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Sketch the parabola (y-3)^(2)=8(x+3) and state the vertex, focus, and directrix.

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Firstly, we should know that the equation given is in the standard form of a parabola that opens to the right or left which states as (y - k)² = 4p(x - h).

Here, if we compare this with our parabola equation (y - 3)² = 8(x + 3), we find:

- The vertex (h, k) = (-3, 3). This is the maximum or minimum point of the parabola.
- The distance from the vertex to the focus or from the vertex to the directrix is p = 8/4 = 2 units away from the vertex.

Let's first find the focus. The focus will be 2 units to the right of the vertex (since the parabola opens to the right). So, if we add 2 to the x-coordinate of the vertex, we get the x-coordinate of the focus. And y-coordinate will remain same. Hence the focus of the parabola is (-1, 3).

Next is the directrix. The directrix is a vertical line that would be 2 units to the left of the vertex (since it's opposite to the focus from the vertex). If we subtract 2 from the x-coordinate of the vertex, we will get x = -5 (which is our directrix).

Finally, to sketch the parabola we use these values. Mark the vertex (-3, 3) and the focus (-1, 3). Sketch the directrix as a vertical line at x = -5. The parabola would be symmetric with respect to a line passing through the focus and the vertex, and it will wrap around the focus and away from the directrix.

User Alejandro Silva
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