Algebra 2 — Semester B Practice

Question 1 of 10

TEKS 6M-6P Hard

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

A x = 5

B x = 2

C x = −2

D no hole

Explanation

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

Question 2 of 10

TEKS 5A-5C Medium

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

A 1/40

B 1/8

C 1/16

D 1/4

Explanation

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

Question 3 of 10

TEKS 7A-7I Medium Diagram

The graph shown most likely belongs to which polynomial?

A Odd degree, negative leading coefficient

B A line

C Odd-degree polynomial with positive leading coefficient

D Even-degree polynomial

Explanation

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

Question 4 of 10

TEKS 5A-5C Medium Diagram

Which equation matches this exponential graph?

A y = x²

B y = log₂(x)

C y = (1/2)ˣ (decay)

D y = 2ˣ (growth)

Explanation

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

Question 5 of 10

TEKS 5A-5C Medium

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

A 3

B log(625)

C log(133)

D 4

Explanation

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

Question 6 of 10

TEKS 6M-6P Easy Diagram

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

A Absolute value

B Polynomial

C Linear function

D Rational function

Explanation

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

Question 7 of 10

TEKS 6M-6P Easy Diagram

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

A A V-shape

B A parabola opening up

C A two-branch hyperbola in quadrants I and III

D A 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 8 of 10

TEKS 5A-5C Hard

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

A ≈ $6,750

B ≈ $7,500

C ≈ $5,300

D ≈ $6,500

Explanation

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

Question 9 of 10

TEKS 5A-5C Medium

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

A P(1 + r)^t

B Pe^r/t

C Pe^rt

D P(rt)

Explanation

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

Question 10 of 10

TEKS 5A-5C Easy Diagram

Which graph shows exponential decay?

A A (curve rising)

B B (curve falling toward x-axis)

C Both

D Neither

Explanation

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

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