12. A parabola has vertex $$ (3,-1) $$, a horizontal axis of symmetry and passes through the point $$ (-3,5) $$. Determine the equation of the parabola. A grid is provided for rough work.

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To find the equation of the parabola with a vertex at $$ (3, -1) $$, a horizontal axis of symmetry, and passing through the point $$ (-3, 5) $$, we can follow these steps:

1. **Identify the standard form of the equation**: Since the parabola has a horizontal axis of symmetry, its equation can be expressed in the form:
   $$
   (y - k) = a(x - h)^2
   $$
   where $$ (h, k) $$ is the vertex of the parabola. Here, $$ h = 3 $$ and $$ k = -1 $$.

2. **Substituting the vertex into the equation**: We substitute $$ h $$ and $$ k $$ into the equation:
   $$
   (y + 1) = a(x - 3)^2
   $$

3. **Using the point on the parabola**: We know the parabola passes through the point $$ (-3, 5) $$. We can substitute $$ x = -3 $$ and $$ y = 5 $$ into the equation to find $$ a $$:
   $$
   5 + 1 = a(-3 - 3)^2
   $$
   Simplifying this gives:
   $$
   6 = a(-6)^2
   $$
   $$
   6 = 36a
   $$
   $$
   a = \frac{6}{36} = \frac{1}{6}
   $$

4. **Writing the final equation**: Now that we have $$ a $$, we can write the equation of the parabola:
   $$
   (y + 1) = \frac{1}{6}(x - 3)^2
   $$

5. **Rearranging to standard form**: To express it in a more standard form, we can rearrange it:
   $$
   y + 1 = \frac{1}{6}(x - 3)^2
   $$
   $$
   y = \frac{1}{6}(x - 3)^2 - 1
   $$

Thus, the equation of the parabola is:
$$
y = \frac{1}{6}(x - 3)^2 - 1
$$
The equation of the parabola is $$ y = \frac{1}{6}(x - 3)^2 - 1 $$.

A parabola has vertex , a horizontal axis of symmetry and passes through the point .
Determine the equation of the parabola. A grid is provided for rough work.

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A parabola has vertex , a horizontal axis of symmetry and passes through the point . Determine the equation of the parabola. A grid is provided for rough work.