pts (or 40 pts with steps)

Teacher Answer Key • Linear Transformations Page 4 of 4

Key Teaching Insight: When an equation is written in the form \(y = a(x - h)\), the horizontal shift \(h\) is inside parentheses. If written as \(y = ax + c\), the horizontal shift is \(h = -c/a\). Have students test the \(x\)-intercept by substituting \(y = 0 \implies 3x + 12 = 0 \implies x = -4\).

Linear Transformations • Teacher Challenge Guide Page 1 of 2

Part II • Advanced Problems & Facilitation

Constraint Puzzles & Questioning Guide

Abstract constraint tasks, invariant point reasoning, and teacher debrief framework.

Teacher Guide

Pacing & Questioning

Challenge 4

Invariant Points: Can Shifts Cancel Out?

Abstract Proof

Task: The parent function \(f(x) = x\) passes directly through the origin \((0, 0)\). Suppose a transformed linear function undergoes a slope stretch \(a\), a horizontal shift \(h\), and a vertical shift \(k\): \[ g(x) = a(x - h) + k \]

Question for Students: Under what condition does the transformed function \(g(x)\) still pass through the origin \((0, 0)\) even when both \(h \neq 0\) and \(k \neq 0\)?

Mathematical Derivation: For \((0, 0)\) to lie on \(g(x)\): \[ g(0) = 0 \implies a(0 - h) + k = 0 \implies -ah + k = 0 \implies \mathbf{k = ah} \] Conclusion: If the vertical shift equals the slope multiplied by the horizontal shift, the line passes through the origin despite both non-zero shifts!

Challenge 5

Mystery Function Deduction (No Graph Provided)

Synthesis

A mystery linear function \(M(x)\) is created by transforming \(f(x) = x\) under three hidden conditions:

Condition 1: Reflected across the \(x\)-axis (\(m < 0\)) and compressed to half steepness (\(|m| = 1/2\)).

Condition 2: Its \(x\)-intercept is at the point \((6, 0)\).

Condition 3: Must be written in both transformation form and standard slope-intercept form.

Solution: Slope \(m = -1/2\). Using intercept \((6, 0)\): \(0 = -\frac{1}{2}(6) + b \implies b = 3\).

\(M(x) = -\frac{1}{2}(x - 6) \iff M(x) = -\frac{1}{2}x + 3\)

Teacher Questioning & Facilitation Framework

Probing Questions during Group Work:

• "If I double the steepness of a line, what happens to the distance between its intercepts?"
• "Can a reflection across the x-axis look identical to a reflection across the y-axis for a line? Why?"
• "What is the only line that cannot be represented in \(y = mx + b\) transformation form? (Vertical line)"

Mastery Criteria for Higher-Order Tasks:

• Distinguishes between \(a(x - h) + k\) and \(ax + b\).
• Explains why horizontal shifts alter the \(y\)-intercept when \(m \neq 1\).
• Connects geometric motion (sliding, tilting) to algebraic operations (addition, multiplication).

Linear Transformations • Teacher Challenge Guide Page 2 of 2