Pre-Calculus — Semester A Practice

Question 1 of 10

TEKS 1A-1GEasy Diagram

The graph below shows two functions: f(x) (blue line) and g(x) (red parabola). Other than the origin, at which value of x are f(x) = g(x)?

Ax = 5

Bx = −2

Cx = 2

Dx = 0 only (the origin is the only solution)

Explanation

Two functions are equal at a point of intersection of their graphs. The two curves cross twice — once at the origin and once around x = 2. Reading the graph, the non-origin intersection is at x = 2. TEKS 1.D: using graphs to solve f(x) = g(x).

Question 2 of 10

TEKS 2A-2JEasy Diagram

The function shown below has 180° rotational symmetry about the origin. Which property does f(x) satisfy?

Af(−x) = f(x) — that would make the function even

Bf(x) is constant — the graph is a flat horizontal line

Cf(0) = 1

Df(−x) = −f(x) — the function is ODD

Explanation

Origin symmetry ⇔ odd function: every (a, b) on the graph has (−a, −b) also on the graph, i.e., f(−a) = −f(a). Examples: x, x³, sin(x), tan(x). All odd functions defined at 0 satisfy f(0) = 0, not 1.

Question 3 of 10

TEKS 3A-3DEasy Diagram

The graph below shows a polynomial P(x). How many REAL zeros does P(x) have?

Explanation

Real zeros = number of x-axis CROSSINGS (the values where P(x) = 0). The graph crosses the x-axis at three distinct points → 3 real zeros. (Be careful: a parabola "touching" the x-axis is a single zero of multiplicity 2; here we have 3 clear crossings.)

Question 4 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?

AOnly Student A is correct because formulas are required

BStudent C is wrong because tables can never replace a formula

CAll three correctly represent the same pattern in valid forms

DOnly Student B is correct because recursive forms are more rigorous

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 1A-1GEasy

A student is solving (x + 2)² = 25. They take the square root of both sides and write x + 2 = 5, then conclude x = 3. What did they miss?

ANothing — x = 3 is the only solution

BThey should multiply both sides by 2 first

CThe equation has no real solution

DThey should write x + 2 = ±5, giving x = 3 OR x = −7

Explanation

When taking the square root of both sides of a squared expression, you must include both roots: x + 2 = ±5. Solving each: x = 3 OR x = −7. Verify: (−7 + 2)² = (−5)² = 25 ✓. Forgetting the negative root loses half your solutions in any squared equation.

Question 6 of 10

TEKS 3E-3IEasy Diagram

The graph below shows an exponential function passing through (0, 1), growing rapidly to the right, and approaching 0 as x → −∞. Which function is shown?

Af(x) = log₂(x)

Bf(x) = x²

Cf(x) = 2^x

Df(x) = −2^x

Explanation

Exponential growth f(x) = a^x with a > 1: passes through (0, 1), increases rapidly for positive x, approaches 0 (never touches) as x → −∞. The logarithm log₂(x) has the opposite shape; −2^x negates everything; the parabola x² has symmetry about the y-axis and doesn't approach 0.

Question 7 of 10

TEKS 3A-3DMedium Diagram

The graph below has a vertical asymptote at x = 1 (red dashed) and a horizontal asymptote at y = 0. Which function could this be?

Af(x) = (x − 1)²

Bf(x) = 1 / (x − 1)

Cf(x) = (x − 1) / (x + 1)

Df(x) = x / (x − 1)

Explanation

Vertical asymptote at x = 1: denominator must be 0 at x = 1 and not cancel. Only f(x) = 1 / (x − 1) satisfies this. Horizontal asymptote at y = 0: requires deg(numerator) < deg(denominator). It has constant (degree 0) over linear (degree 1) → HA at y = 0 ✓. The forms (x − 1) / (x + 1) and x / (x − 1) have HA at y = 1 (matching leading coefficients); (x − 1)² is a parabola, no asymptotes.

Question 8 of 10

TEKS 2A-2JEasy Diagram

The function shown below is symmetric about the y-axis. Which algebraic property does f(x) satisfy?

Af(−x) = −f(x) — the function is ODD

Bf(x) = f(x + 1) — the function is periodic with period 1

Cf(−x) = f(x) — the function is EVEN

Df(x) = −f(x) — the function is identically zero

Explanation

Symmetry about the y-axis ⇔ even function: for every (a, b) on the graph, (−a, b) is also on the graph, which is the algebraic statement f(−a) = f(a). Classic examples: x², |x|, cos(x). The property f(−x) = −f(x) describes rotational symmetry about the origin (odd functions like x³); f(x) = f(x + 1) describes a periodic pattern (like sin/cos repeating), unrelated to y-axis symmetry.

Question 9 of 10

TEKS 2A-2JEasy

Find the domain of f(x) = √(x − 4).

A(−∞, ∞)

B(4, ∞)

C[4, ∞)

D(−∞, 4)

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: writing (4, ∞) excludes 4, but √0 is defined. Use parentheses only when the endpoint is NOT in the domain.

Question 10 of 10

TEKS 3A-3DMedium

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 ✓.

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