Building Functions Worksheet Answer Key – Algebra 1

Problem 1 Gym Membership Rule

Fixed fee plus rate times months: M(t) = 30 + 18t. M(6) = 30 + 108 = 138.

M(t) = 30 + 18t; $138

Problem 2 Rising Pool Level

Recursive: D(0) = 12 and D(n) = D(n − 1) + 2.5. Explicit: D(n) = 12 + 2.5n.

D(3) = 12 + 7.5 = 19.5, matching the last bar.

D(n) = 12 + 2.5n; 19.5 inches

Problem 3 Steps to a Taxi Fare

C(m) = 2.75m + 4. C(8) = 22 + 4 = 26.

C(m) = 2.75m + 4; $26

Problem 4 Constant Rate in a Table

Each step of 2 in x adds 6 to y, so the rate is 6 ÷ 2 = 3 per unit. The value at x = 0 is 5.

y = 3x + 5

Problem 5 Add Two Functions

(f + g)(x) = 5x − 3 + x² + 4 = x² + 5x + 1. At x = 3: 9 + 15 + 1 = 25.

Check: f(3) = 12, g(3) = 13, and 12 + 13 = 25 ✓.

x² + 5x + 1; 25

Problem 6 Subtract Two Functions

(f − g)(x) = 5x − 3 − (x² + 4) = −x² + 5x − 7. At x = 3: −9 + 15 − 7 = −1.

Check: f(3) − g(3) = 12 − 13 = −1 ✓.

−x² + 5x − 7; −1

Problem 7 Multiply Two Functions

(x + 7)(2x − 5) = 2x² − 5x + 14x − 35 = 2x² + 9x − 35.

At x = 2: 8 + 18 − 35 = −9. Check: f(2) · g(2) = 9 · (−1) = −9 ✓.

2x² + 9x − 35; −9

Problem 8 Profit from Revenue and Cost

P(x) = 14x − (210 + 6x) = 8x − 210. P(40) = 320 − 210 = 110.

Profit needs 8x > 210, so x > 26.25. At x = 26 profit is −2 and at x = 27 it is 6.

P(x) = 8x − 210; $110; 27 hoodies

Problem 9 Inverse of a Linear Function

y = 4x − 7, so 4x = y + 7 and x = y + 74. So f⁻¹(x) = x + 74.

Check: f(2) = 1 and f⁻¹(1) = 84 = 2 ✓.

f⁻¹(x) = x + 74

Problem 10 Inverse with a Negative Slope

y = −2x + 10, so 2x = 10 − y and x = 10 − y2.

f⁻¹(x) = 10 − x2 = 5 − x2. Check: f(1) = 8 and f⁻¹(8) = 22 = 1 ✓.

f⁻¹(x) = 10 − x2

Problem 11 Solve f(x) = c

5x + 12 = 47, so 5x = 35 and x = 7.

Check: f(7) = 35 + 12 = 47 ✓.

x = 7

Problem 12 Convert Temperatures

F − 32 = 95C, so C = 59(F − 32).

77°F: 59(45) = 25. 14°F: 59(−18) = −10.

C = 59(F − 32); 25°C and −10°C

Problem 13 Reflect a Graph

An inverse swaps coordinates: (0, −4) → (−4, 0) and (2, 0) → (0, 2). The slope through them is (2 − 0) ÷ (0 − (−4)) = 12, and the y-intercept is 2.

(−4, 0) and (0, 2); f⁻¹(x) = x2 + 2

Problem 14 Area and Perimeter Rules

Area: (3x + 2)(x + 5) = 3x² + 15x + 2x + 10 = 3x² + 17x + 10.

Perimeter: 2(3x + 2) + 2(x + 5) = 6x + 4 + 2x + 10 = 8x + 14.

A(x) = 3x² + 17x + 10; P(x) = 8x + 14

Problem 15 Savings Plan, Recursive to Explicit

Each month adds 45, so S(n) = 250 + 45n.

Solve 250 + 45n ≥ 1000: 45n ≥ 750, so n ≥ 16.67. At n = 16 the balance is 970; at n = 17 it is 1,015.

S(n) = 250 + 45n; month 17

Problem 16 Unequal Steps, Same Rate

Rates: (22 − 14) ÷ 2 = 4, (34 − 22) ÷ 3 = 4, (50 − 34) ÷ 4 = 4. The rate is 4 per unit each time.

Use y = 4x + b with (1, 14): 14 = 4 + b, so b = 10. Check (10, 50): 40 + 10 = 50 ✓.

y = 4x + 10

Problem 17 Break-Even Point

P(x) = 15x − 200 − 7x = 8x − 200. Break-even: 8x − 200 = 0, so x = 25.

P(40) = 320 − 200 = 120.

P(x) = 8x − 200; 25 keychains; $120

Problem 18 Inverse in Context

C = 2.75m + 4, so m = C − 42.75. This is the inverse function.

For $26: 22 ÷ 2.75 = 8 miles. For $15: 11 ÷ 2.75 = 4 miles.

m = C − 42.75; 8 miles and 4 miles

Problem 19 Discount Then Shipping

T(p) = 0.85p + 8. Solve T = 0.85p + 8: p = T − 80.85.

For T = 110: 102 ÷ 0.85 = 120. Check: 0.85(120) + 8 = 110 ✓.

T(p) = 0.85p + 8; the item costs $120

Problem 20 Fill and Leak

V(t) = 20 + 8t − 3t = 20 + 5t.

Solve 20 + 5t = 100: 5t = 80, so t = 16.

V(t) = 20 + 5t; after 16 hours