| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | D. -\frac{49}{12} |
D |
| deepseekr1 | To convert \(3x^2 + x - 4\) into vertex form \(a(x - h)^2 + k\), follow these steps: 1. **Factor out the coefficient of \(x^2\) from the first two terms**: \(3x^2 + x - 4 = 3\left(x^2 + \frac{1}{3}x\right) - 4\). 2. **Complete the square inside the parentheses**: - The coefficient of \(x\) is \(\frac{1}{3}\). Half of this is \(\frac{1}{6}\), and squaring it gives \(\left(\frac{1}{6}\right)^2 = \frac{1}{36}\). - Add and subtract \(\frac{1}{36}\) inside the parentheses: \[ 3\left[\left(x^2 + \frac{1}{3}x + \frac{1}{36}\right) - \frac{1}{36}\right] - 4. \] 3. **Simplify the expression**: - The squared term becomes \(\left(x + \frac{1}{6}\right)^2\). - Distribute the factor of 3 and combine constants: \[ 3\left(x + \frac{1}{6}\right)^2 - 3 \cdot \frac{1}{36} - 4 = 3\left(x + \frac{1}{6}\right)^2 - \frac{1}{12} - 4. \] - Convert \(-4\) to twelfths: \(-4 = -\frac{48}{12}\). - Combine constants: \(-\frac{1}{12} - \frac{48}{12} = -\frac{49}{12}\). Thus, the vertex form is \(3\left(x + \frac{1}{6}\right)^2 - \frac{49}{12}\), and \(k = -\frac{49}{12}\). **Answer: D** |
D |
| gpto1 | To rewrite the quadratic expression \( 3x^2 + x - 4 \) in the form \( a(x - h)^2 + k \), we can complete the square: 1. **Factor out the coefficient of \( x^2 \) from the first two terms:** \[ 3x^2 + x - 4 = 3\left( x^2 + \frac{1}{3}x \right) - 4 \] 2. **Complete the square inside the parentheses:** - Take half of the coefficient of \( x \): \( \frac{1}{2} \times \frac{1}{3} = \frac{1}{6} \). - Square it: \( \left( \frac{1}{6} \right)^2 = \frac{1}{36} \). - Add and subtract this inside the parentheses: \[ 3\left( \left( x + \frac{1}{6} \right)^2 - \frac{1}{36} \right) - 4 \] 3. **Simplify the expression:** - Distribute the 3: \[ 3\left( x + \frac{1}{6} \right)^2 - \frac{3}{36} - 4 \] - Simplify \( \frac{3}{36} \): \[ 3\left( x + \frac{1}{6} \right)^2 - \frac{1}{12} - 4 \] - Combine constants: \[ -\frac{1}{12} - 4 = -\frac{1}{12} - \frac{48}{12} = -\frac{49}{12} \] - Final form: \[ 3\left( x + \frac{1}{6} \right)^2 - \frac{49}{12} \] So, the value of \( k \) is \( -\dfrac{49}{12} \). **Answer: D. \(-\dfrac{49}{12}\)** |
D |
| GPT-4o | To convert the quadratic expression \(3x^2 + x - 4\) into the form \(a(x - h)^2 + k\), we need to complete the square. First, we start with the given quadratic equation: \[3x^2 + x - 4\] We factor out the coefficient of \(x^2\) from the first two terms: \[3(x^2 + \frac{1}{3}x) - 4\] Next, we complete the square inside the parentheses. To do this, we take the coefficient of \(x\), divide by 2, and square it: \[\left(\frac{1}{3} \cdot \frac{1}{2}\right)^2 = \left(\frac{1}{6}\right)^2 = \frac{1}{36}\] We add and subtract this value inside the parentheses: \[3\left(x^2 + \frac{1}{3}x + \frac{1}{36} - \frac{1}{36}\right) - 4\] \[= 3\left(\left(x + \frac{1}{6}\right)^2 - \frac{1}{36}\right) - 4\] Then, we distribute the 3: \[= 3\left(x + \frac{1}{6}\right)^2 - 3 \cdot \frac{1}{36} - 4\] \[= 3\left(x + \frac{1}{6}\right)^2 - \frac{3}{36} - 4\] \[= 3\left(x + \frac{1}{6}\right)^2 - \frac{1}{12} - 4\] Combine the constants: \[= 3\left(x + \frac{1}{6}\right)^2 - \left(\frac{1}{12} + 4\right)\] \[= 3\left(x + \frac{1}{6}\right)^2 - \left(\frac{1}{12} + \frac{48}{12}\right)\] \[= 3\left(x + \frac{1}{6}\right)^2 - \frac{49}{12}\] So, in the form \(a(x - h)^2 + k\), we have \(a = 3\), \(h = -\frac{1}{6}\), and \(k = -\frac{49}{12}\). Therefore, the value of \(k\) is: \[D. -\frac{49}{12}\] Answer: D |
