A vertical asymptote occurs where the denominator is zero: x − 2 = 0 gives x = 2.

Exponential y = bˣ has y(0) = 1 and y(1) = b = 3.

Rational Root Theorem: ±(factor of 5)/(factor of 2). Constants ±1, ±5; leading ±1, ±2. So ±1, ±5, ±1/2, ±5/2.

Two unknowns, so eliminate one first. Dividing the two model equations, P₀e^(6k)/P₀e^(2k) = 1800/500, gives e^(4k) = 3.6 by the quotient rule for exponents — the exponent gap is 6 − 2 = 4 hours. So k = ln(3.6)/4 ≈ 0.3202. Now go backwards from the t = 2 reading: P₀ = 500/e^(2k) = 500/e^(0.6404) = 500/√3.6 ≈ 263.5, so 264 cells. The most tempting wrong value is 326, which comes from dividing ln(3.6) by 6 instead of 4 — treating 1800 as the count 6 hours after the first reading rather than 6 hours after t = 0. Another trap is 949, obtained by multiplying 500 by e^(2k) instead of dividing; going from t = 2 back to t = 0 must undo growth, not apply more of it. The value 139 divides 500 by the whole factor 3.6, using the four-hour ratio as if it covered only the two hours from t = 0 to t = 2.

A hole occurs where a factor cancels in both numerator and denominator; a vertical asymptote occurs where a denominator factor does NOT cancel. The denominator factors as (x − 3)(x − 2). Since the hole is at x = 3, the numerator must contain (x − 3), and since x = 2 is an asymptote, the numerator must not contain (x − 2). The one remaining x-intercept at x = −4 supplies the factor (x + 4), so the numerator is (x − 3)(x + 4) = x² + x − 12, i.e. b = 1 and c = −12. Cancelling gives the reduced function (x + 4)/(x − 2) for x ≠ 3, and the hole's height is (3 + 4)/(3 − 2) = 7, placing the hole at (3, 7). The most tempting wrong answer puts the hole on the x-axis at height 0: students see the numerator vanish at x = 3 and conclude the point sits on the axis. But the denominator vanishes there too — the 0/0 form must be removed by cancelling FIRST, and the reduced expression is not zero at x = 3. Height −1 comes from misreading the intercept x = −4 as the factor (x − 4), giving (3 − 4)/(3 − 2) = −1, and the point at x = 2 swaps which feature belongs to the hole and which to the asymptote.

By the Factor Theorem, (x − 2) is a factor exactly when f(2) = 0. So 8 + 4k − 8 − 12 = 0, giving 4k = 12 and k = 3. Now f(x) = x³ + 3x² − 4x − 12. Synthetic division by the ROOT 2 on the coefficients 1, 3, −4, −12 gives 1, 5, 6, remainder 0, so f(x) = (x − 2)(x² + 5x + 6) = (x − 2)(x + 2)(x + 3). The other two zeros are −2 and −3, and their sum is −5 (this also matches the quotient quadratic's −b/a = −5/1). The tempting wrong answer is −1: that comes from running synthetic division with −2 instead of 2, i.e. using the divisor's constant term rather than the root of x − 2 = 0. That row gives 1, 1, −6, whose roots sum to −1. Sum 5 flips the sign of −b/a, and k = −3 solves 4k = 12 with a sign slip. Always divide by the value that makes the factor zero.

log₂(32) asks "2 to what power equals 32?" Since 2⁵ = 32, the answer is 5.

VA at x = 1 (denominator = x − 1) and HA at y = 0 (numerator degree < denominator degree). Matches f(x) = 1/(x − 1).

Group: (x² − 6x) + (y² + 4y) = 12. Complete: (x−3)² + (y+2)² = 12 + 9 + 4 = 25. Circle, center (3,−2), r=5.