Answer: $b = 3$

Substitute x = 2 into the expression: $f(2) = (2)^2 + b(2) + 10 = 4 + 2b + 10 = 14 + 2b$. Solve for b: Set $14 + 2b = 20$, which gives $2b = 6$, so $b = 3$. Why distractors fail: Option A ($b = 1$) gives $f(2) = 4 + 2 + 10 = 16 \neq 20$. Option C ($b = 5$) gives $f(2) = 4 + 10 + 10 = 24 \neq 20$. Option D ($b = -3$) gives $f(2) = 4 - 6 + 10 = 8 \neq 20$.

Answer: In $(-5)^2$, the base is $-5$ so the result is $25$; in $-5^2$, the exponent applies only to $5$, giving $-(25) = -25$.

Parentheses determine the base: When parentheses enclose the negative sign, as in $(-5)^2$, the entire quantity $-5$ is the base: $(-5) \times (-5) = 25$. Without parentheses, negation is applied last: In $-5^2$, exponentiation has higher precedence than the unary negative, so it means $-(5^2) = -25$. Why distractors fail: Option A claims they are equal, ignoring the order of operations. Option C reverses which expression is positive and which is negative. Option D references an irrelevant property.

Answer: Both approaches are valid and give the same result; Approach 1 is more efficient for a single evaluation.

Both methods are mathematically equivalent: Substitution can be performed on any algebraically equivalent form of an expression. Both vertex form and standard form yield the same value. Efficiency comparison: Approach 1 requires fewer arithmetic steps for a single evaluation, making it more efficient. Approach 2 might be useful if evaluating at many values. Why distractors fail: Options A and B each claim only one approach is valid, which is false. Option D incorrectly claims Approach 2 is always preferred; in fact, direct substitution in vertex form is often less error-prone for single evaluations.

Answer: $-7$

Substitute x = 0: $g(0) = 3(0)^2 + 2(0) - 7 = 0 + 0 - 7 = -7$. Key insight: When $x = 0$, the $x^2$ and $x$ terms vanish, leaving only the constant term $c = -7$. Why distractors fail: Option A ($0$) forgets the constant term. Option B ($-2$) may come from misadding. Option D ($3$) confuses the leading coefficient with the output.

Answer: $0$, $12$, $0$

Evaluate P(1): $P(1) = -3(1)^2 + 18(1) - 15 = -3 + 18 - 15 = 0$. Evaluate P(3): $P(3) = -3(9) + 18(3) - 15 = -27 + 54 - 15 = 12$. Evaluate P(5): $P(5) = -3(25) + 18(5) - 15 = -75 + 90 - 15 = 0$. Why distractors fail: Option B incorrectly computes $P(5) = -12$ (likely an arithmetic error with $-75 + 90 = 15$, then $15 - 15 \neq -12$). Option C gets $P(1) = 6$ from a sign error. Option D computes $P(3) = 18$ by omitting $-27$.

Answer: The student wrote $-2^2$ instead of $(-2)^2$, giving $-4$ instead of $4$.

Identify the error: The student wrote $-2^2 = -4$. In standard mathematical notation, $-2^2 = -(2^2) = -4$. However, when substituting $x = -2$ into $x^2$, the correct expression is $(-2)^2 = 4$. Correct computation: $f(-2) = (-2)^2 - 6(-2) + 8 = 4 + 12 + 8 = 24$. Why distractors fail: Option A is incorrect because $-6(-2) = 12$ was computed correctly. Option C is wrong because the student did include $8$. Option D is irrelevant — factoring is not required to evaluate an expression.

Answer: $-10$

Substitute x = -1 with parentheses: $f(-1) = -2(-1)^2 + 5(-1) - 3$. Compute each term carefully: $(-1)^2 = 1$, so $-2(1) = -2$. Then $5(-1) = -5$. So $f(-1) = -2 - 5 - 3 = -10$. Why distractors fail: Option B ($0$) may result from computing $(-1)^2 = -1$, giving $-2(-1) = 2$, then $2 - 5 - 3 = -6$… no, that gives $-6$. $0$ arises from additional sign errors. Option C ($-6$) comes from $(-1)^2 = -1$ yielding $2 - 5 - 3 = -6$. Option D ($4$) comes from treating $-2(-1)^2$ as $-2(-1) = 2$ and $5(-1)$ as $5$.

