Polynomial Power Review

3(a) Model Justification [2 Points]

Because the input values are equally spaced by \(\Delta x = 2\), and the second finite differences are non-zero and constant (\(\Delta^2 f = 8\)), the average rate of change changes at a constant rate. Therefore, \(f\) is best modeled by a quadratic function.

3(b) Explicit Quadratic Formula [2 Points]

For consecutive inputs spaced by \(h = 2\), the second difference is \(2a \cdot h^2 = 2a(4) = 8a\).
Setting \(8a = 8 \implies \mathbf{a = 1}\).
Then \(f(x) = x^2 + bx + c\). Using \(f(1) = 1 + b + c = 3 \implies b + c = 2\).
Using \(f(3) = 9 + 3b + c = 11 \implies 3b + c = 2\).
Subtracting equations: \(2b = 0 \implies \mathbf{b = 0}\), so \(\mathbf{c = 2}\).
Formula: \(\mathbf{f(x) = x^2 + 2}\). (Check: \(f(5) = 25+2=27\); \(f(7) = 49+2=51\); \(f(9) = 81+2=83\) ✓).

1.4 Topic 1.4: Extrema, Turning Points & Concavity

[3 Points]

4(a) Rate of Change Increasing (Concave Up Intervals) [1 Point]

The rate of change is increasing where the graph is concave up. From the inflection points at \(x = \pm 1.15\), this occurs on: \( (-\infty, -1.15) \) and \( (1.15, \infty) \) (or \((-\infty, -1.15] \cup [1.15, \infty)\))

4(b) Decreasing and Concave Down [1 Point]

Function \(p\) is decreasing on \((-\infty, -2)\) and \((0, 2)\). Concave down on \((-1.15, 1.15)\).
The intersection where it is both decreasing and concave down is: \( (0, 1.15) \)

4(c) Maximum Extrema & Inflection for Degree \(n=5\) [1 Point]

Maximum turning points (extrema) = \( n - 1 = 5 - 1 = \mathbf{4} \).
Maximum points of inflection = \( n - 2 = 5 - 2 = \mathbf{3} \).

AP Precalculus Teacher Key • Unit 1 Page 2 of 4 Scoring Rubric • Topics 1.3–1.4

AP Precalculus Teacher Scoring Guide

Solutions: Topics 1.5 & 1.6

[7 Points Total]

1.5 Topic 1.5: Complex Zeros Solutions

[4 Points]

5(a) Zeros of \(P(x)\) [1 Point]

Because coefficients are real, non-real complex zeros occur in conjugate pairs (Complex Conjugate Theorem). Since \(2 - 3i\) is a zero, \(2 + 3i\) is also a zero.
Full list of 4 zeros: \(\mathbf{2 - 3i, \; 2 + 3i, \; -1 \text{ (multiplicity 2)}}\).

5(b) Real Factorization [1 Point]

Irreducible quadratic: \([x - (2 - 3i)][x - (2 + 3i)] = (x - 2)^2 - (3i)^2 = x^2 - 4x + 13\).
With root \(-1\) (mult 2) and lead coefficient 2: \(\mathbf{P(x) = 2(x + 1)^2(x^2 - 4x + 13)}\).

6(a) & 6(b) Factoring \(Q(x) = x^4 + 3x^2 - 28\) [2 Points]

Let \(u = x^2 \implies u^2 + 3u - 28 = (u + 7)(u - 4) = (x^2 + 7)(x^2 - 4)\).
• Over \(\mathbb{R}\): \(\mathbf{Q(x) = (x - 2)(x + 2)(x^2 + 7)}\) (\(x^2+7\) is irreducible over reals).
• Over \(\mathbb{C}\): \(\mathbf{Q(x) = (x - 2)(x + 2)(x - i\sqrt{7})(x + i\sqrt{7})}\).

