To determine if a parabolic equation can be written with the given three points \((0, 3)\), \((-3, 2)\), and \((3, 4)\), we can use the standard form of a parabolic equation, which is \(y = ax^2 + bx + c\). Since we have three points, we can set up a system of equations to find the values of \(a\), \(b\), and \(c\).
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Using the point \((0, 3)\): \[ 3 = a(0)^2 + b(0) + c \Rightarrow c = 3. \]
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Using the point \((-3, 2)\): \[ 2 = a(-3)^2 + b(-3) + 3 \Rightarrow 2 = 9a - 3b + 3 \Rightarrow 9a - 3b = -1 \quad \text{(Eq. 1)}. \]
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Using the point \((3, 4)\): \[ 4 = a(3)^2 + b(3) + 3 \Rightarrow 4 = 9a + 3b + 3 \Rightarrow 9a + 3b = 1 \quad \text{(Eq. 2)}. \]
Now we can solve the system of equations formed by Eq. 1 and Eq. 2:
From Eq. 1: \[ 9a - 3b = -1 \quad \Rightarrow \quad 3b = 9a + 1 \quad \Rightarrow \quad b = 3a + \frac{1}{3}. \]
Substituting \(b\) into Eq. 2: \[ 9a + 3(3a + \frac{1}{3}) = 1 \Rightarrow 9a + 9a + 1 = 1 \Rightarrow 18a + 1 = 1 \Rightarrow 18a = 0 \Rightarrow a = 0. \]
If \(a = 0\), the equation becomes \(y = bx + c\), which is a linear equation. Substituting back, \[ b = 3(0) + \frac{1}{3} = \frac{1}{3} \Rightarrow y = \frac{1}{3}x + 3. \]
Now the potential parabolic equation cannot be formed as \(a = 0\) indicates a linear function rather than a quadratic one.
Thus, we can conclude:
This is not possible. Since \(a = 0\), only higher degree equations can be written.
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