Function Transformations Worksheet Answer Key – Algebra 2

Problem 1 Describing shifts

a) Up 5.

b) Right 4 (the shift is opposite the sign inside).

c) Left 6.

d) Down 2.

Problem 2 Writing an equation for a shift

Left 3 replaces x with x + 3. Down 6 subtracts 6.

y = |x + 3| − 6

Problem 3 Evaluating transformed functions

3f(16) = 3 · 4 = 12; 3f(25) = 3 · 5 = 15; 3f(36) = 3 · 6 = 18.

12, 15, 18

Problem 4 Equation from a graph

The vertex moved from (0, 0) to (2, 1): right 2 and up 1.

g(x) = |x − 2| + 1

Problem 5 Even, odd, or neither

a) f(−x) = x⁶ − 4x² = f(x): even.

b) g(−x) = −x⁵ + 9x = −g(x): odd.

c) h(−x) = x² + 3x, which is neither h(x) nor −h(x): neither.

Problem 6 Reflections

g(x) = −√x has the point (4, −2): reflection across the x-axis.

h(x) = √(−x) has the point (−4, 2): reflection across the y-axis.

(4, −2) and (−4, 2)

Problem 7 Stretches and shrinks

The factor 12 shrinks vertically by half, the minus sign reflects across the x-axis, and +3 shifts up 3.

g(4) = –12(16) + 3 = −8 + 3 = −5.

Reflect, shrink by 12, up 3; g(4) = −5

Problem 8 Several transformations at once

Reflection across the x-axis, vertical stretch by 2, left 1, up 4.

The vertex (0, 0) moves to (−1, 4).

Vertex (−1, 4)

Problem 9 Graphing a square root function

The parent √x starts at (0, 0), so the graph starts at (2, 1). Points: x = 3 gives 2, x = 6 gives 3, x = 11 gives 4.

Start (2, 1); domain x ≥ 2; range y ≥ 1

Problem 10 Graphing a cube root function

The parent passes through (0, 0), (1, 1), (−1, −1), (8, 2), (−8, −2). Shift left 1 and down 2: (−1, −2), (0, −1), (−2, −3), (7, 0), (−9, −4).

Domain: all real numbers; range: all real numbers

Problem 11 An exponential transformation

Right 1 and up 4. The asymptote y = 0 moves up to y = 4.

g(0) = 5⁻¹ + 4 = 15 + 4 = 215.

Right 1, up 4; y = 4; (0, 215)

Problem 12 Domain and range

Domain: x − 4 ≥ 0, so x ≥ 4. The root is at least 0, so −√(x − 4) ≤ 0 and g(x) ≤ 5.

Domain x ≥ 4; range y ≤ 5

Problem 13 Transforming a polygonal graph

g shifts every point right 1 and up 2: (−1, 3), (1, 5), (3, 2), (5, 4).

h changes the sign of every y-coordinate: (−2, −1), (0, −3), (2, 0), (4, −2).

g: (−1, 3), (1, 5), (3, 2), (5, 4); h: (−2, −1), (0, −3), (2, 0), (4, −2)

Problem 14 Equation of a parabola from its graph

h = −2 and k = −3, so g(x) = a(x + 2)² − 3. At (0, 1): 1 = 4a − 3, so a = 1.

g(x) = (x + 2)² − 3

Problem 15 Proving even and odd

a) f(−x) = −x³(−x)² + 4 = –x³x² + 4 = −f(x): odd.

b) g(−x) = |−x| + (−x)⁴ = |x| + x⁴ = g(x): even.

c) h(−x) = x⁴ − x³. This is not h(x) = x⁴ + x³, and not −h(x): neither.

Problem 16 Three launches of a ball

Second ball: shift up 20; the vertex moves from (2, 64) to (2, 84), so the maximum is 84 feet.

Third ball: shift right 0.5; the vertex moves to (2.5, 64), so the maximum is still 64 feet, reached 0.5 second later.

First 64 ft at 2 s; second 84 ft at 2 s; third 64 ft at 2.5 s

Problem 17 Delayed video views

A 3-day delay is a shift right 3: V₂(t) = 500 · 2^{t − 3}.

On day 5: 500 · 2² = 500 · 4 = 2,000.

With 800 extra views, add 800: V₃(t) = 500 · 2^{t − 3} + 800.

V₂(t) = 500 · 2^{t − 3}; 2,000 views; V₃(t) = 500 · 2^{t − 3} + 800

Problem 18 Weekly donations

Vertical stretch by 2, right 1, up 3.

D(10) = 2√9 + 3 = 9. D(5) = 2√4 + 3 = 7.

2√(w − 1) + 3 = 11 gives √(w − 1) = 4, so w − 1 = 16 and w = 17.

D(10) = 9; D(5) = 7; week 17

Problem 19 Writing and using a transformed function

g(x) = −3|x − 2| − 1.

g(−1) = −3(3) − 1 = −10. g(2) = −3(0) − 1 = −1. g(5) = −3(3) − 1 = −10.

The maximum value is the vertex height −1, which is below 0, so the graph never reaches the x-axis.

g(x) = −3|x − 2| − 1; −10, −1, −10; no x-intercepts

Problem 20 A horizontal change that is really vertical

g(x) = √(4x − 12) = √(4(x − 3)) = 2√(x − 3).

This is f shifted right 3 and stretched vertically by 2. The domain needs 4x − 12 ≥ 0, so x ≥ 3.

g(x) = 2√(x − 3); domain x ≥ 3