Algebra 2 — Semester B Practice

Question 1 of 10

TEKS 6M-6PEasy Diagram

Which graph corresponds to f(x) = 1/x?

AA parabola opening up

BA two-branch hyperbola in quadrants I and III

CA V-shape

DA line through the origin

Explanation

f(x) = 1/x has two branches: positive x → positive y (Q I), negative x → negative y (Q III), with asymptotes at the axes.

Question 2 of 10

TEKS 6M-6PEasy Diagram

For the function whose graph approaches the dashed lines, what type of function is this most likely?

AAbsolute value

BPolynomial

CRational function

DLinear function

Explanation

Both vertical and horizontal asymptotes are characteristic of rational functions where degrees of numerator and denominator are similar.

Question 3 of 10

TEKS 5A-5CMedium

Use log properties to simplify: log(8) + log(125).

Blog(133)

Dlog(625)

Explanation

log(a) + log(b) = log(ab). log(8) + log(125) = log(1000) = 3 (assuming log base 10).

Question 4 of 10

TEKS 5A-5CEasy Diagram

Which graph shows exponential decay?

AA (curve rising)

BBoth

CNeither

DB (curve falling toward x-axis)

Explanation

Exponential decay: starts high, falls toward zero. Graph B matches; graph A is exponential growth.

Question 5 of 10

TEKS 5A-5CMedium

Which represents continuous compound interest of $P at rate r for t years?

AP(1 + r)^t

BP(rt)

CPe^rt

DPe^r/t

Explanation

Continuous compounding uses A = Pe^(rt). The discrete annual formula is P(1 + r)^t.

Question 6 of 10

TEKS 7A-7IMedium Diagram

The graph shown most likely belongs to which polynomial?

AOdd-degree polynomial with positive leading coefficient

BA line

CEven-degree polynomial

DOdd degree, negative leading coefficient

Explanation

Left end goes up (+∞), right end goes down (−∞). That signature is odd degree, negative leading coefficient.

Question 7 of 10

TEKS 5A-5CHard

$5,000 invested at 6% continuously for 5 years (use A = Pe^rt):

A≈ $6,750

B≈ $7,500

C≈ $6,500

D≈ $5,300

Explanation

A = 5000 · e^(0.06·5) = 5000 · e^0.3 ≈ 5000 · 1.3499 ≈ $6,749.

Question 8 of 10

TEKS 5A-5CMedium

A radioactive isotope has a half-life of 10 years. What fraction remains after 40 years?

A1/40

B1/16

C1/4

D1/8

Explanation

40 / 10 = 4 half-lives. (1/2)⁴ = 1/16.

Question 9 of 10

TEKS 5A-5CMedium Diagram

Which equation matches this exponential graph?

Ay = log₂(x)

By = 2ˣ (growth)

Cy = (1/2)ˣ (decay)

Dy = x²

Explanation

Curve approaches 0 as x → −∞ and grows rapidly as x increases → exponential growth.

Question 10 of 10

TEKS 6M-6PHard

For f(x) = (x + 2) / [(x + 2)(x − 5)], where is the hole?

Ano hole

Bx = −2

Cx = 5

Dx = 2

Explanation

(x + 2) cancels → simplifies to 1/(x − 5). The cancelled factor (x+2 = 0 at x = −2) creates a hole.

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