| 模型 | 模型答案 | 正确答案 |
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| 金声玉亮2.0 | To find the polynomial \(f(t) = a + bt + ct^2\) that passes through the points \((1, -1)\), \((2, 3)\), and \((3, 13)\), we can set up a system of equations using these points. Each point \((x, y)\) gives us an equation when we substitute \(x\) for \(t\) and \(y\) for \(f(t)\). 1. For \((1, -1)\), we have: \[a + b(1) + c(1)^2 = -1\] 2. For \((2, 3)\), we have: \[a + b(2) + c(2)^2 = 3\] 3. For \((3, 13)\), we have: \[a + b(3) + c(3)^2 = 13\] These equations simplify to: 1. \(a + b + c = -1\) 2. \(a + 2b + 4c = 3\) 3. \(a + 3b + 9c = 13\) We can solve this system of equations for \(a\), \(b\), and \(c\). Let's subtract the first equation from the second and third equations to eliminate \(a\): 1. \(a + b + c = -1\) 2. \(a + 2b + 4c = 3\), subtracting the first equation gives us \(b + 3c = 4\) 3. \(a + 3b + 9c = 13\), subtracting the first equation gives us \(2b + 8c = 14\) Now we have: 1. \(b + 3c = 4\) 2. \(2b + 8c = 14\) We can multiply the first equation by 2 and subtract it from the second equation to eliminate \(b\): 1. \(2b + 6c = 8\) 2. \(2b + 8c = 14\), subtracting the multiplied first equation gives us \(2c = 6\) So, \(c = 3\). Substituting \(c = 3\) in \(b + 3c = 4\), we get: \[b + 3(3) = 4\] \[b + 9 = 4\] \[b = -5\] Now we know \(b = -5\) and \(c = 3\), and we can substitute these values back into \(a + b + c = -1\) to find \(a\): \[a - 5 + 3 = -1\] \[a - 2 = -1\] \[a = 1\] So, the polynomial \(f(t) = 1 - 5t + 3t^2\). Finally, to find \(f(-1)\), we substitute \(-1\) for \(t\): \[f(-1) = 1 - 5(-1) + 3(-1)^2\] \[f(-1) = 1 + 5 + 3\] \[f(-1) = 9\] |
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| deepseekr1 | Given the polynomial \( f(t) = a + bt + ct^2 \) that passes through the points \((1, -1)\), \((2, 3)\), and \((3, 13)\), we need to determine the coefficients \(a\), \(b\), and \(c\) and then find \(f(-1)\). First, we set up the system of equations based on the given points: 1. For the point \((1, -1)\): \[ a + b + c = -1 \] 2. For the point \((2, 3)\): \[ a + 2b + 4c = 3 \] 3. For the point \((3, 13)\): \[ a + 3b + 9c = 13 \] Subtracting the first equation from the second equation: \[ (a + 2b + 4c) - (a + b + c) = 3 - (-1) \implies b + 3c = 4 \] Subtracting the second equation from the third equation: \[ (a + 3b + 9c) - (a + 2b + 4c) = 13 - 3 \implies b + 5c = 10 \] We now solve the system of equations: 1. \( b + 3c = 4 \) 2. \( b + 5c = 10 \) Subtracting these two equations: \[ (b + 5c) - (b + 3c) = 10 - 4 \implies 2c = 6 \implies c = 3 \] Substituting \( c = 3 \) back into \( b + 3c = 4 \): \[ b + 3(3) = 4 \implies b + 9 = 4 \implies b = -5 \] Substituting \( b = -5 \) and \( c = 3 \) back into the first equation: \[ a - 5 + 3 = -1 \implies a - 2 = -1 \implies a = 1 \] Thus, the polynomial is \( f(t) = 1 - 5t + 3t^2 \). To find \( f(-1) \): \[ f(-1) = 1 - 5(-1) + 3(-1)^2 = 1 + 5 + 3 = 9 \] The value of \( f(-1) \) is \(\boxed{9}\). |
