Answer: The $x$-coordinate of the vertex

Connect the formula to vertex form: In $a(x - h)^2 + k$, the value $h$ is the $x$-coordinate of the vertex, and it equals $-b/(2a)$ where $b$ is the coefficient of $x$ in standard form. Why the correct answer works: By definition, $h = -b/(2a)$ locates the horizontal position of the vertex, which is the $x$-coordinate. Why distractors fail: Option A describes $k$, not $h$. Option C is $a$, which is independent of this formula. Option D is $b^2 - 4ac$, a different quantity entirely.

Answer: $2x^2 - 16x + 29$

Expand the vertex form: $2(x - 4)^2 - 3 = 2(x^2 - 8x + 16) - 3 = 2x^2 - 16x + 32 - 3 = 2x^2 - 16x + 29$. Why the correct answer works: Direct expansion confirms $2x^2 - 16x + 29$. Why distractors fail: Option B uses $-8x$ instead of $-16x$ (forgot to multiply the middle term by $2$). Option C has $32 + 3 = 35$ instead of $32 - 3 = 29$. Option D has a positive $16x$ (sign error).

Answer: $-(x - 3)^2 + 4$

Factor out the leading coefficient: Factor $-1$ from the first two terms: $-(x^2 - 6x) - 5$. Complete the square inside: Half of $-6$ is $-3$, and $(-3)^2 = 9$. Add and subtract $9$ inside: $-[(x^2 - 6x + 9) - 9] - 5 = -[(x - 3)^2 - 9] - 5$. Distribute and simplify: $-(x - 3)^2 + 9 - 5 = -(x - 3)^2 + 4$. Why distractors fail: Option B uses $(x + 3)$ instead of $(x - 3)$, a sign error inside the square. Option C has $-4$ instead of $+4$. Option D drops the negative leading coefficient.

Answer: $b = 4$

Determine when k is an integer: For $x^2 + bx + 10$, $k = 10 - (b/2)^2 = 10 - b^2/4$. This is an integer when $b^2/4$ is an integer, i.e., when $b$ is even. Check each option: $b = 3$: $k = 10 - 9/4 = 31/4$ (fraction). $b = 4$: $k = 10 - 4 = 6$ (integer ✓). $b = 5$: $k = 10 - 25/4 = 15/4$ (fraction). $b = 7$: $k = 10 - 49/4 = -9/4$ (fraction). Why distractors fail: Options A, C, and D all have odd values of $b$, which produce $b^2/4$ as a non-integer, making $k$ fractional.

Answer: They are equivalent representations of the same expression; vertex form highlights the vertex, while standard form highlights the coefficients.

Understand equivalent forms: Both forms represent the same quadratic expression. Completing the square converts standard form to vertex form, and expanding vertex form returns standard form. Why the correct answer works: The two forms are algebraically equivalent. Each emphasizes different features: standard form shows coefficients $a$, $b$, $c$, while vertex form reveals the vertex $(h, k)$. Why distractors fail: Option A is false—they describe the same expression. Option B is false—the equivalence is exact, not approximate. Option D is false—vertex form can always be expanded back to standard form.

Answer: $2(x + 2)^2 - 7$

Factor out the leading coefficient: Factor $2$ from the first two terms: $2(x^2 + 4x) + 1$. Complete the square inside: Half of $4$ is $2$, and $2^2 = 4$. Add and subtract $4$ inside: $2(x^2 + 4x + 4 - 4) + 1 = 2[(x + 2)^2 - 4] + 1$. Distribute and simplify: $2(x + 2)^2 - 8 + 1 = 2(x + 2)^2 - 7$. Why distractors fail: Option A forgets to subtract the compensating term ($-8$). Option C uses $4$ instead of $2$ inside the square (forgot to halve). Option D subtracts only $4$ instead of $2 \times 4 = 8$ (forgot to distribute).

Answer: The first student is correct; the classmate forgot to distribute the $2$.

Compute step by step: $2(x^2 + 6x) + 20$. Half of $6$ is $3$, $3^2 = 9$. $2(x^2 + 6x + 9 - 9) + 20 = 2(x + 3)^2 - 18 + 20 = 2(x + 3)^2 + 2$. Why the first student is correct: The result $2(x + 3)^2 + 2$ matches the computation. Expanding confirms: $2(x^2 + 6x + 9) + 2 = 2x^2 + 12x + 18 + 2 = 2x^2 + 12x + 20$. ✓ Diagnose the classmate's error: The classmate subtracted only $9$ instead of $2 \times 9 = 18$: $20 - 9 = 11$. This is the most common mistake when $a \neq 1$.

