\(f\left(\frac{16}{9}\right) = 4\left(\frac{16}{9}\right) - 5 = \frac{64}{9} - \frac{45}{9} = \mathbf{\frac{19}{9}}\)
(b) Domain restrictions (denom = 0):
\(x^2 - 16 = 0 \implies (x - 4)(x + 4) = 0 \implies \mathbf{x \ne \pm 4}\)
Key Misconception: Forgetting the negative root \(x = -4\) in the domain.
Final Answer: \(f(g(5)) = \mathbf{\frac{19}{9}}\) ; Domain: \(\mathbf{x \ne \pm 4}\)
TSIA2 Mathematics Solution Manual • Educator Copy • College Readiness Score Conversion: 950+ = College Level
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Teacher Solution Guide TSIA2 Mathematics • Full Solutions & Rubric
Official Scoring Key • Problems 5–8
Problems 5–8: Vertex Optimization, Intersections, Work-Rate & Growth
Advanced Placement Track
Calculators Not Allowed on TSI Section
Non-Calculator Strategy: Factor prior to expanding Use \(x = -b/(2a)\) for vertex Clear denominators with LCD Total: 16 Pts (This Section)
5 Quadratic Optimization
4 Pts
Revenue: \(R(p) = -15p^2 + 450p\)
(a) Find vertex price \(p\):
\(a = -15,\ b = 450\)
\(p = -\frac{b}{2a} = -\frac{450}{2(-15)} = -\frac{450}{-30} = \mathbf{15}\) ($)
(b) Find maximum revenue \(R(15)\):
\(R(15) = -15(15^2) + 450(15) = -15(225) + 6,750\)
\(= -3,375 + 6,750 = \mathbf{\$3,375}\)
Key Misconception: Finding the roots (\(p = 0, 30\)) instead of the vertex.
Final Answer: Price \(p = \mathbf{\$15}\) ; Max Rev = \(\mathbf{\$3,375}\)
6 Nonlinear System Solution
4 Pts
System: \(y = x^2 - 3x - 4\) & \(y = 2x + 2\)
Step 1 (Equate equations):
\(x^2 - 3x - 4 = 2x + 2 \implies x^2 - 5x - 6 = 0\)
Factor: \((x - 6)(x + 1) = 0 \implies x_1 = 6,\ x_2 = -1\)
Step 2 (Find \(y\)-values via line):
\(y(6) = 2(6) + 2 = 14 \implies \mathbf{(6, 14)}\)
\(y(-1) = 2(-1) + 2 = 0 \implies \mathbf{(-1, 0)}\)
Key Misconception: Stopping at \(x\)-values and not finding full coordinate pairs.
Final Answer: \(\mathbf{(6, 14)}\) and \(\mathbf{(-1, 0)}\)
7 Rational Work-Rate Model
4 Pts
Rate Model: \(\frac{1}{6} + \frac{1}{t} = \frac{1}{4}\)
Step 1 (Isolate rate of B):
\(\frac{1}{t} = \frac{1}{4} - \frac{1}{6}\)
LCD of 4 and 6 is 12:
\(\frac{1}{t} = \frac{3}{12} - \frac{2}{12} = \frac{1}{12}\)
Step 2 (Solve for \(t\)):
\(\frac{1}{t} = \frac{1}{12} \implies \mathbf{t = 12}\text{ hours}\)
Key Misconception: Subtracting times directly: \(6 - 4 = 2\text{ hrs}\) (incorrect).
Final Answer: Scanner B Alone: \(t = \mathbf{12\text{ hours}}\)
8 Exponential Growth Model
4 Pts
Model: \(N(t) = 300 \cdot 2^{kt}\), \(N(2) = 2,400\)
(a) Solve for constant \(k\):
\(2,400 = 300 \cdot 2^{2k} \implies 8 = 2^{2k}\)
\(2^3 = 2^{2k} \implies 2k = 3 \implies \mathbf{k = 1.5}\quad\text{(or } 3/2\text{)}\)
(b) Evaluate at \(t = 4\):
\(N(4) = 300 \cdot 2^{1.5 \cdot 4} = 300 \cdot 2^6\)
\(N(4) = 300 \cdot 64 = \mathbf{19,200\text{ cells}}\)
Key Misconception: Multiplying base 300 by 2 before applying the exponent.
Final Answer: \(k = \mathbf{1.5}\) ; \(N(4) = \mathbf{19,200\text{ cells}}\)
Score Classification:
28–32 Pts: College Ready (CRC 950+) 22–27 Pts: Corequisite Placement < 22 Pts: Diagnostic Intervention
TSIA2 Mathematics Solution Manual • Educator Copy • College Readiness Benchmark: Advanced Track
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