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| gpto1 | To find \( f(-1) \) for the polynomial \( f(t) = a + b t + c t^2 \) that passes through the points \((1, -1)\), \((2, 3)\), and \((3, 13)\), we'll follow these steps: **1. Set up equations using the given points.** For each point \((t, f(t))\), substitute \( t \) and \( f(t) \) into the polynomial: - **Point (1, -1):** \[ f(1) = a + b(1) + c(1)^2 = -1 \] Simplifies to: \[ a + b + c = -1 \quad \text{(Equation 1)} \] - **Point (2, 3):** \[ f(2) = a + b(2) + c(2)^2 = 3 \] Simplifies to: \[ a + 2b + 4c = 3 \quad \text{(Equation 2)} \] - **Point (3, 13):** \[ f(3) = a + b(3) + c(3)^2 = 13 \] Simplifies to: \[ a + 3b + 9c = 13 \quad \text{(Equation 3)} \] **2. Solve the system of equations to find \( a \), \( b \), and \( c \).** First, express \( a \) in terms of \( b \) and \( c \) using Equation 1: \[ a = -1 - b - c \] **Substitute \( a \) into Equations 2 and 3:** - **Equation 2:** \begin{align*} (-1 - b - c) + 2b + 4c & = 3 \\ -1 - b - c + 2b + 4c & = 3 \\ (-1) + ( -b + 2b ) + ( -c + 4c ) & = 3 \\ -1 + b + 3c & = 3 \\ b + 3c & = 4 \quad \text{(Equation 2a)} \end{align*} - **Equation 3:** \begin{align*} (-1 - b - c) + 3b + 9c & = 13 \\ -1 - b - c + 3b + 9c & = 13 \\ (-1) + ( -b + 3b ) + ( -c + 9c ) & = 13 \\ -1 + 2b + 8c & = 13 \\ 2b + 8c & = 14 \quad \text{(Equation 3a)} \end{align*} **Now, solve for \( b \) and \( c \) using Equations 2a and 3a.** First, solve Equation 2a for \( b \): \[ b = 4 - 3c \] Substitute \( b \) into Equation 3a: \begin{align*} 2(4 - 3c) + 8c & = 14 \\ 8 - 6c + 8c & = 14 \\ 8 + 2c & = 14 \\ 2c & = 6 \\ c & = 3 \end{align*} Now, find \( b \) using \( c \): \begin{align*} b & = 4 - 3c \\ b & = 4 - 3(3) \\ b & = 4 - 9 \\ b & = -5 \end{align*} Finally, find \( a \) using \( b \) and \( c \): \begin{align*} a & = -1 - b - c \\ a & = -1 - (-5) - 3 \\ a & = -1 + 5 - 3 \\ a & = 1 \end{align*} **3. Write the specific form of the polynomial.** Now that we have \( a \), \( b \), and \( c \): \[ f(t) = 1 - 5t + 3t^2 \] **4. Calculate \( f(-1) \).** Substitute \( t = -1 \) into the polynomial: \begin{align*} f(-1) & = 1 - 5(-1) + 3(-1)^2 \\ & = 1 + 5 + 3(1) \\ & = 1 + 5 + 3 \\ & = 9 \end{align*} **Answer:** \[ f(-1) = \boxed{9} \] |
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| GPT-4o | To determine \( f(-1) \) for the polynomial \( f(t) = a + bt + ct^2 \) passing through the points \((1, -1)\), \((2, 3)\), and \((3, 13)\), we need to find the coefficients \( a \), \( b \), and \( c \). First, we set up a system of equations using the given points: 1. For the point \((1, -1)\): \[ f(1) = a + b(1) + c(1)^2 = -1 \implies a + b + c = -1 \] 2. For the point \((2, 3)\): \[ f(2) = a + b(2) + c(2)^2 = 3 \implies a + 2b + 4c = 3 \] 3. For the point \((3, 13)\): \[ f(3) = a + b(3) + c(3)^2 = 13 \implies a + 3b + 9c = 13 \] Now we have the following system of linear equations: \[ \begin{cases} a + b + c = -1 \\ a + 2b + 4c = 3 \\ a + 3b + 9c = 13 \end{cases} \] We solve this system step by step. First, subtract the first equation from the second equation: \[ (a + 2b + 4c) - (a + b + c) = 3 - (-1) \implies b + 3c = 4 \quad \text{(Equation 4)} \] Next, subtract the second equation from the third equation: \[ (a + 3b + 9c) - (a + 2b + 4c) = 13 - 3 \implies b + 5c = 10 \quad \text{(Equation 5)} \] Now, we have two new equations: \[ \begin{cases} b + 