To find the equations that represent the parabola given the vertex and directrix, we can use the properties of parabolas.
Given the vertex at \((-1, 2)\) and the directrix \(y = 0\), we can determine the orientation and the equation of the parabola.
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Orientation: Since the vertex is above the directrix (the vertex \(y\) value is 2, while the directrix \(y\) value is 0), the parabola opens upwards.
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Standard Form: The standard form of a parabola that opens upwards is given by:
\[ (x - h)^2 = 4p(y - k) \]
where \((h, k)\) is the vertex and \(p\) is the distance from the vertex to the directrix. In this case:
- \((h, k) = (-1, 2)\)
- The distance \(p\) is \(2\) (from \(2\) to \(0\)).
Thus, \(p = 2\), and \(4p = 8\).
Substituting in the values, we have:
\[ (x + 1)^2 = 8(y - 2) \]
Rearranging gives us:
\[ (x + 1)^2 = 8y - 16 \quad \text{or} \quad (x + 1)^2 - 8y + 16 = 0 \]
This shows us the standard form of our parabola with the vertex at \((-1, 2)\) and a directrix at \(y = 0\).
Now, we analyze the answer options provided:
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Option 1: \(4y = x^2 + 2x + 5\)
- Rearranging gives \(y = \frac{1}{4}(x^2 + 2x + 5)\). This does not fit our vertex or directrix.
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Option 2: \(8y = x^2 - 4x - 4\)
- Rearranging gives \(y = \frac{1}{8}(x^2 - 4x - 4)\). This also does not fit our vertex or directrix.
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Option 3: \(8y = x^2 + 2x + 17\)
- Rearranging gives \(y = \frac{1}{8}(x^2 + 2x + 17)\). This also does not fit our vertex or directrix.
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Option 4: \((x + 1)^2 = 4(y - 2)\)
- Rearranging gives \((x + 1)^2 = 8y - 16\), which confirms that this option is exactly of the expected form of the parabola.
Therefore, the only option that represents the parabola is:
Option 4: \((x + 1)^2 = 4(y - 2)\).
Thus, the only correct answer is option 4.