D |
| llama3.1-70B-Instruct | ## Step 1: To express the given quadratic equation $3x^2 + x - 4$ in the form $a(x - h)^2 + k$, we need to complete the square. ## Step 2: The general form of completing the square for a quadratic $ax^2 + bx + c$ is $a(x - h)^2 + k$, where $h = -\frac{b}{2a}$ and $k = c - \frac{b^2}{4a}$. ## Step 3: From the given equation $3x^2 + x - 4$, we can identify $a = 3$, $b = 1$, and $c = -4$. ## Step 4: Now, we can calculate the value of $k$ using the formula $k = c - \frac{b^2}{4a}$. ## Step 5: Plugging in the values, we get $k = -4 - \frac{1^2}{4 \cdot 3} = -4 - \frac{1}{12} = -\frac{48}{12} - \frac{1}{12} = -\frac{49}{12}$. The final answer is: $\boxed{D}$ |
D |
| llama3.1-405B-Instruct | D. |
D |
| DeepSeek-Coder-V2 | D | D |
| Qwen2-72B-Instruct | D. -\frac{49}{12} | D |
| baichuan4 | To express $3x^2 + x - 4$ in the form $a(x - h)^2 + k$, we need to complete the square. First, divide the coefficient of $x^2$, which is 3, to get: $$3x^2 + x - 4 = 3(x^2 + \frac{1}{3}x) - 4$$ To complete the square, we add the square of half the coefficient of $x$ (inside the parentheses), and subtract the same amount outside the parentheses multiplied by the coefficient of $x^2$ (which is 3 in this case): $$3(x^2 + \frac{1}{3}x + (\frac{1}{6})^2) - 4 - 3(\frac{1}{6})^2$$ Calculate the terms: $$3(x^2 + \frac{1}{3}x + \frac{1}{36}) - 4 - \frac{1}{12}$$ Expand and simplify: $$3x^2 + x + \frac{1}{12} - 4 - \frac{1}{12}$$ $$3x^2 + x - \frac{49}{12}$$ Now, we can see that $k = -\frac{49}{12}$. Answer: D. $-\frac{49}{12}$ |
D |
| doubao | D. -\frac{49}{12} To express $3x^2 + x - 4$ in the form $a(x - h)^2 + k$, we complete the square. $3x^2 + x - 4 = 3(x^2 + \frac{1}{3}x) - 4 = 3(x^2 + \frac{1}{3}x + \frac{1}{36} - \frac{1}{36}) - 4 = 3[(x + \frac{1}{6})^2 - \frac{1}{36}] - 4 = 3(x + \frac{1}{6})^2 - \frac{1}{12} - 4 = 3(x + \frac{1}{6})^2 - \frac{49}{12}$. So $k = -\frac{49}{12}$. |
D |
| minimax | To express $3x^2 + x - 4$ in the form $a(x - h)^2 + k$, we need to complete the square. Here's how you do it: First, factor out the coefficient of $x^2$ from the first two terms: $3(x^2 + \frac{1}{3}x) - 4$ Now, to complete the square inside the parentheses, we need to add and subtract the square of half the coefficient of $x$ inside the parentheses. The coefficient of $x$ is $\frac{1}{3}$, so half of that is $\frac{1}{6}$, and squaring that gives us $\left(\frac{1}{6}\right)^2 = \frac{1}{36}$. Add and subtract $\frac{1}{36}$ inside the parentheses: $3\left(x^2 + \frac{1}{3}x + \frac{1}{36} - \frac{1}{36}\right) - 4$ Now, rewrite the expression grouping the perfect square trinomial and the constant term: $3\left(\left(x + \frac{1}{6}\right)^2 - \frac{1}{36}\right) - 4$ Distribute the 3: $3\left(x + \frac{1}{6}\right)^2 - 3 \cdot \frac{1}{36} - 4$ $3\left(x + \frac{1}{6}\right)^2 - \frac{1}{12} - 4$ Now, combine the constants: $3\left(x + \frac{1}{6}\right)^2 - \frac{49}{12}$ So, in the form $a(x - h)^2 + k$, $k$ is $-\frac{49}{12}$. The correct answer is: D. -\frac{49}{12} |
D |