3c = 4 \\ b + 5c = 10 \end{cases} \] Subtract Equation 4 from Equation 5: \[ (b + 5c) - (b + 3c) = 10 - 4 \implies 2c = 6 \implies c = 3 \] Substitute \( c = 3 \) back into Equation 4: \[ b + 3(3) = 4 \implies b + 9 = 4 \implies b = -5 \] Finally, substitute \( b = -5 \) and \( c = 3 \) back into the first equation to find \( a \): \[ a + (-5) + 3 = -1 \implies a - 5 + 3 = -1 \implies a - 2 = -1 \implies a = 1 \] Thus, the coefficients are \( a = 1 \), \( b = -5 \), and \( c = 3 \). The polynomial is: \[ f(t) = 1 - 5t + 3t^2 \] Now we find \( f(-1) \): \[ f(-1) = 1 - 5(-1) + 3(-1)^2 = 1 + 5 + 3 = 9 \] Therefore, \( f(-1) = 9 \). |
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| llama3.1-70B-Instruct | To find the value of f(-1), we need to first find the coefficients a, b, and c of the polynomial f(t) = a + bt + ct^2. We are given that the graph of f(t) passes through the points (1, -1), (2, 3), and (3, 13). This means that we can substitute these values into the equation f(t) = a + bt + ct^2 to get a system of three equations: -1 = a + b + c 3 = a + 2b + 4c 13 = a + 3b + 9c We can solve this system of equations to find the values of a, b, and c. First, we can subtract the first equation from the second equation to get: 4 = b + 3c Next, we can subtract the second equation from the third equation to get: 10 = b + 5c Now, we can subtract the equation 4 = b + 3c from the equation 10 = b + 5c to get: 6 = 2c Dividing both sides by 2, we get: c = 3 Now that we have found c, we can substitute this value back into the equation 4 = b + 3c to get: 4 = b + 3(3) 4 = b + 9 b = -5 Finally, we can substitute the values of b and c back into the first equation -1 = a + b + c to get: -1 = a + (-5) + 3 -1 = a - 2 a = -1 + 2 a = 1 Now that we have found the values of a, b, and c, we can substitute these values into the equation f(t) = a + bt + ct^2 to get: f(t) = 1 - 5t + 3t^2 To find f(-1), we can substitute t = -1 into this equation: f(-1) = 1 - 5(-1) + 3(-1)^2 f(-1) = 1 + 5 + 3 f(-1) = 9 Therefore, the value of f(-1) is 9. |
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| llama3.1-405B-Instruct | To find the value of f(-1), we need to find the coefficients a, b, and c of the polynomial f(t) = a + bt + ct^2. We are given that the graph of f(t) passes through the points (1, -1), (2, 3), and (3, 13). This means that we can substitute these values of t and f(t) into the equation to get three equations: 1. f(1) = a + b + c = -1 2. f(2) = a + 2b + 4c = 3 3. f(3) = a + 3b + 9c = 13 Now, we can solve this system of equations to find the values of a, b, and c. First, we can subtract equation (1) from equation (2) to get: b + 3c = 4 Next, we can subtract equation (2) from equation (3) to get: b + 5c = 10 Now, we can subtract the first equation from the second equation to get: 2c = 6 c = 3 Now that we have found c, we can substitute this value back into one of the earlier equations to find b. Let's use the equation b + 3c = 4: b + 3(3) = 4 b + 9 = 4 b = -5 Now that we have found b and c, we can substitute these values back into one of the original equations to find a. Let's use the equation a + b + c = -1: a + (-5) + 3 = -1 a - 2 = -1 a = 1 Now that we have found the values of a, b, and c, we can substitute these values into the original equation to find f(-1): f(-1) = a + b(-1) + c(-1)^2 f(-1) = 1 + (-5)(-1) + 3(-1)^2 f(-1) = 1 + 5 + 3 f(-1) = 9 Therefore, the value of f(-1) is 9. |