Answer: To preserve the value of the original expression while creating a perfect square trinomial

Understand the purpose of the technique: Adding and subtracting the same quantity is equivalent to adding zero. This lets us restructure the expression without changing its value. Why the correct answer works: The goal is to form a perfect square trinomial while keeping the expression equivalent to the original—exactly what adding and subtracting $(b/2)^2$ achieves. Why distractors fail: Option A: the result is still quadratic, not linear. Option C: the constant term is adjusted, not eliminated. Option D: the leading coefficient is not changed by this process.

Answer: The first student is correct.

Perform the computation: Factor out $3$: $3(x^2 + 6x) + 10$. Half of $6$ is $3$, $3^2 = 9$. So $3(x^2 + 6x + 9 - 9) + 10 = 3(x + 3)^2 - 27 + 10 = 3(x + 3)^2 - 17$. Why the first student is correct: The computation confirms $3(x + 3)^2 - 17$, matching the first student's answer. Why the second student erred: The second student likely subtracted only $9$ instead of $3 \times 9 = 27$, forgetting to distribute the factor of $3$ when removing the compensating term: $10 - 9 = 1$ instead of $10 - 27 = -17$.

Answer: Factor out $3$ from the $x^2$ and $x$ terms

Recognize when a ≠ 1: When the leading coefficient $a \neq 1$, you must first factor $a$ out of the $x^2$ and $x$ terms before completing the square inside. Why the correct answer works: Factoring out $3$ gives $3(x^2 + 4x) + 7$, which sets up the expression so you can complete the square on $x^2 + 4x$ inside the parentheses. Why distractors fail: Option A skips the factoring step, which leads to errors when $a \neq 1$. Option C treats this as an equation rather than an expression. Option D changes the value of the expression unless you are solving an equation set equal to zero.

Answer: Yes, since $h = -b/(2a)$ and expanding gives $k = c - a \cdot b^2/(4a^2) = c - b^2/(4a)$.

Check the formula algebraically: Expand $a(x - h)^2 + k = ax^2 - 2ahx + ah^2 + k$. Matching with $ax^2 + bx + c$: $b = -2ah$ and $c = ah^2 + k$, so $k = c - ah^2$. Verify with h = -b/(2a): $ah^2 = a \cdot b^2/(4a^2) = b^2/(4a)$. So $k = c - b^2/(4a)$. The formula $k = c - ah^2$ is valid. Why distractors fail: Option A has the wrong sign. Option B uses $b^2/(2a)$ instead of $b^2/(4a)$. Option D is false because $k$ can be computed both by substitution and by formula.

Answer: The minimum value is $k$, occurring at $x = h$.

Analyze the structure: Since $a > 0$, the term $a(x - h)^2 \geq 0$ for all $x$, and it equals $0$ when $x = h$. Determine the minimum: The minimum value of $a(x - h)^2 + k$ is $0 + k = k$, achieved when $x = h$. Why distractors fail: Option A swaps $h$ and $k$. Option C incorrectly claims the minimum is $0$ (it ignores $k$). Option D is incorrect because a positive $a$ guarantees a minimum exists.

Answer: $3(x - 2)^2 + 3$; minimum value is $3$

Factor out the leading coefficient: $3(x^2 - 4x) + 15$. Complete the square: Half of $-4$ is $-2$, $(-2)^2 = 4$. So $3(x^2 - 4x + 4 - 4) + 15 = 3(x - 2)^2 - 12 + 15 = 3(x - 2)^2 + 3$. Determine the minimum: Since $3 > 0$, the minimum occurs at $x = 2$ and equals $k = 3$. Why distractors fail: Option B does not subtract the compensating $12$. Option C subtracts $18$ instead of $12$. Option D uses $-6$ inside instead of $-2$ (forgot to halve after factoring).

Answer: $(x - 1)^2 + 4$

Complete the square: Half of $-2$ is $-1$, and $(-1)^2 = 1$. So $x^2 - 2x + 1 - 1 + 5 = (x - 1)^2 + 4$. Why the correct answer works: $(x - 1)^2 + 4$ expands to $x^2 - 2x + 1 + 4 = x^2 - 2x + 5$, matching the original. Why distractors fail: Option A does not subtract $1$ from the constant (gives $x^2 - 2x + 6$). Option B uses $-2$ instead of $-1$ (forgot to halve). Option D has $(x + 1)^2$, a sign error.

Answer: Add and subtract $(b/2)^2$ inside the expression

Recall the halve-and-square procedure: To complete the square for $x^2 + bx$, take half of $b$, square it to get $(b/2)^2$, then add and subtract this value so the expression's value does not change. Why the correct answer works: Adding and subtracting $(b/2)^2$ creates a perfect square trinomial $x^2 + bx + (b/2)^2 = (x + b/2)^2$ while the subtracted $(b/2)^2$ compensates to preserve equality. Why distractors fail: Option A adds $b^2$ without halving or subtracting, changing the expression's value. Option C multiplies, which is not part of the procedure. Option D introduces an $x^2$ term that changes the degree of the expression.

