• Fluency with both algebraic and geometric criteria (Vertical Line Test).
Algebra II • Relations & Functions Answer Key Page 1 of 3
Algebra II Answer Key: Relations & Functions | Domain, Range & Function Notation
Part 2: 12 pts | Part 3: 10 pts
5. Continuous Function \( f(x) \) Domain & Range 6 Pts
Left endpoint: solid dot at \( (-4, -2) \). Right endpoint: open circle at \( (4, \approx 2) \). Global maximum: peak reaches \( y = 3 \).
Interval Notation (3 pts total)
Domain: [-4, 4) (1.5 pts: bracket at -4, paren at 4)
Range: [-2, 3] (1.5 pts: min -2, absolute max 3)
Inequality Notation (3 pts total)
Domain: -4 ≤ x < 4 (1.5 pts)
Range: -2 ≤ y ≤ 3 (1.5 pts)
6. Algebraic Domain Restrictions 6 Pts
a) Rational: \( g(x) = \frac{2x+5}{x^2 - 4x - 12} \) [3 pts]
1. Set denominator \( = 0 \): \( x^2 - 4x - 12 = 0 \)
2. Factor: \( (x - 6)(x + 2) = 0 \implies x = 6, \; x = -2 \)
3. Domain excludes zeros of denominator: \( x \neq -2, \; 6 \)
Answer: \( (-\infty, -2) \cup (-2, 6) \cup (6, \infty) \)
b) Radical: \( h(x) = \sqrt{3x - 15} \) [3 pts]
1. Set radicand \( \ge 0 \): \( 3x - 15 \ge 0 \)
2. Add 15: \( 3x \ge 15 \)
3. Divide by 3: \( x \ge 5 \)
Answer: \( [5, \infty) \)
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7. Evaluate \( f(-3) \) 3 Pts
\( f(x) = -2x^2 + 3x - 1 \)
f(-3) = -2(-3)² + 3(-3) - 1
= -2(9) - 9 - 1
= -18 - 9 - 1
= -28
\( f(-3) = \) -28
8. \( 2q(-1) - p(5) \) 3 Pts
\( p(x)=4-3x, \; q(x)=|2x-5| \)
q(-1) = |2(-1)-5| = |-7| = 7
2q(-1) = 2(7) = 14
p(5) = 4 - 3(5) = -11
14 - (-11) = 14 + 11 = 25
Result = 25
9. Solve for \( x \) 4 Pts
\( p(x) = 19 \implies 4 - 3x = 19 \)
4 - 3x = 19
-3x = 19 - 4
-3x = 15
x = 15 / -3 = -5
\( x = \) -5
Common Algebra II Evaluation Errors: In Problem 7, squaring negative numbers without parentheses (\(-3^2 \neq 9\) vs \((-3)^2 = 9\)). In Problem 8, dropping the double negative (\(14 - (-11) = 25\), not \(14 - 11 = 3\)).
Algebra II • Relations & Functions Answer Key Page 2 of 3
Algebra II Answer Key: Relations & Functions | Operations, Compositions & Modeling
Part 4: 12 pts | Part 5: 10 pts
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10. \( (f - g)(x) \) 3 Pts
(f - g)(x) = (2x - 3) - (x² - 4x + 1)
= 2x - 3 - x² + 4x - 1
= -x² + 6x - 4
Answer: -x² + 6x - 4
11. \( (g \circ f)(x) \) 4 Pts
g(f(x)) = (2x - 3)² - 4(2x - 3) + 1
= (4x² - 12x + 9) - 8x + 12 + 1
= 4x² - 20x + 22
Answer: 4x² - 20x + 22
12. \( (f \circ g)(-2) \) 2 Pts
1. g(-2) = (-2)² - 4(-2) + 1 = 4 + 8 + 1 = 13
2. f(13) = 2(13) - 3 = 26 - 3 = 23
Value: 23
13. Quotient Domain Restrictions 3 Pts
• \( h(x) \) denominator zero: \( x + 2 = 0 \implies x = -2 \)
• \( h(x) = 0 \) (entire denominator): \( x - 1 = 0 \implies x = 1 \)
Restrictions: \( x \neq -2, \; 1 \)
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14. Supply Chain Manufacturing Model 10 Pts
a) Composite Cost Function \( C(N(t)) \) [3 pts]
Substitute \( N(t) = 45t \) into \( C(n) = 120n + 3500 \):
\( C(N(t)) = 120(45t) + 3500 = 5400t + 3500 \)
b) Evaluate & Context [4 pts]
\( C(N(8)) = 5400(8) + 3500 = \$46,700 \) (2 pts)
Meaning with units: (2 pts)
"Operating the factory for 8 hours produces 360 units at a total production cost of $46,700."
c) Practical Domain & Range [3 pts]
Domain: [0, 12] (hours)
Range: [3500, 68300] ($)
\( C(0) = \$3,500 \); \( C(12) = 5400(12)+3500 = \$68,300 \)
Suggested Grading Scale: A: 54-60 (90%+) • B: 48-53 (80-89%) • C: 42-47 (70-79%) • D: 36-41 (60-69%)
Pass Standard: 42+ / 60
Algebra II • Relations & Functions Answer Key Page 3 of 3 • End of Guide