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

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

A Use the Rule of 72: doubling at rate r% takes ≈ 72/r years, so 4 years implies r ≈ 18%
B Compute the cube root of the answer
C Verify the answer is positive
D Verify the answer is an integer
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 2 of 10
TEKS 3A-3D Medium

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

A VA at x = 2; hole at x = 3
B No vertical asymptote; holes at x = 2 and x = 3
C VAs at x = 2 and x = 3; no holes
D VA at x = 3; hole at x = 2
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 3 of 10
TEKS 2A-2J Easy

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 4 of 10
TEKS 1A-1G Easy

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?

A Student C is wrong because tables can never replace a formula
B Only Student A is correct because formulas are required
C Only Student B is correct because recursive forms are more rigorous
D All 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 5 of 10
TEKS 3A-3D Easy

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

A x = 4, x = i, x = −i (treating i as real)
B x = −4 only
C x = 4, x = 1, x = −1
D x = 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 6 of 10
TEKS 3E-3I Medium

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 $8,000
B $5,600
C $9,111
D $50,000
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).
Question 7 of 10
TEKS 2A-2J Medium

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

A 2
B 0
C 9
D 5
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 8 of 10
TEKS 3A-3D Hard

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

A y = 3
B y = 0
C y = 1
D No 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 9 of 10
TEKS 2A-2J Hard

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.

A f(1) = 3, and f IS continuous because the left-hand limit (1+2=3) equals the function value (3·1=3)
B f(1) is undefined because x = 1 is on the boundary
C f(1) = 3, and f is NOT continuous because piecewise functions are never continuous at their boundaries
D f(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
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 10 of 10
TEKS 1A-1G Medium

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?

A Reject t = 2 because the smaller solution is always the answer
B Average the two: t = (2 + (−4))/2 = −1 second
C Reject t = −4 (negative time has no physical meaning here); report t = 2 seconds
D Report 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."

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