Holes and Limits Assessment

Watch For: Sign errors when factoring \((2x+1)(x-4)\).

3. \(h(x) = \dfrac{x^3 - 2x^2 - 4x + 8}{x^2 - 4}\)

4 Points Total

Step 1: Factor numerator by grouping:

\(x^2(x - 2) - 4(x - 2) = (x^2 - 4)(x - 2) = (x - 2)^2(x + 2)\)

Denominator: \((x - 2)(x + 2)\). Both denominator factors cancel out!
Simplified form: \(\widetilde{h}(x) = x - 2\) for \(x \ne 2, -2\).

Hole 1: at \(x = 2 \implies y = 2 - 2 = 0 \implies\) \((2, 0)\) (x-intercept hole!)
Hole 2: at \(x = -2 \implies y = -2 - 2 = -4 \implies\) \((-2, -4)\)
Vertical Asymptotes: NONE (multiplicities in num \(\ge\) denom).

Scoring Breakdown:

• [1 pt] Correct factoring by grouping to \((x-2)^2(x+2)\)
• [1 pt] Identifying Hole 1 at \((2, 0)\)
• [1 pt] Identifying Hole 2 at \((-2, -4)\)
• [1 pt] Explaining why there are NO vertical asymptotes

AP Takeaway: A hole on the x-axis occurs when numerator has higher multiplicity than denominator.

AP Precalculus Teacher Key • Solutions Page 1 of 3 Advanced Holes & Discontinuities

Part II: Graphical Limit Analysis Solutions Grading Keys for Graphical Questions 4 & 5

4. Solution & Scoring for Graph \(y = f(x)\) 4 Points Total

Graph Analysis:

• Hole at: \((3, 4)\)

• Vertical Asymptote: \(x = 1\)

• Horizontal Asymptote: \(y = 2\)

Underlying Function:
\(f(x) = \dfrac{2(x-3)(x+1)}{(x-3)(x-1)} = \dfrac{2x+2}{x-1} \quad (x \ne 3)\)

(a) Two-Sided Limit & Function Value [1.5 pts]:

\(\lim\limits_{x \to 3} f(x) = 4\)

\(f(3)\) does not exist (DNE / undefined) because there is an open circle at \(x = 3\).

(b) Limit Arithmetic Evaluation [1.5 pts]:

\(\lim\limits_{x \to 3} \left[ (f(x))^2 - 2f(x) \right] = (4)^2 - 2(4) = 16 - 8 = \mathbf{8}\)

Apply limit properties: \([\lim f]^2 - 2[\lim f]\).

(c) Continuous Extension Value [1 pt]:

\(k = 4\)

For continuity, \(F(3) = \lim\limits_{x \to 3} f(x) \implies k = 4\).

5. Solution & Scoring for Graph \(y = g(x)\) 4 Points Total

Graph Analysis:

• Hole at: \((-2, -2)\)

• Vertical Asymptote: \(x = 1\)

• Horizontal Asymptote: \(y = -1\)

Underlying Function:
\(g(x) = \dfrac{-(x+2)(x-4)}{(x+2)(x-1)} = \dfrac{4-x}{x-1} \quad (x \ne -2)\)

(a) Removable Discontinuity Limits [1.5 pts]:

\(\lim\limits_{x \to -2^-} g(x) = -2\)

\(\lim\limits_{x \to -2^+} g(x) = -2\)

\(\lim\limits_{x \to -2} g(x) = -2\) • Hole: \((-2, -2)\)

(b) Non-Removable (VA) Behavior [1.5 pts]:

\(\lim\limits_{x \to 1^-} g(x) = -\infty\)

\(\lim\limits_{x \to 1^+} g(x) = +\infty\)

Contrasts finite limit at hole with infinite limit at asymptote.

(c) Transformed Limit Evaluation [1 pt]:

\(\lim\limits_{x \to -2} \dfrac{1}{[g(x)]^2} = \dfrac{1}{(-2)^2} = \mathbf{\dfrac{1}{4}}\) (or 0.25)

AP Precalculus Teacher Key • Solutions Page 2 of 3 Graphical Limits

Part III & IV: Tables & AP Parameter FRQ Solutions Grading Keys for Questions 6, 7, and 8

6. Solution for Table: \(r(x) = \dfrac{2x^2 - 5x + 2}{x^2 - 4} = \dfrac{2x - 1}{x + 2} \quad (x \ne 2)\)

4 Points Total

Hole Location [2 pts]: \((2, \frac{3}{4})\) or \((2, 0.75)\)

Exact Limit [2 pts]: \(\lim\limits_{x \to 2} r(x) = \frac{3}{4}\)

7. Solution for Table: \(k(x) = \dfrac{x^3 - 8}{x^2 - 2x} = \dfrac{x^2 + 2x + 4}{x} \quad (x \ne 2)\)

4 Points Total

Hole Location [2 pts]: \((2, 6)\)

Difference of Cubes Limit [2 pts]: \(\lim\limits_{x \to 2} k(x) = 6\)

8. Solution for AP Parameter FRQ Challenge

4 Points Total

(a) Finding Parameter \(k\) [1 pt]:

Denominator has zero at \(x = 3\) since \(3^2 + 3 - 12 = 0\). For a hole, numerator must also vanish at \(x = 3\):
\(2(3)^2 + 3k - 15 = 0 \implies 18 + 3k - 15 = 0 \implies 3k = -3 \implies \mathbf{k = -1}\).

(b) Hole Coordinates & Limit [1 pt]:

\(P(x) = \dfrac{(2x + 5)(x - 3)}{(x + 4)(x - 3)} = \dfrac{2x + 5}{x + 4}\).
\(\lim\limits_{x \to 3} P(x) = \dfrac{2(3) + 5}{3 + 4} = \mathbf{\dfrac{11}{7}} \implies\) Hole: \((3, \frac{11}{7})\).

(c) Vertical Asymptote Limits at \(x = -4\) [1 pt]:

\(\lim\limits_{x \to -4^-} P(x) = \dfrac{-3}{0^-} = \mathbf{+\infty}\)

\(\lim\limits_{x \to -4^+} P(x) = \dfrac{-3}{0^+} = \mathbf{-\infty}\)

(d) Continuous Extension Function [1 pt]:

\(H(x) = \begin{cases} P(x), & x \ne 3 \\ \mathbf{\dfrac{11}{7}}, & x = 3 \end{cases}\)

Defines the hole's limit value as the function value at \(x = 3\).

AP Precalculus Teacher Key • Solutions Page 3 of 3 Scoring Rubric Complete