Pre-Calculus — Semester A
Free Practice · 10 Questions · 20 min
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Question 1 of 10
TEKS 1A-1GMedium

A quadratic projectile-motion problem produces two algebraic solutions: t = 2 seconds and t = −4 seconds. The problem asks when the ball hits the ground after being hit. Which response demonstrates the BEST mathematical reasoning?

AReject t = −4 (negative time has no physical meaning here); report t = 2 seconds
BReject t = 2 because the smaller solution is always the answer
CAverage the two: t = (2 + (−4))/2 = −1 second
DReport both because they are algebraically valid
Explanation
A complete solution requires checking that the algebraic answer fits the physical context. Both t = 2 and t = −4 are algebraic roots, but negative time has no physical meaning when the timer starts at t = 0 — so only t = 2 is contextually valid. Common mistakes: (C) there's no rule that smaller is always correct; (D) averaging roots gives the parabola's axis of symmetry, which is irrelevant to "when does it hit the ground."
Question 2 of 10
TEKS 2A-2JMedium

Given f(x) = x² + 1 and g(x) = 2x − 3, find (f ∘ g)(2).

A0
B9
C5
D2
Explanation
Composition (f ∘ g)(x) = f(g(x)) — apply g first, then f. Step 1: g(2) = 2(2) − 3 = 1. Step 2: f(1) = 1² + 1 = 2. Common mistake (C): computing (g ∘ f)(2) instead — f(2) = 5. Order matters: in f ∘ g, the function written second (g) is applied first.
Question 3 of 10
TEKS 1A-1GMedium

A student computes that an investment doubles in 4 years. Before re-checking the algebra, what is the BEST quick reasonableness check?

AVerify the answer is an integer
BVerify the answer is positive
CCompute the cube root of the answer
DUse the Rule of 72: doubling at rate r% takes ≈ 72/r years, so 4 years implies r ≈ 18%
Explanation
Reasonableness checks (TEKS 1.G) tie the answer to the context. The Rule of 72 says doubling time ≈ 72 ÷ r%. So 4 years → r ≈ 72/4 = 18% — plausible for an aggressive/risky investment but not a savings account. If the problem says "standard savings at 2%," 4 years is unreasonable (should be ~36 years), flagging a calculation error. Positive/integer checks don't constrain the answer to the *context*; cube root is irrelevant here.
Question 4 of 10
TEKS 2A-2JEasy

Find the domain of f(x) = √(x − 4).

A[4, ∞)
B(−∞, 4)
C(4, ∞)
D(−∞, ∞)
Explanation
The expression inside a square root must be ≥ 0 for a real output. So x − 4 ≥ 0 → x ≥ 4 → interval [4, ∞). The bracket [ at 4 includes x = 4 because √(4 − 4) = √0 = 0 is a real number. Common mistake (D): writing (4, ∞) excludes 4, but √0 is defined. Use parentheses only when the endpoint is NOT in the domain.
Question 5 of 10
TEKS 3A-3DHard

Find the horizontal asymptote of R(x) = (3x² − x + 5) / (x² + 2x − 7).

Ay = 0
By = 3
Cy = 1
DNo horizontal asymptote (oblique asymptote instead)
Explanation
Horizontal asymptote rules: (1) deg(num) < deg(denom) → HA at y = 0. (2) deg(num) = deg(denom) → HA at y = ratio of leading coefficients. (3) deg(num) > deg(denom) → no HA (oblique if exactly one degree higher). Here both have degree 2 (case 2). Leading coefficients: 3 and 1. So HA: y = 3. Verify intuitively: as |x| → ∞, lower-order terms become insignificant and R(x) ≈ 3x²/x² = 3.
Question 6 of 10
TEKS 2A-2JHard

A piecewise function is defined as:

f(x) = x + 2 if x < 1
f(x) = 3x if x ≥ 1

Find f(1) and determine whether f is continuous at x = 1.

