To determine whether it is possible to fit a parabolic equation (in the form \(y = ax^2 + bx + c\)) to three given points \((0,3)\), \((-3,2)\), and \((3,4)\), we can substitute these points into the equation to create a system of equations.
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For the point \((0,3)\): \[ 3 = a(0)^2 + b(0) + c \implies c = 3 \]
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For the point \((-3,2)\): \[ 2 = a(-3)^2 + b(-3) + c \implies 2 = 9a - 3b + 3 \] Simplifying gives: \[ 9a - 3b = -1 \quad \text{(Equation 1)} \]
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For the point \((3,4)\): \[ 4 = a(3)^2 + b(3) + c \implies 4 = 9a + 3b + 3 \] Simplifying gives: \[ 9a + 3b = 1 \quad \text{(Equation 2)} \]
Now we have two equations:
- \(9a - 3b = -1\) (Equation 1)
- \(9a + 3b = 1\) (Equation 2)
We can add these two equations to eliminate \(b\): \[ (9a - 3b) + (9a + 3b) = -1 + 1 \] This gives: \[ 18a = 0 \implies a = 0 \]
Substituting \(a = 0\) back into either equation will provide \(b\). Using Equation 2: \[ 9(0) + 3b = 1 \implies 3b = 1 \implies b = \frac{1}{3} \]
Thus:
- \(a = 0\)
- \(b = \frac{1}{3}\)
- \(c = 3\)
The resulting equation is: \[ y = 0x^2 + \frac{1}{3}x + 3 \implies y = \frac{1}{3}x + 3 \]
This is a linear equation, meaning that, while we can fit these points with a linear function, we cannot fit them to a quadratic function (parabola). Therefore, the concluding statement is:
This is not possible. Since \(a = 0\), the result is a linear equation, \(y = \frac{1}{3}x + 3\).
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