1.6 Topic 1.6: End Behavior Limits Solutions

[3 Points]

7. Dominant Term Limit Evaluations [1.5 Points]

\(f(x)\): \(-3x^5\) \( \lim_{x \to -\infty} f(x) = \mathbf{+\infty} \)
\( \lim_{x \to \infty} f(x) = \mathbf{-\infty} \)

\(g(x)\): \(4x^6\) \( \lim_{x \to -\infty} g(x) = \mathbf{+\infty} \)
\( \lim_{x \to \infty} g(x) = \mathbf{+\infty} \)

\(h(x)\): \(4x^4\) \( \lim_{x \to -\infty} h(x) = \mathbf{+\infty} \)
\( \lim_{x \to \infty} h(x) = \mathbf{+\infty} \)

8(a) & 8(b) Limit Deductions [1.5 Points]

(a) Both ends point in same direction (\(-\infty\)) \(\implies\) even degree. Both limits negative \(\implies\) negative leading coefficient (\(a_n < 0\)).
(b) No. Degree is even. By Fundamental Theorem of Algebra, total zeros (with multiplicity) is even. Non-real complex roots occur in pairs, so sum of multiplicities of real zeros must be even. Thus, \(R(x)\) cannot have exactly 3 simple real zeros.

AP Precalculus Teacher Key • Unit 1 Page 3 of 4 Scoring Rubric • Topics 1.5–1.6

FRQ Scoring Rubric

Unit 1 Cumulative FRQ Solutions & Rubric

AP Free Response Standards Total: 6 Points

Part (a): Average Rate of Change on \([0,4]\) and Meaning 2 Points

\( \text{AROC} = \frac{C(4) - C(0)}{4 - 0} = \frac{66 - 42}{4} = \frac{24}{4} = \mathbf{6\ \mu\text{g/m}^3\text{ per hour}} \).

Interpretation: Over the first 4 hours of the flight, the aerosol pollutant concentration increased at an average rate of 6 micrograms per cubic meter per hour.

Point Breakdown: 1 pt for correct value (\(6\)); 1 pt for interpretation with correct units in context.

Part (b): Rate Comparison and Concavity Deduction 2 Points

• AROC on \([0,2]\): \( \frac{58 - 42}{2 - 0} = \frac{16}{2} = \mathbf{8\ \mu\text{g/m}^3/\text{hr}} \)

• AROC on \([2,4]\): \( \frac{66 - 58}{4 - 2} = \frac{8}{2} = \mathbf{4\ \mu\text{g/m}^3/\text{hr}} \)

Deduction: Because consecutive average rates of change over equal intervals are decreasing (\(8 > 4\)), the rate of change of pollutant concentration is decreasing. This implies that the graph of \(C(t)\) is concave down on \([0,4]\).

Point Breakdown: 1 pt for consecutive rate values; 1 pt for conclusion of decreasing rate with concavity down.

Part (c): Quadratic Model Vertex & Context 1 Point

The vertex form \( M(t) = a(t - h)^2 + k \) gives vertex at \(\mathbf{(5, 66.5)}\).

Significance: At \(t = 5\) hours into the flight, the drone records its maximum pollutant concentration of \(66.5\ \mu\text{g/m}^3\).

Part (d): Polynomial End Behavior & Limit Rigor 1 Point

The dominant term of \(W(t)\) is \(-0.1t^4\). As \(|t| \to \infty\), \(W(t) \approx -0.1t^4\).

Because the degree 4 is even and the leading coefficient \(-0.1 < 0\):

\( \lim_{t \to -\infty} W(t) = -\infty \) \( \lim_{t \to \infty} W(t) = -\infty \)

Scoring: 1 pt for correct limits with reference to the leading term \(-0.1t^4\).

Key AP Misconceptions to Address During Review:

• Confusing rate value with rate change: A function can be increasing while its rate is decreasing (e.g., \(8 > 0\), but decreasing from \(8\) to \(4\)).
• Missing units: On AROC questions, omitting units of measure (\(\mu\text{g/m}^3/\text{hr}\)) loses the communication point.
• Even/Odd function vs. Even/Odd degree: Remind students that end behavior depends on the polynomial degree parity, not symmetry across the axes.

AP Precalculus Teacher Key • Unit 1 FRQ Page 4 of 4 Scoring Rubric • Section II FRQ