Af(1) is undefined because x = 1 is on the boundary
Bf(1) = 1 + 2 = 3, but the function jumps to 3·1 = 3 from the right (which is the same), so it's continuous by coincidence only
Cf(1) = 3, and f is NOT continuous because piecewise functions are never continuous at their boundaries
Df(1) = 3, and f IS continuous because the left-hand limit (1+2=3) equals the function value (3·1=3)
Explanation
x = 1 falls into the second piece (x ≥ 1), so f(1) = 3(1) = 3. Continuity check: left-hand limit from the first piece: lim(x→1⁻) (x + 2) = 3. Right-hand limit and value: f(1) = 3. Since all three agree, f is continuous at x = 1. Key rule: a piecewise function is continuous at a boundary iff both one-sided limits agree with the function's value there. (A) is a common myth — piecewise functions CAN be continuous when pieces match.
Question 7 of 10
TEKS 3A-3DMedium

Identify the vertical asymptote and any holes of R(x) = (x² − 4) / (x² − 5x + 6).

AVA at x = 3; hole at x = 2
BNo vertical asymptote; holes at x = 2 and x = 3
CVA at x = 2; hole at x = 3
DVAs at x = 2 and x = 3; no holes
Explanation
Factor both: x² − 4 = (x−2)(x+2); x² − 5x + 6 = (x−2)(x−3). Simplify: R(x) = (x+2)/(x−3), with the restriction x ≠ 2. The cancelled (x−2) leaves a hole at x = 2 (function undefined there, but simplified form is). The remaining (x−3) in the denominator gives a vertical asymptote at x = 3 (simplified form blows up). Rule: cancelled factors → holes; uncancelled denominator factors → vertical asymptotes.
Question 8 of 10
TEKS 3A-3DEasy

Find all REAL zeros of P(x) = x³ − 4x² + x − 4 by factoring.

Ax = 4, x = 1, x = −1
Bx = 4, x = i, x = −i (treating i as real)
Cx = 4 only
Dx = −4 only
Explanation
Factor by grouping: P(x) = x²(x − 4) + 1(x − 4) = (x − 4)(x² + 1). Solving each factor: x − 4 = 0 → x = 4; x² + 1 = 0 → x² = −1, which has NO real solutions (only complex i and −i). The question asks for REAL zeros, so the answer is x = 4 only. Verify: P(4) = 64 − 64 + 4 − 4 = 0 ✓.
Question 9 of 10
TEKS 1A-1GEasy

Three students each model the same growth pattern 1, 2, 4, 8, 16, …

Student A writes f(n) = 2n−1. Student B writes f(1) = 1, f(n) = 2·f(n−1). Student C draws a table. Which statement is TRUE?

AStudent C is wrong because tables can never replace a formula
BOnly Student B is correct because recursive forms are more rigorous
COnly Student A is correct because formulas are required
DAll three correctly represent the same pattern in valid forms
Explanation
Multiple representations of the same function — explicit formula, recursive rule, table, or graph — are all valid mathematical communication (TEKS 1.D–1.E). Student A's explicit form gives f(1)=1, f(2)=2, f(3)=4, … ✓. Student B's recursive rule generates the same sequence step by step. A complete table also encodes the pattern. Choose whichever representation makes the problem easiest.
Question 10 of 10
TEKS 3E-3IMedium

A bank account pays 6% annual interest compounded continuously. If $5,000 is deposited, the balance after t years is A(t) = 5000 · e0.06t. To the nearest dollar, what is the balance after 10 years?

A$9,111
B$50,000
C$5,600
D$8,000
Explanation
Continuous compounding: A = P·e^(rt). Here P = $5,000, r = 0.06, t = 10. Compute: 0.06·10 = 0.6, then e^0.6 ≈ 1.8221, then 5000·1.8221 ≈ $9,111. Common mistakes: (A) uses a flawed simple-interest estimate; (B) round guess; (D) multiplies principal by t (confuses compound growth with linear growth). Insight: even moderate rates over long horizons nearly double the principal due to the exponential nature of e^(rt).

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