Answer: $f(x) = 2x^2 - 2x + 6$

Test Option A at x = -1 and x = 2: $f(-1) = 2(-1)^2 - 2(-1) + 6 = 2 + 2 + 6 = 10$ ✓. $f(2) = 2(4) - 2(2) + 6 = 8 - 4 + 6 = 10$ ✓. Test remaining options: Option B: $f(-1) = 1 - 1 + 8 = 8 \neq 10$ ✗. Option C: $f(-1) = 2 + 2 + 8 = 12 \neq 10$ ✗. Option D: $f(-1) = 3 - 3 + 4 = 4 \neq 10$ ✗. Why only Option A works: Only $f(x) = 2x^2 - 2x + 6$ produces an output of $10$ for both $x = -1$ and $x = 2$. The other options fail at least one of the two conditions.

Answer: Evaluate $f(x)$ at small positive integers starting from $1$, stopping at the first output of $0$

Reasoning about efficiency: For a quadratic with small integer coefficients, the roots (if they are integers) are likely to be small integers. Testing $f(1) = 2$, $f(2) = 0$ finds a root in just two evaluations. Why the correct answer works: Option B is efficient because it starts from small values and stops as soon as a zero is found, minimizing unnecessary work. Why distractors fail: Option A evaluates unnecessarily many values without a stopping criterion. Option C starts with an irrelevant large value. Option D incorrectly assumes negative inputs are more likely to be roots.

Answer: Student A is correct; Student B computed $-2(-2)$ as $-4$ instead of $+4$.

Compute the correct value: $g(-2) = 3(-2)^2 - 2(-2) + 4 = 3(4) - (-4) + 4 = 12 + 4 + 4 = 20$. Identify Student B's error: Student B likely computed $-2(-2)$ as $-4$ (forgetting that a negative times a negative is positive), arriving at $12 - 4 + 4 = 12$. Why distractors fail: Option A claims Student A squared incorrectly, but $20$ is the correct answer. Option C suggests a missing constant, but both values differ from $16$ (which is $20 - 4$). Option D is incorrect because $20$ is the right answer.

Answer: $b = 1$

Substitute x = 3 into the expression: $g(3) = 2(3)^2 + b(3) - 4 = 2(9) + 3b - 4 = 18 + 3b - 4 = 14 + 3b$. Solve for b: Set $14 + 3b = 17$, which gives $3b = 3$, so $b = 1$. Why distractors fail: Option B ($b = -1$) gives $g(3) = 14 - 3 = 11 \neq 17$. Option C ($b = 3$) gives $g(3) = 14 + 9 = 23 \neq 17$. Option D ($b = 7$) gives $g(3) = 14 + 21 = 35 \neq 17$.

Answer: 15

Substitute x = 2: Replace every $x$ with $2$: $f(2) = (2)^2 + 3(2) + 5$. Compute each term: $(2)^2 = 4$, $3(2) = 6$, and the constant is $5$. So $f(2) = 4 + 6 + 5 = 15$. Why distractors fail: Option A (13) results from computing $3(2)$ as $5$. Option C (11) results from omitting the constant term. Option D (17) results from squaring incorrectly as $(2)^2 = 6$.

Answer: Yes, because both $f(0)$ and $f(4)$ equal $4$.

Evaluate f(0): $f(0) = (0)^2 - 4(0) + 4 = 4$. Evaluate f(4): $f(4) = (4)^2 - 4(4) + 4 = 16 - 16 + 4 = 4$. Explain the symmetry: Both values equal $4$. This occurs because $0$ and $4$ are equidistant from the axis of symmetry at $x = 2$. Why distractors fail: Option A is incorrect on both counts: $f(0) = 4$, not an arbitrary value, and $f(4) = 4$, not $0$. Neither output is zero. Option C is false — quadratic functions can produce equal outputs for two different inputs, as demonstrated here. Option D overgeneralizes; symmetry holds only for pairs of inputs equidistant from the axis of symmetry, not for any two arbitrary values.