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| DeepSeek-Coder-V2 | To solve the problem, we need to determine the coefficients \(a\), \(b\), and \(c\) of the polynomial \(f(t) = a + bt + ct^2\) that passes through the given points: \((1, -1)\), \((2, 3)\), and \((3, 13)\). Then, we will use these coefficients to find \(f(-1)\). First, we set up the system of equations by substituting the points into the polynomial: 1. For the point \((1, -1)\): \[ a + b(1) + c(1^2) = -1 \implies a + b + c = -1 \] 2. For the point \((2, 3)\): \[ a + b(2) + c(2^2) = 3 \implies a + 2b + 4c = 3 \] 3. For the point \((3, 13)\): \[ a + b(3) + c(3^2) = 13 \implies a + 3b + 9c = 13 \] We now have the following system of linear equations: \[ \begin{cases} a + b + c = -1 \\ a + 2b + 4c = 3 \\ a + 3b + 9c = 13 \end{cases} \] Next, we solve this system step by step. First, subtract the first equation from the second: \[ (a + 2b + 4c) - (a + b + c) = 3 - (-1) \implies b + 3c = 4 \] Then, subtract the second equation from the third: \[ (a + 3b + 9c) - (a + 2b + 4c) = 13 - 3 \implies b + 5c = 10 \] Now we have a simpler system: \[ \begin{cases} b + 3c = 4 \\ b + 5c = 10 \end{cases} \] Subtract the first new equation from the second new equation: \[ (b + 5c) - (b + 3c) = 10 - 4 \implies 2c = 6 \implies c = 3 \] Substitute \(c = 3\) back into \(b + 3c = 4\): \[ b + 3(3) = 4 \implies b + 9 = 4 \implies b = -5 \] Finally, substitute \(b = -5\) and \(c = 3\) back into \(a + b + c = -1\): \[ a - 5 + 3 = -1 \implies a - 2 = -1 \implies a = 1 \] Now we have the coefficients \(a = 1\), \(b = -5\), and \(c = 3\). The polynomial is: \[ f(t) = 1 - 5t + 3t^2 \] To find \(f(-1)\), substitute \(t = -1\) into the polynomial: \[ f(-1) = 1 - 5(-1) + 3(-1)^2 = 1 + 5 + 3 = 9 \] Thus, the value of \(f(-1)\) is: \[ \boxed{9} \] |
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| Qwen2-72B-Instruct | To find the polynomial \(f(t) = a + bt + ct^2\) that passes through the points \((1, -1)\), \((2, 3)\), and \((3, 13)\), we can set up a system of equations using these points. Each point \((x, y)\) gives us an equation when we substitute \(x\) for \(t\) and \(y\) for \(f(t)\). 1. For \((1, -1)\), we have: \[a + b(1) + c(1)^2 = -1\] 2. For \((2, 3)\), we have: \[a + b(2) + c(2)^2 = 3\] 3. For \((3, 13)\), we have: \[a + b(3) + c(3)^2 = 13\] These equations simplify to: 1. \(a + b + c = -1\) 2. \(a + 2b + 4c = 3\) 3. \(a + 3b + 9c = 13\) We can solve this system of equations for \(a\), \(b\), and \(c\). Let's subtract the first equation from the second and third equations to eliminate \(a\): 1. \(a + b + c = -1\) 2. \(a + 2b + 4c = 3\), subtracting the first equation gives us \(b + 3c = 4\) 3. \(a + 3b + 9c = 13\), subtracting the first equation gives us \(2b + 8c = 14\) Now we have: 1. \(b + 3c = 4\) 2. \(2b + 8c = 14\) We can multiply the first equation by 2 and subtract it from the second equation to eliminate \(b\): 1. \(2b + 