Answer: $h = 6$, $k = 4$

Halve b: $h = b/2 = 12/2 = 6$. Compute k: $k = c - (b/2)^2 = 40 - 36 = 4$. Verify: $(x + 6)^2 + 4 = x^2 + 12x + 36 + 4 = x^2 + 12x + 40$. ✓ Why distractors fail: Option B uses $b$ and $c$ directly without computation. Option C has a sign error on $k$. Option D adds $36 + 40$ instead of subtracting.

Answer: Maximum value $8$ at $x = -1$

Interpret the leading coefficient: Since $a = -2 < 0$, the parabola opens downward, so the vertex is a maximum. Read the vertex from vertex form: $-2(x + 1)^2 + 8 = -2(x - (-1))^2 + 8$, so $h = -1$ and $k = 8$. The maximum value is $8$ at $x = -1$. Why distractors fail: Option A has $x = 1$ instead of $x = -1$ (sign error reading $h$ from $(x + 1)$). Option C reverses $h$ and $k$. Option D is wrong because a negative $a$ guarantees a maximum exists.

Answer: $a(x - h)^2 + k$

Recall the definition of vertex form: Vertex form is defined as $a(x - h)^2 + k$, where $(h, k)$ is the vertex of the parabola. Why the correct answer works: Option B matches the standard definition of vertex form with parameters $a$, $h$, and $k$. Why distractors fail: Option A is standard form. Option C is factored form. Option D has an incorrect structure that does not generalize to all quadratics.

Answer: $(2, 5)$

Complete the square: Half of $-4$ is $-2$, and $(-2)^2 = 4$. So $x^2 - 4x + 4 - 4 + 9 = (x - 2)^2 + 5$. Read the vertex: From $(x - 2)^2 + 5$, the vertex is $(2, 5)$. Why distractors fail: Option A reads $b$ and $c$ directly as coordinates without completing the square. Option B has the wrong sign for the $x$-coordinate. Option D has the wrong sign for $k$.

Answer: The student forgot to subtract the added value; adding $4$ inside the parentheses adds $5 \times 4 = 20$ to the expression, which must be subtracted.

Trace the student's work: The student correctly factored $5$: $5(x^2 + 4x) + 3$, and correctly identified that $4$ should be added inside. However, adding $4$ inside parentheses multiplied by $5$ adds $20$ total. Identify the missing step: The student must subtract $20$ outside: $5(x + 2)^2 - 20 + 3 = 5(x + 2)^2 - 17$. The student wrote $5(x + 2)^2 + 3$, omitting the subtraction of $20$. Why distractors fail: Option A is wrong because the student correctly halved $4$ (the coefficient of $x$ inside the parentheses), not $20$. Option C is wrong because factoring $5$ out is the correct first step. Option D is wrong because $(x + 2)$ is correctly signed.

Answer: $(x + 5/2)^2 - 13/4$

Apply the halve-and-square step with an odd coefficient: Half of $5$ is $5/2$, and $(5/2)^2 = 25/4$. Add and subtract $25/4$: $x^2 + 5x + 25/4 - 25/4 + 3$. Simplify the constant: $(x + 5/2)^2 + (3 - 25/4) = (x + 5/2)^2 + (12/4 - 25/4) = (x + 5/2)^2 - 13/4$. Why distractors fail: Option B has a positive $13/4$ instead of negative. Option C uses $5$ instead of $5/2$ (forgot to halve). Option D computes $3 + 25/4$ subtracted rather than $3 - 25/4$.

Answer: Completing the square works for any $a \neq 0$; when $a \neq 1$, you first factor $a$ from the $x^2$ and $x$ terms before completing the square inside.

Clarify the scope of the technique: Completing the square is a general algebraic technique that works for any quadratic expression with $a \neq 0$. Why the correct answer works: When $a \neq 1$, we factor $a$ out of the first two terms, then complete the square on the resulting expression with leading coefficient 1 inside the parentheses. Why distractors fail: Option A incorrectly limits the method to $a = 1$. Option C incorrectly restricts $a$ to positive integers; it works for any nonzero real $a$. Option D is false; $b$ can be any real number, though fractions may arise when $b$ is odd.

