Function Transformations Worksheet Answer Key – Algebra 1

Problem 1 Shift Up

Adding 5 outside the function raises every point 5 units. The vertex (0, 0) moves to (0, 5).

Up 5 units; vertex (0, 5)

Problem 2 Shift Right

Subtracting 4 inside the parentheses moves the graph 4 units to the right. The vertex is (4, 0).

Right 4 units; vertex (4, 0)

Problem 3 Absolute Value, Shifted Down

The −3 is outside the absolute value, so the V moves down 3. The vertex moves from (0, 0) to (0, −3).

Down 3 units; vertex (0, −3)

Problem 4 Absolute Value, Shifted Left

x + 2 = x − (−2), so the V moves 2 units left. The vertex is (−2, 0).

Left 2 units; vertex (−2, 0)

Problem 5 Evaluate an Absolute Value Function

f(0) = |−7| = 7. f(3) = |−1| = 1. f(5) = |3| = 3. f(−1) = |−9| = 9.

7, 1, 3, 9

Problem 6 Evaluate a Piecewise Function

−3 < 0, so use x² − 1: 9 − 1 = 8. 0 ≥ 0, so use 3x + 2: 2. 4 ≥ 0: 3(4) + 2 = 14.

f(−3) = 8, f(0) = 2, f(4) = 14

Problem 7 Stretch, Shrink and Flip

1. Every y-value is tripled: a vertical stretch by 3 (narrower).
2. Every y-value is divided by 4: a vertical shrink (wider).
3. Every y-value changes sign: a reflection over the x-axis.

(a) stretch by 3; (b) shrink by 14; (c) reflection over the x-axis

Problem 8 Combine Several Changes

Right 3, stretch by 2, flip over the x-axis, up 5. The vertex is (3, 5). Check: g(3) = 5 and g(1) = −2(4) + 5 = −3.

Because a = −2 is negative, the parabola opens down.

Vertex (3, 5); opens down

Problem 9 Equation from an Absolute Value Graph

The vertex is (3, −2), so h = 3 and k = −2. Use (5, 2): 2 = a|5 − 3| − 2, so 2a = 4 and a = 2.

Check (6, 4): 2|6 − 3| − 2 = 4 ✓.

y = 2|x − 3| − 2

Problem 10 Table for f(x − 2) + 1

g(0) = f(−2) + 1 = 4 + 1 = 5. g(2) = f(0) + 1 = 1. g(4) = f(2) + 1 = 5.

The vertex (2, 1) is the lowest point.

(0, 5), (2, 1), (4, 5)

Problem 11 Read a Piecewise Graph

The left piece is y = x + 1: f(−1) = 0 and f(1) = 2. At x = 2 the closed circle is at (2, 4), so f(2) = 4. The right piece is y = −x + 6: f(3) = 3.

The open circle at (2, 3) belongs to the left piece and is not included.

0, 2, 4, 3; open at (2, 3), closed at (2, 4)

Problem 12 Absolute Value Equation

|x − 4| + 1 = 7, so |x − 4| = 6. Then x − 4 = 6 or x − 4 = −6.

x = 10 or x = −2. Both are 6 units from 4 on the number line.

x = 10 or x = −2

Problem 13 Parking Fee Steps

1 hour is the right end of the first step, a closed circle at (1, 4): $4.

2.5 hours lies in the step 2 < h ≤ 3: $8. 3.2 hours lies in the step 3 < h ≤ 4: $10.

$4, $8, $10

Problem 14 Taxi Fare Rule

C(m) = 3.50 + 2.25m for 0 ≤ m ≤ 4, and C(m) = 12.50 + 1.75(m − 4) for m > 4. (At 4 miles the fare is 3.50 + 9 = 12.50.)

C(2) = 3.50 + 4.50 = 8. C(10) = 12.50 + 1.75(6) = 23.

$8.00 for 2 miles; $23.00 for 10 miles

Problem 15 Move a Point

1. Same x, y + 4: (3, 9).
2. Shift right 2: (5, 5).
3. Flip the y-value: (3, −5).
4. Triple the y-value: (3, 15).

(a) (3, 9); (b) (5, 5); (c) (3, −5); (d) (3, 15)

Problem 16 Write the Transformed Parabola

Left 4: (x + 4)². Reflect: −(x + 4)². Up 7: y = −(x + 4)² + 7.

At x = −2: −(2)² + 7 = 3.

y = −(x + 4)² + 7; y = 3

Problem 17 Absolute Value as Piecewise

The inside x − 3 is negative when x < 3 and non-negative when x ≥ 3. So f(x) = −(x − 3) = 3 − x for x < 3, and f(x) = x − 3 for x ≥ 3.

x = −1: 3 − (−1) = 4 = |−4| ✓. x = 6: 6 − 3 = 3 = |3| ✓.

f(x) = 3 − x for x < 3; f(x) = x − 3 for x ≥ 3

Problem 18 Electricity Bill

C(k) = 0.12k for 0 ≤ k ≤ 500, and C(k) = 60 + 0.18(k − 500) for k > 500. At 500 kWh both rules give $60, so the graph has no gap.

C(430) = 0.12(430) = 51.60. C(760) = 60 + 0.18(260) = 60 + 46.80 = 106.80.

$51.60 and $106.80

Problem 19 Describe a Transformed Absolute Value

Left 1, stretch by 3, flip over the x-axis, up 2. The vertex is (−1, 2) and the V opens down.

g(−4) = −3|−3| + 2 = −9 + 2 = −7. g(2) = −3|3| + 2 = −7. The values match because −4 and 2 are both 3 units from −1.

Vertex (−1, 2); g(−4) = −7 and g(2) = −7

Problem 20 Robot on a Track

The vertex is (4, 0), so h = 4 and k = 0. At t = 0: 6 = a|0 − 4| = 4a, so a = 1.5.

Check t = 8: 1.5(4) = 6 ✓. The vertex means that the robot reaches the dock after 4 seconds, and the speed is 1.5 feet per second.

d = 1.5|t − 4|