6c = 8\) 2. \(2b + 8c = 14\), subtracting the multiplied first equation gives us \(2c = 6\) So, \(c = 3\). Substituting \(c = 3\) in \(b + 3c = 4\), we get: \[b + 3(3) = 4\] \[b + 9 = 4\] \[b = -5\] Now we know \(b = -5\) and \(c = 3\), and we can substitute these values back into \(a + b + c = -1\) to find \(a\): \[a - 5 + 3 = -1\] \[a - 2 = -1\] \[a = 1\] So, the polynomial \(f(t) = 1 - 5t + 3t^2\). Finally, to find \(f(-1)\), we substitute \(-1\) for \(t\): \[f(-1) = 1 - 5(-1) + 3(-1)^2\] \[f(-1) = 1 + 5 + 3\] \[f(-1) = 9\] |
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| baichuan4 | To find the value of \(f(-1)\), we first need to determine the coefficients \(a\), \(b\), and \(c\) of the polynomial function \(f(t) = a + bt + ct^2\). We can do this by substituting the given points through which the graph passes and solving the resulting system of equations. Given points: 1. \((1, -1)\): When \(t = 1\), \(f(t) = -1\) 2. \((2, 3)\): When \(t = 2\), \(f(t) = 3\) 3. \((3, 13)\): When \(t = 3\), \(f(t) = 13\) Substituting these points into the equation \(f(t) = a + bt + ct^2\), we get the following system of equations: 1. \(a + b + c = -1\) 2. \(a + 2b + 4c = 3\) 3. \(a + 3b + 9c = 13\) Now, we need to solve this system of equations for \(a\), \(b\), and \(c\). From the first equation, we have: \[a = -1 - b - c\] Substitute this into the second equation: \[-1 - b - c + 2b + 4c = 3\] \[b + 3c = 4\] Substitute \(a\) into the third equation: \[-1 - b - c + 3b + 9c = 13\] \[2b + 8c = 14\] \[b + 4c = 7\] Now, we have two equations: 1. \(b + 3c = 4\) 2. \(b + 4c = 7\) Subtracting the first equation from the second gives: \[c = 3\] Substituting \(c = 3\) into the first equation: \[b + 3(3) = 4\] \[b = -5\] Finally, substituting \(b = -5\) and \(c = 3\) into the equation for \(a\): \[a = -1 - (-5) - 3\] \[a = 1\] Now that we have the coefficients \(a = 1\), \(b = -5\), and \(c = 3\), we can determine \(f(-1)\): \[f(-1) = a + b(-1) + c(-1)^2\] \[f(-1) = 1 - 5 + 3\] \[f(-1) = -1\] So, the value of \(f(-1)\) is \(-1\). |
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| doubao | Substituting the points into the equation: For (1, -1): -1 = a + b + c For (2, 3): 3 = a + 2b + 4c For (3, 13): 13 = a + 3b + 9c Solve this system of equations: From the first equation, a = -1 - b - c. Substituting this into the second equation: 3 = -1 - b - c + 2b + 4c, simplifies to b + 3c = 4. Substituting a = -1 - b - c into the third equation: 13 = -1 - b - c + 3b + 9c, simplifies to 2b + 8c = 14. Multiply the equation b + 3c = 4 by 2: 2b + 6c = 8. Subtracting this from 2b + 8c = 14 gives 2c = 6, so c = 3. Substituting c = 3 into b + 3c = 4 gives b = -5. Substituting b = -5 and c = 3 into a = -1 - b - c gives a = 1. So the polynomial is f(t) = 1 - 5t + 3t^2. To find f(-1), substitute t = -1 into the polynomial: f(-1) = 1 - 5(-1) + 3(-1)^2 = 1 + 5 + 3 = 9. |
