To find the equation of the parabola, we start with the focus \((0, -3)\) and directrix \(y = 1\). The parabola opens downwards because the focus is below the directrix.
The standard equation of a parabola with a vertical axis of symmetry is given by: \[ (x - h)^2 = 4p(y - k) \] where the vertex \((h, k)\) is halfway between the focus and directrix, and \(p\) is the distance from the vertex to the focus.
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Finding the vertex: The focus is at \((0, -3)\), and the directrix is at \(y = 1\). The vertex is the midpoint between the focus and directrix: \[ k = \frac{-3 + 1}{2} = \frac{-2}{2} = -1 \quad \text{and} \quad h = 0 \] Therefore, the vertex is \((0, -1)\).
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Finding \(p\): The distance from the vertex to the focus (or directrix) is: \[ p = -1 - (-3) = 2 \] Since the parabola opens downwards, we take \(p = -2\) (negative value).
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Equation of the parabola: Plugging \(h = 0\), \(k = -1\), and \(p = -2\) into the equation: \[ (x - 0)^2 = 4(-2)(y + 1) \] Simplifying, we get: \[ x^2 = -8(y + 1) \] Rearranging gives: \[ y = -\frac{1}{8}x^2 - 1 \]
Now we can calculate the missing \(y\)-values for the given \(x\)-values using this equation.
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When \(x = 0\): \[ y = -\frac{1}{8}(0)^2 - 1 = -1 \]
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For \(x = -1\): \[ y = -\frac{1}{8}(-1)^2 - 1 = -\frac{1}{8} - 1 = -\frac{1}{8} - \frac{8}{8} = -\frac{9}{8} \]
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To find the missing \(x\)-value for \(y = -\frac{4}{3}\): We set up the equation: \[ -\frac{4}{3} = -\frac{1}{8}x^2 - 1 \] Rearranging gives: \[ -\frac{4}{3} + 1 = -\frac{1}{8}x^2 \] Simplifying gives: \[ -\frac{4}{3} + \frac{3}{3} = -\frac{1}{8}x^2 \implies -\frac{1}{3} = -\frac{1}{8}x^2 \] Eliminating the negative sign gives: \[ \frac{1}{3} = \frac{1}{8}x^2 \] Multiplying both sides by 24: \[ 8 = 3x^2 \implies x^2 = \frac{8}{3} \implies x = \pm \sqrt{\frac{8}{3}} = \pm \frac{2\sqrt{2}}{\sqrt{3}} = \pm \frac{2\sqrt{6}}{3} \]
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Therefore, the table is filled as follows:
\[ \begin{tabular}{|c|c|} \hline \( x \)-value & \( y \)-value \ \hline 2 & -4 \ \hline 0 & -1 \ \hline -1 & -\frac{9}{8} \ \hline \frac{2\sqrt{6}}{3} & -\frac{4}{3} \ \hline \end{tabular} \]
This gives the completed table with the missing \(y\)-values filled in.