To find the equation of the parabola given the focus at \((0, -3)\) and the directrix at \(y = 1\), we can determine its orientation and vertex.
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Determine the vertex: The vertex of the parabola is located halfway between the focus and directrix. The focus is at \((0, -3)\) and the directrix is a horizontal line at \(y = 1\).
The y-coordinate of the vertex can be found by averaging the focus and the directrix: \[ \text{Vertex } y = \frac{-3 + 1}{2} = \frac{-2}{2} = -1 \] The x-coordinate of the vertex is the same as that of the focus (since the focus and directrix are both vertical): \[ \text{Vertex } = (0, -1) \]
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Determine the equation of the parabola: Since the parabola opens downward (the focus is below the directrix), we can use the standard form for a vertical parabola: \[ (x - h)^2 = 4p(y - k) \] where \((h, k)\) is the vertex and \(p\) is the distance from the vertex to the focus.
From the vertex \((0, -1)\) to the focus \((0, -3)\), we see: \[ p = -3 - (-1) = -2 \quad (\text{since } p \text{ is negative for downward opening}) \]
So, the equation becomes: \[ (x - 0)^2 = 4(-2)(y - (-1)) \] Simplifying this gives: \[ x^2 = -8(y + 1) \]
Rearranging: \[ y = -\frac{1}{8}x^2 - 1 \]
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Populate the table values: Now, we can plug the x-values into the equation to find the y-values.
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For \(x = 2\): \[ y = -\frac{1}{8}(2^2) - 1 = -\frac{1}{8}(4) - 1 = -\frac{4}{8} - 1 = -\frac{1}{2} - 1 = -\frac{3}{2} \]
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For \(x = 0\): \[ y = -\frac{1}{8}(0^2) - 1 = -1 \]
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For \(x = -4\): \[ y = -\frac{1}{8}(-4^2) - 1 = -\frac{1}{8}(16) - 1 = -2 - 1 = -3 \]
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For \(x = -1\): \[ y = -\frac{1}{8}(-1^2) - 1 = -\frac{1}{8}(1) - 1 = -\frac{1}{8} - 1 = -\frac{1}{8} - \frac{8}{8} = -\frac{9}{8} \]
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For \(x = 74\): \[ y = -\frac{1}{8}(74^2) - 1 = -\frac{1}{8}(5476) - 1 = -684.5 - 1 = -685.5 \]
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For \(x = -4\) (again): The corresponding value is already calculated as \(y = -3\).
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Now the table would look like this:
\[ \begin{array}{|c|c|} \hline \text{x-value} & \text{y-value} \ \hline 2 & -\frac{3}{2} \ 0 & -1 \ -4 & -3 \ -1 & -\frac{9}{8} \ 74 & -685.5 \ -4 & -3 \ \hline \end{array} \]
Feel free to ask if you need any clarification or further assistance!