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| minimax | To solve this problem, we need to find the coefficients a, b, and c of the polynomial f(t) = a + bt + ct^2. We know that the graph of this polynomial passes through the points (1, -1), (2, 3), and (3, 13). This means that when we substitute the t-values into the polynomial, we should get the corresponding y-values. Let's set up a system of equations using the given points: For the point (1, -1): a + b(1) + c(1)^2 = -1 a + b + c = -1 (Equation 1) For the point (2, 3): a + b(2) + c(2)^2 = 3 a + 2b + 4c = 3 (Equation 2) For the point (3, 13): a + b(3) + c(3)^2 = 13 a + 3b + 9c = 13 (Equation 3) Now we have a system of three equations with three unknowns. We can solve this system using various methods such as substitution, elimination, or matrix operations. For simplicity, let's use the elimination method. First, we can subtract Equation 1 from Equation 2 and Equation 3 to eliminate a: (Equation 2) - (Equation 1): (a + 2b + 4c) - (a + b + c) = 3 - (-1) a + 2b + 4c - a - b - c = 3 + 1 b + 3c = 4 (Equation 4) (Equation 3) - (Equation 1): (a + 3b + 9c) - (a + b + c) = 13 - (-1) a + 3b + 9c - a - b - c = 13 + 1 2b + 8c = 14 (Equation 5) Now we can divide Equation 5 by 2 to simplify it: 2b + 8c = 14 b + 4c = 7 (Equation 6) Now we have two equations with two unknowns (Equation 4 and Equation 6): b + 3c = 4 (Equation 4) b + 4c = 7 (Equation 6) Subtract Equation 4 from Equation 6 to solve for c: (Equation 6) - (Equation 4): (b + 4c) - (b + 3c) = 7 - 4 b + 4c - b - 3c = 7 - 4 c = 3 Now that we have c, we can substitute it back into Equation 4 to find b: b + 3(3) = 4 b + 9 = 4 b = 4 - 9 b = -5 Finally, we can substitute b and c back into Equation 1 to find a: a - 5 + 3 = -1 a - 2 = -1 a = -1 + 2 a = 1 So the coefficients of the polynomial are a = 1, b = -5, and c = 3. The polynomial is: f(t) = 1 - 5t + 3t^2 Now we can find f(-1): f(-1) = 1 - 5(-1) + 3(-1)^2 f(-1) = 1 + 5 + 3 f(-1) = 9 Therefore, f(-1) equals 9. |
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| kimi | To find the values of \(a\), \(b\), and \(c\) in the polynomial \(f(t) = a + bt + ct^2\), we can use the given points \((1, -1)\), \((2, 3)\), and \((3, 13)\) to set up a system of equations. 1. For the point \((1, -1)\): \[ f(1) = a + b(1) + c(1)^2 = -1 \] \[ a + b + c = -1 \] (Equation 1) 2. For the point \((2, 3)\): \[ f(2) = a + b(2) + c(2)^2 = 3 \] \[ a + 2b + 4c = 3 \] (Equation 2) 3. For the point \((3, 13)\): \[ f(3) = a + b(3) + c(3)^2 = 13 \] \[ a + 3b + 9c = 13 \] (Equation 3) Now we have a system of three equations with three unknowns: \[ \begin{align*} a + b + c &= -1 \quad \text{(Equation 1)} \\ a + 2b + 4c &= 3 \quad \text{(Equation 2)} \\ a + 3b + 9c &= 13 \quad \text{(Equation 3)} \end{align*} \] We can solve this system using substitution or elimination. Let's use elimination: Subtract Equation 1 from Equation 2: \[ (a + 2b + 4c) - (a + b + c) = 3 - (-1) \] \[ b + 3c = 4 \] (Equation 4) Subtract Equation 2 from Equation 3: \[ (a + 3b + 9c) - (a + 2b + 4c) = 13 - 3 \] \[ b + 5c = 10 \] (Equation 5) Now subtract Equation 4 from Equation 5: \[ (b + 5c) - (b + 3c) = 10 - 4 \] \[ 2c = 6 \] \[ c = 3 \] Now that we have \(c\), we can find \(b\) using Equation 4: \[ b + 3(3) = 4 \] \[ b + 9 = 4 \] \[ b = 4 - 9 \] \[ b = -5 \] Finally, we can find \(a\) using Equation 1: \[ a + (-5) + 3 = -1 \] \[ a - 5 + 3 = -1 \] \[ a - 2 = -1 \] \[ a = -1 + 2 \] \[ a = 1 \] So the polynomial is: \[ f(t) = 1 - 5t + 3t^2 \] Now we can find \(f(-1)\): \[ f(-1) = 1 - 5(-1) + 3(-1)^2 \] \[ f(-1) = 1 + 5 + 3 \] \[ f(-1) = 9 \] Thus, \(f(-1) = 9\). |
