Select the graph that represents the funct $$ -x^{2}+2 x+8 $$

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1. The function is 
   $$
   y = -x^2 + 2x + 8.
   $$
   Since the coefficient of $$ x^2 $$ is negative ($$ -1 $$), the parabola opens downward.

2. To find the vertex, we use the formula for the $$ x $$-coordinate:
   $$
   x = -\frac{b}{2a} = -\frac{2}{2(-1)} = 1.
   $$
   Substitute $$ x = 1 $$ into the function to find the $$ y $$-coordinate:
   $$
   y = -1^2 + 2(1) + 8 = -1 + 2 + 8 = 9.
   $$
   Thus, the vertex is at $$ (1, 9) $$.

3. Find the $$ x $$-intercepts by setting $$ y = 0 $$:
   $$
   -x^2 + 2x + 8 = 0.
   $$
   Multiply by $$-1$$ to simplify:
   $$
   x^2 - 2x - 8 = 0.
   $$
   Factor or use the quadratic formula:
   $$
   x = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-8)}}{2(1)} 
   = \frac{2 \pm \sqrt{4 + 32}}{2}
   = \frac{2 \pm \sqrt{36}}{2}
   = \frac{2 \pm 6}{2}.
   $$
   This yields:
   $$
   x = \frac{8}{2} = 4 \quad \text{and} \quad x = \frac{-4}{2} = -2.
   $$
   So, the $$ x $$-intercepts are $$ (-2, 0) $$ and $$ (4, 0) $$.

4. The $$ y $$-intercept is found by setting $$ x = 0 $$:
   $$
   y = -0^2 + 2(0) + 8 = 8.
   $$
   Thus, the $$ y $$-intercept is $$ (0, 8) $$.

5. Summary:  
   - The parabola opens downward.  
   - Vertex at $$ (1, 9) $$ (the maximum point).  
   - $$ x $$-intercepts at $$ (-2, 0) $$ and $$ (4, 0) $$.  
   - $$ y $$-intercept at $$ (0, 8) $$.

The graph representing the function should have the vertex at $$ (1, 9) $$, pass through $$ (0, 8) $$ for the $$ y $$-intercept, and cross the $$ x $$-axis at $$ (-2, 0) $$ and $$ (4, 0) $$, with the parabola opening downward.
The graph of $$ y = -x^2 + 2x + 8 $$ is a downward-opening parabola with its vertex at $$ (1, 9) $$. It crosses the $$ y $$-axis at $$ (0, 8) $$ and the $$ x $$-axis at $$ (-2, 0) $$ and $$ (4, 0) $$.

Select the graph that represents the funct

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