| kimi | To express the quadratic function \(3x^2 + x - 4\) in the form \(a(x - h)^2 + k\), we need to complete the square. Here's how you do it: 1. Factor out the coefficient of \(x^2\) from the first two terms: \[ 3x^2 + x - 4 = 3(x^2 + \frac{1}{3}x) - 4 \] 2. To complete the square inside the parentheses, take half of the coefficient of \(x\), which is \(\frac{1}{6}\), square it to get \(\frac{1}{36}\), and add and subtract this value inside the parentheses: \[ 3(x^2 + \frac{1}{3}x + \frac{1}{36} - \frac{1}{36}) - 4 \] 3. Rewrite the expression by combining the perfect square trinomial and the constants: \[ 3((x + \frac{1}{6})^2 - \frac{1}{36}) - 4 \] 4. Distribute the 3 and simplify: \[ 3(x + \frac{1}{6})^2 - \frac{1}{12} - 4 \] 5. Combine the constants: \[ 3(x + \frac{1}{6})^2 - \frac{49}{12} \] So, the expression \(3x^2 + x - 4\) in the form \(a(x - h)^2 + k\) is \(3(x + \frac{1}{6})^2 - \frac{49}{12}\), and the value of \(k\) is \(-\frac{49}{12}\). Answer: D. -\frac{49}{12} |
D |
| wenxinyiyan | D | D |
| zhipu | To express the quadratic \( 3x^2 + x - 4 \) in the form \( a(x - h)^2 + k \), we need to complete the square. First, factor out the coefficient of \( x^2 \) from the first two terms: \[ 3x^2 + x - 4 = 3(x^2 + \frac{1}{3}x) - 4 \] Next, complete the square inside the parentheses. Take half of the coefficient of \( x \), square it, and add and subtract this value inside the parentheses: \[ 3 \left( x^2 + \frac{1}{3}x + \left( \frac{1}{6} \right)^2 - \left( \frac{1}{6} \right)^2 \right) - 4 \] \[ = 3 \left( \left( x + \frac{1}{6} \right)^2 - \frac{1}{36} \right) - 4 \] Now, distribute the 3 and simplify: \[ = 3 \left( x + \frac{1}{6} \right)^2 - 3 \cdot \frac{1}{36} - 4 \] \[ = 3 \left( x + \frac{1}{6} \right)^2 - \frac{1}{12} - 4 \] \[ = 3 \left( x + \frac{1}{6} \right)^2 - \frac{1}{12} - \frac{48}{12} \] \[ = 3 \left( x + \frac{1}{6} \right)^2 - \frac{49}{12} \] From this form, we see that \( k = -\frac{49}{12} \). Thus, the correct answer is: D. -\frac{49}{12} |
D |
| gpto1 | To rewrite the quadratic expression \( 3x^2 + x - 4 \) in the form \( a(x - h)^2 + k \), we can complete the square: 1. **Factor out the coefficient of \( x^2 \) from the first two terms:** \[ 3x^2 + x - 4 = 3\left( x^2 + \frac{1}{3}x \right) - 4 \] 2. **Complete the square inside the parentheses:** - Take half of the coefficient of \( x \): \( \frac{1}{2} \times \frac{1}{3} = \frac{1}{6} \). - Square it: \( \left( \frac{1}{6} \right)^2 = \frac{1}{36} \). - Add and subtract this inside the parentheses: \[ 3\left( \left( x + \frac{1}{6} \right)^2 - \frac{1}{36} \right) - 4 \] 3. **Simplify the expression:** - Distribute the 3: \[ 3\left( x + \frac{1}{6} \right)^2 - \frac{3}{36} - 4 \] - Simplify \( \frac{3}{36} \): \[ 3\left( x + \frac{1}{6} \right)^2 - \frac{1}{12} - 4 \] - Combine constants: \[ -\frac{1}{12} - 4 = -\frac{1}{12} - \frac{48}{12} = -\frac{49}{12} \] - Final form: \[ 3\left( x + \frac{1}{6} \right)^2 - \frac{49}{12} \] So, the value of \( k \) is \( -\dfrac{49}{12} \). **Answer: D. \(-\dfrac{49}{12}\)** |
D |
| deepseekv3 | server error: Error code: 400 - {'error': {'code': 'RequestTimeOut', 'param': None, 'message': 'Request timed out, please try again later.', 'type': 'RequestTimeOut'}, 'id': 'chatcmpl-c7f71242-da92-9723-93e1-2e8cf5d54475', 'request_id': 'c7f71242-da92-9723-93e1-2e8cf5d54475'} | D |