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| wenxinyiyan | 首先，我们根据题目信息，知道多项式$f(t) = a + bt + ct^2$经过点$(1, -1)$，$(2, 3)$，和$(3, 13)$。 这意味着当$t=1$时，$f(t)=-1$；当$t=2$时，$f(t)=3$；当$t=3$时，$f(t)=13$。 我们可以根据这些信息设立以下三个方程： 1. $a + b(1) + c(1^2) = -1$ 2. $a + b(2) + c(2^2) = 3$ 3. $a + b(3) + c(3^2) = 13$ 这样我们得到了一个三元一次方程组： $\begin{cases} a + b + c = -1, \\ a + 2b + 4c = 3, \\ a + 3b + 9c = 13. \end{cases}$ 解这个方程组，我们得到： $\begin{cases} a = 1, \\ b = -5, \\ c = 3. \end{cases}$ 因此，多项式$f(t)$可以表示为$f(t) = 1 - 5t + 3t^2$。 最后，为了找到$f(-1)$的值，我们将$t = -1$代入多项式$f(t)$： $f(-1) = 1 - 5(-1) + 3(-1)^2 = 1 + 5 + 3 = 9.$ 所以，$f(-1) = 9$。 |
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| zhipu | To find the polynomial \( f(t) = a + bt + ct^2 \) that passes through the points (1, -1), (2, 3), and (3, 13), we can set up a system of equations based on these points and solve for the coefficients \( a \), \( b \), and \( c \). For the point (1, -1): \[ f(1) = a + b(1) + c(1)^2 = -1 \] \[ a + b + c = -1 \] (Equation 1) For the point (2, 3): \[ f(2) = a + b(2) + c(2)^2 = 3 \] \[ a + 2b + 4c = 3 \] (Equation 2) For the point (3, 13): \[ f(3) = a + b(3) + c(3)^2 = 13 \] \[ a + 3b + 9c = 13 \] (Equation 3) Now we have a system of linear equations: \[ \begin{cases} a + b + c = -1 \\ a + 2b + 4c = 3 \\ a + 3b + 9c = 13 \end{cases} \] We can solve this system using either substitution, elimination, or matrix methods. Here, we'll use the elimination method. Subtract Equation 1 from Equation 2: \[ (a + 2b + 4c) - (a + b + c) = 3 - (-1) \] \[ b + 3c = 4 \] (Equation 4) Subtract Equation 2 from Equation 3: \[ (a + 3b + 9c) - (a + 2b + 4c) = 13 - 3 \] \[ b + 5c = 10 \] (Equation 5) Now subtract Equation 4 from Equation 5: \[ (b + 5c) - (b + 3c) = 10 - 4 \] \[ 2c = 6 \] \[ c = 3 \] Substitute \( c = 3 \) back into Equation 4: \[ b + 3(3) = 4 \] \[ b + 9 = 4 \] \[ b = -5 \] Substitute \( b = -5 \) and \( c = 3 \) back into Equation 1: \[ a - 5 + 3 = -1 \] \[ a - 2 = -1 \] \[ a = 1 \] So the polynomial is: \[ f(t) = 1 - 5t + 3t^2 \] Now we need to find \( f(-1) \): \[ f(-1) = 1 - 5(-1) + 3(-1)^2 \] \[ f(-1) = 1 + 5 + 3 \] \[ f(-1) = 9 \] Therefore, \( f(-1) = 9 \). |
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| gpto1 | To find \( f(-1) \) for the polynomial \( f(t) = a + b t + c t^2 \) that passes through the points \((1, -1)\), \((2, 3)\), and \((3, 13)\), we'll follow these steps: **1. Set up equations using the given points.** For each point \((t, f(t))\), substitute \( t \) and \( f(t) \) into the polynomial: - **Point (1, -1):** \[ f(1) = a + b(1) + c(1)^2 = -1 \] Simplifies to: \[ a + b + c = -1 \quad \text{(Equation 1)} \] - **Point (2, 3):** \[ f(2) = a + b(2) + c(2)^2 = 3 \] Simplifies to: \[ a + 2b + 4c = 3 \quad \text{(Equation 2)} \] - **Point (3, 13):** \[ f(3) = a + b(3) + c(3)^2 = 13 \] Simplifies to: \[ a + 3b + 9c = 13 \quad \text{(Equation 3)} \] **2. Solve the system of equations to find \( a \), \( b \), and \( c \).