Answer: $(x - 7)^2 + 1$; vertex $(7, 1)$

Apply the halve-and-square step: Half of $-14$ is $-7$, and $(-7)^2 = 49$. So $x^2 - 14x + 49 - 49 + 50 = (x - 7)^2 + 1$. Read the vertex: From $(x - 7)^2 + 1$, the vertex is $(h, k) = (7, 1)$. Why distractors fail: Option B has $k = -1$ instead of $+1$ (sign error in $-49 + 50$). Option C has $(x + 7)^2$ instead of $(x - 7)^2$ (sign error when halving). Option D fails to halve the coefficient and does not adjust the constant.

Answer: $-4(x - 2)^2 + 9$; vertex $(2, 9)$

Factor out the leading coefficient: $-4(x^2 - 4x) - 7$. Complete the square inside: Half of $-4$ is $-2$, $(-2)^2 = 4$. $-4(x^2 - 4x + 4 - 4) - 7 = -4[(x - 2)^2 - 4] - 7$. Distribute and simplify: $-4(x - 2)^2 + 16 - 7 = -4(x - 2)^2 + 9$. The vertex is $(2, 9)$. Why distractors fail: Option B has $(x + 2)$ instead of $(x - 2)$. Option C subtracts $16$ instead of adding it (distributing $-4 \times -4 = +16$, not $-16$). Option D adds $16 + 16 = 32$ incorrectly.

Answer: The vertex (minimum or maximum point) of the parabola

Identify what vertex form reveals: The vertex form $a(x - h)^2 + k$ is specifically designed so that $(h, k)$ can be read directly as the vertex. Why the correct answer works: The vertex is the minimum point when $a > 0$ or the maximum point when $a < 0$, and it is located at $(h, k)$. Why distractors fail: Option A: $x$-intercepts are found by setting the expression to zero, not read from $(h, k)$. Option B: the $y$-intercept is found by substituting $x = 0$. Option D: while $x = h$ is the axis of symmetry, $(h, k)$ as a point represents the vertex, not the discriminant.

Answer: $25$

Apply the halve-and-square step: To complete the square, take the coefficient of $x$, which is $10$, halve it to get $5$, then square it: $5^2 = 25$. Why 25 is correct: We add and subtract $25$ so that $x^2 + 10x + 25$ becomes the perfect square $(x + 5)^2$. Why distractors fail: Option A ($5$) is only the halved value before squaring. Option C ($10$) is the original coefficient, not halved or squared. Option D ($100$) squares the coefficient without halving first.

Answer: Substitute $x = 0$ into the original and both forms

Identify an efficient verification strategy: Substituting a specific value like $x = 0$ into the original and each candidate form quickly reveals which is equivalent. Apply the check: Original at $x=0$: $0 + 0 + 30 = 30$. Student A: $(0+5)^2 + 5 = 30$. ✓ Student B: $(0+5)^2 + 55 = 80$. ✗ So Student A is correct. Why distractors fail: Option B only checks that $h$ is correct; both students have the same $h$, so this does not distinguish them. Option C and D are true for both forms and do not help differentiate.

Answer: Yes, this is correct.

Verify by completing the square: Half of $8$ is $4$, and $4^2 = 16$. So $x^2 + 8x + 16 - 16 + 20 = (x + 4)^2 + 4$. Verify by expanding: $(x + 4)^2 + 4 = x^2 + 8x + 16 + 4 = x^2 + 8x + 20$. This matches the original expression. Why distractors fail: Option A gives $(x+4)^2 - 4$, which expands to $x^2 + 8x + 12 \neq x^2 + 8x + 20$. Option B does not halve the coefficient. Option D adds $16 + 20$ instead of computing $-16 + 20 = 4$.

Answer: $(x + 3)^2 - 7$

Halve-and-square step: The coefficient of $x$ is $6$. Half of $6$ is $3$, and $3^2 = 9$. Add and subtract $9$: $x^2 + 6x + 9 - 9 + 2$. Write as a perfect square: $(x + 3)^2 - 9 + 2 = (x + 3)^2 - 7$. Why distractors fail: Option B has the wrong sign on the constant ($+7$ instead of $-7$). Option C uses $6$ instead of $3$ (forgot to halve). Option D subtracts $9 + 2 = 11$ incorrectly instead of $9 - 2 = 7$.

Answer: $4$

Factor and complete the square: Factor $4$: $4(x^2 - 6x) + 40$. Half of $-6$ is $-3$, and $(-3)^2 = 9$. So $4(x^2 - 6x + 9 - 9) + 40 = 4(x - 3)^2 - 36 + 40$. Find k: $k = -36 + 40 = 4$. Why distractors fail: Option B has a sign error ($-4$ instead of $+4$). Option C subtracts only $9$ instead of $4 \times 9 = 36$: $40 - 9 = 31$. Option D may come from $40 - 24 = 16$, an incorrect subtraction.