Answer: $-16$

Understand the notation: $-x^2$ means $-(x^2)$, not $(-x)^2$. The negative sign is not part of the base being squared. Evaluate step by step: First compute $x^2 = (-4)^2 = 16$. Then apply the leading negative: $-16$. Why distractors fail: Option A ($16$) results from interpreting $-x^2$ as $(-x)^2 = (-(-4))^2 = 16$. Options C and D confuse squaring with doubling.

Answer: Replace the variable with the given number, then compute the numerical result using order of operations

Definition of evaluation: Evaluating an expression means substituting a specific number for the variable and simplifying to obtain a single numerical value. Why the correct answer works: Option B captures both essential steps: substitution and computation following order of operations. Why distractors fail: Option A describes solving the equation, not evaluating. Option C describes factoring. Option D describes analyzing the leading coefficient's sign, which is unrelated to evaluation.

Answer: Replace $x$ with $(-3)$, using parentheses, to get $2(-3)^2 - 4(-3) + 1$

Identify the substitution principle: When evaluating a quadratic expression, the first step is always to replace the variable with the given value, using parentheses around the substituted number to preserve sign accuracy. Why the correct answer works: Option B correctly shows $2(-3)^2 - 4(-3) + 1$, where parentheses around $-3$ ensure the negative sign is included in the squaring operation. Why distractors fail: Option A incorrectly multiplies before squaring, violating order of operations. Option C computes $(-3)^2$ as $-9$ instead of $9$. Option D tries to simplify coefficients before substitution, which is not valid.

Answer: $0$

Substitute x = 3/2: $f\left(\frac{3}{2}\right) = 4\left(\frac{3}{2}\right)^2 - 12\left(\frac{3}{2}\right) + 9$. Compute each term: $\left(\frac{3}{2}\right)^2 = \frac{9}{4}$, so $4 \cdot \frac{9}{4} = 9$. Then $12 \cdot \frac{3}{2} = 18$. Thus $f\left(\frac{3}{2}\right) = 9 - 18 + 9 = 0$. Why distractors fail: Option B ($3$) could result from a computation error in the middle term. Option C ($9$) comes from forgetting to subtract $18$. Option D ($-9$) results from a sign error in the final addition.

Answer: $f(x) = x^2 + 2x + 3$

Use f(0) = 3 to find c: $f(0) = a(0)^2 + b(0) + c = c = 3$. Use f(1) = 6 and f(-1) = 2: $f(1) = a + b + 3 = 6 \Rightarrow a + b = 3$. $f(-1) = a - b + 3 = 2 \Rightarrow a - b = -1$. Solve the system: Adding: $2a = 2 \Rightarrow a = 1$. Then $b = 2$. So $f(x) = x^2 + 2x + 3$. Verify and check distractors: Option A: $f(0)=3$, $f(1)=6$, $f(-1)=1-2+3=2$ ✓. Option B: $f(-1)=1+2+3=6 \neq 2$. Option C: $f(1)=2+1+3=6$ ✓ but $f(-1)=2-1+3=4 \neq 2$. Option D: $f(1)=-1+2+3=4 \neq 6$.

Answer: $16$ metres

Substitute t = 3: $h(3) = -5(3)^2 + 20(3) + 1$. Compute each term: $-5(9) = -45$, $20(3) = 60$, constant $= 1$. So $h(3) = -45 + 60 + 1 = 16$. Why distractors fail: Option B ($6$) omits the constant term. Option C ($-14$) results from computing $-5(3)^2$ as $-5(3) \times 3 = -75$ (an arithmetic slip). Option D ($26$) results from incorrectly using $-5(3)^2 = -15$ instead of $-45$.