** First, express \( a \) in terms of \( b \) and \( c \) using Equation 1: \[ a = -1 - b - c \] **Substitute \( a \) into Equations 2 and 3:** - **Equation 2:** \begin{align*} (-1 - b - c) + 2b + 4c & = 3 \\ -1 - b - c + 2b + 4c & = 3 \\ (-1) + ( -b + 2b ) + ( -c + 4c ) & = 3 \\ -1 + b + 3c & = 3 \\ b + 3c & = 4 \quad \text{(Equation 2a)} \end{align*} - **Equation 3:** \begin{align*} (-1 - b - c) + 3b + 9c & = 13 \\ -1 - b - c + 3b + 9c & = 13 \\ (-1) + ( -b + 3b ) + ( -c + 9c ) & = 13 \\ -1 + 2b + 8c & = 13 \\ 2b + 8c & = 14 \quad \text{(Equation 3a)} \end{align*} **Now, solve for \( b \) and \( c \) using Equations 2a and 3a.** First, solve Equation 2a for \( b \): \[ b = 4 - 3c \] Substitute \( b \) into Equation 3a: \begin{align*} 2(4 - 3c) + 8c & = 14 \\ 8 - 6c + 8c & = 14 \\ 8 + 2c & = 14 \\ 2c & = 6 \\ c & = 3 \end{align*} Now, find \( b \) using \( c \): \begin{align*} b & = 4 - 3c \\ b & = 4 - 3(3) \\ b & = 4 - 9 \\ b & = -5 \end{align*} Finally, find \( a \) using \( b \) and \( c \): \begin{align*} a & = -1 - b - c \\ a & = -1 - (-5) - 3 \\ a & = -1 + 5 - 3 \\ a & = 1 \end{align*} **3. Write the specific form of the polynomial.** Now that we have \( a \), \( b \), and \( c \): \[ f(t) = 1 - 5t + 3t^2 \] **4. Calculate \( f(-1) \).** Substitute \( t = -1 \) into the polynomial: \begin{align*} f(-1) & = 1 - 5(-1) + 3(-1)^2 \\ & = 1 + 5 + 3(1) \\ & = 1 + 5 + 3 \\ & = 9 \end{align*} **Answer:** \[ f(-1) = \boxed{9} \] |
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| deepseekv3 | To find the polynomial \( f(t) = a + bt + ct^2 \) that passes through the points \((1, -1)\), \((2, 3)\), and \((3, 13)\), we can set up a system of equations based on these points and solve for the coefficients \(a\), \(b\), and \(c\). Once we have the coefficients, we can determine \(f(-1)\). ### Step 1: Set up the system of equations For each point \((t, f(t))\), substitute the values into the polynomial: 1. For \((1, -1)\): \[ a + b(1) + c(1)^2 = -1 \implies a + b + c = -1 \] 2. For \((2, 3)\): \[ a + b(2) + c(2)^2 = 3 \implies a + 2b + 4c = 3 \] 3. For \((3, 13)\): \[ a + b(3) + c(3)^2 = 13 \implies a + 3b + 9c = 13 \] ### Step 2: Solve the system of equations We have the following system: \[ \begin{cases} a + b + c = -1 \quad \text{(Equation 1)} \\ a + 2b + 4c = 3 \quad \text{(Equation 2)} \\ a + 3b + 9c = 13 \quad \text{(Equation 3)} \end{cases} \] First, subtract Equation 1 from Equation 2: \[ (a + 2b + 4c) - (a + b + c) = 3 - (-1) \] \[ b + 3c = 4 \quad \text{(Equation 4)} \] Next, subtract Equation 2 from Equation 3: \[ (a + 3b + 9c) - (a + 2b + 4c) = 13 - 3 \] \[ b + 5c = 10 \quad \text{(Equation 5)} \] Now, subtract Equation 4 from Equation 5: \[ (b + 5c) - (b + 3c) = 10 - 4 \] \[ 2c = 6 \implies c = 3 \] Substitute \(c = 3\) into Equation 4: \[ b + 3(3) = 4 \implies b + 9 = 4 \implies b = -5 \] Finally, substitute \(b = -5\) and \(c = 3\) into Equation 1: \[ a + (-5) + 3 = -1 \implies a - 2 = -1 \implies a = 1 \] ### Step 3: Write the polynomial The polynomial is: \[ f(t) = 1 - 5t + 3t^2 \] ### Step 4: Calculate \(f(-1)\) Substitute \(t = -1\) into the polynomial: \[ f(-1) = 1 - 5(-1) + 3(-1)^2 = 1 + 5 + 3 = 9 \] ### Final Answer \[ f(-1) = \boxed{9} \] |
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