Algebra 1 — Semester B Practice

Question 1 of 10

TEKS 12A-12E Easy Calc

For the quadratic function f(x) = x² + 1, find f(−2).

⚠️ Pay close attention to what happens when you square a negative number.

A −5, because the result of any function at a negative input is negative

B 3, because −2 + 1 + 2² = 3

C 5, because (−2)² = 4 and 4 + 1 = 5

D −3, because (−2)² = −4 and −4 + 1 = −3

Explanation

Substitute x = −2 into f(x) = x² + 1: f(−2) = (−2)² + 1 = 4 + 1 = 5.

The critical step: (−2)² = (−2) × (−2) = +4, NOT −4. A negative times a negative is positive.

Common mistake that gets distractor (A): writing −2² instead of (−2)². Without parentheses, −2² is read by order of operations as −(2²) = −(4) = −4. With parentheses (−2)², you square the negative number itself, giving +4. Always write parentheses around a negative input to a power.

Question 2 of 10

TEKS 10A-10E Easy Word Diagram

A rectangular vegetable garden has a length of (x + 7) feet and a width of (x + 3) feet. Which polynomial represents the area of the garden in square feet?

A 2x + 10

B x² + 4x + 21

C x² + 10x + 21

D x² + 21

Explanation

Area = length × width = (x + 7)(x + 3). Use FOIL: x·x + x·3 + 7·x + 7·3 = x² + 3x + 7x + 21 = x² + 10x + 21. Choice A only multiplies the first and last terms (skips the middle). C is the perimeter formula 2(L + W) wrongly applied. D mixes up the middle coefficient.

Question 3 of 10

TEKS 1A-1G Easy Calc

A square garden has an area of 64 square feet. If the side length is x feet, the relationship is x² = 64.

The algebra gives two mathematical solutions, but only one makes physical sense as a side length. Which choice gives BOTH algebraic solutions AND correctly identifies the physical side length?

A Only x = −8, because squaring a negative gives a positive

B x = 32, because half of 64 is the side length

C Only x = 8 — that's the only solution to x² = 64

D x = ±8 algebraically; the garden's side length is +8 ft

Explanation

Taking the square root of both sides gives x = ±√64 = ±8, because both (+8)² = 64 AND (−8)² = 64. In pure algebra you must report both roots. For the physical garden, length can't be negative, so only +8 ft applies in context — but the algebraic answer set is {−8, +8}.

Common mistakes: (A) Dropping the negative root entirely. (D) Dividing 64 by 2 instead of taking a square root.

Tip: when you see x² = N, always write x = ±√N first, then decide which roots fit the real-world context.

Question 4 of 10

TEKS 12A-12E Easy Calc Word

What is the 5th term of geometric sequence: 2, 6, 18, ...?

A 486

B 108

C 54

D 162

Explanation

📌 r = 3. a₅ = 2·3⁴ = 2·81 = 162

Question 5 of 10

TEKS 6A-6C Medium Calc Word Diagram

The parabola in the graph below has a vertex at which point?

A (1, -2)

B (0, 0)

C (2, 1)

D (-1, 2)

Explanation

The vertex is the lowest point of an upward-opening parabola, marked at approximately (1, -2).

Question 6 of 10

TEKS 7A-7C Easy Calc Word

The vertex of y = (x − 3)² + 2 is:

A (2, 3)

B (−3, 2)

C (3, 2)

D (3, −2)

Explanation

📌 Vertex form: y = a(x−h)² + k. Vertex = (h, k) = (3, 2).

Question 7 of 10

TEKS 8A-8B Medium Calc Diagram

For a quadratic equation ax² + bx + c = 0, the discriminant is b² − 4ac. It tells us how many times the related parabola y = ax² + bx + c crosses the x-axis — and therefore how many real solutions the equation has.

The discriminant's sign matches the number of x-axis crossings.

Which statement about D = 0 is TRUE?

A Exactly one real solution — the parabola is tangent to the x-axis

B Two real solutions — the parabola crosses the x-axis twice

C The number of solutions can't be determined from D alone

D No real solutions — the parabola doesn't reach the x-axis

Explanation

The discriminant b² − 4ac counts how many real solutions a quadratic equation has by reflecting how many times the parabola y = ax² + bx + c crosses the x-axis.

The three cases:
• D > 0: parabola crosses the x-axis at TWO different points → 2 real solutions.
• D = 0: parabola is *tangent* to the x-axis — it touches at exactly ONE point (the vertex itself sits on the x-axis) → 1 real solution (often called a *double root*).
• D < 0: parabola sits entirely above or below the x-axis, never touching it → 0 real solutions (the roots are complex / imaginary).

Application: D = 0 is the borderline case useful for problems like "for what value of c does ax² + bx + c = 0 have exactly one solution?" — set b² − 4ac = 0 and solve.

Question 8 of 10

TEKS 7A-7C Easy Calc Word Diagram

Which graph represents a quadratic function?

A Both

B Neither

C B

D A

Explanation

📌 Quadratic = U-shaped parabola. Graph A shows a parabola.
Graph B is a straight line → linear, not quadratic.

Question 9 of 10

TEKS 9A-9C Easy Word

Which represents exponential growth?

A y = x²

B y = 2/x

C y = 2ˣ

D y = 3x + 2

Explanation

📌 Exponential growth: y = a·bˣ where b > 1. y = 2(3)ˣ has b = 3 > 1.

Question 10 of 10

TEKS 11A-11B Easy Calc

A biologist models a bacterial colony where each hour, the population multiplies by x. After 3 hours the multiplier is x³, and after another 4 hours it's x⁴. The combined growth factor across all 7 hours is x³ · x⁴.

Using the product of powers property, simplify x³ · x⁴ to a single power of x.

A 2x⁷ — the two factors double the coefficient

B x¹² — multiply the exponents

C x¹ — subtract the exponents

D x⁷ — add the exponents

Explanation

Product of powers rule: when multiplying powers with the *same base*, add the exponents → x³ · x⁴ = x^(3+4) = x⁷.

Why the distractors are wrong: (B) Multiplying exponents (3×4=12) is the power-of-a-power rule, used only for (x³)⁴ — a single power raised to another power. (C) There's only one base, x, and no separate coefficients to double. (D) Subtracting exponents is the quotient rule (x³ ÷ x⁴), used for division, not multiplication.

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