Select the graph that represents $$ -x^{2}-6 x+27 $$

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1. Write the quadratic function:  
   $$
   y = -x^{2} - 6x + 27.
   $$

2. Factor out $$-1$$ from the quadratic and linear terms:  
   $$
   y = -(x^{2} + 6x) + 27.
   $$

3. Complete the square inside the parentheses. Recall that for $$x^{2}+6x$$ we add and subtract $$\left(\frac{6}{2}\right)^2 = 9$$:  
   $$
   x^{2}+6x = (x^{2}+6x+9) - 9 = (x+3)^2 - 9.
   $$

4. Substitute the completed square back into the equation:  
   $$
   y = -\left[(x+3)^2 - 9\right] + 27.
   $$

5. Distribute the negative sign:  
   $$
   y = -(x+3)^2 + 9 + 27 = -(x+3)^2 + 36.
   $$

6. Identify key features of the parabola from the vertex form $$ y=-(x+3)^2+36 $$:  
   - The vertex is at $$(-3, 36)$$.  
   - The coefficient of $$(x+3)^2$$ is negative, so the parabola opens downward.

7. Find the $$x$$-intercepts by setting $$y=0$$:  
   $$
   0 = -x^{2} - 6x + 27 \quad \Longrightarrow \quad x^{2} + 6x - 27 = 0.
   $$  
   Use the quadratic formula:  
   $$
   x = \frac{-6 \pm \sqrt{6^2 - 4(1)(-27)}}{2(1)} = \frac{-6 \pm \sqrt{36+108}}{2} = \frac{-6 \pm \sqrt{144}}{2}.
   $$  
   $$
   x = \frac{-6 \pm 12}{2}.
   $$  
   Thus,  
   $$
   x = 3 \quad \text{or} \quad x = -9.
   $$

8. Find the $$y$$-intercept by setting $$x=0$$:  
   $$
   y = -0^2 - 6(0) + 27 = 27.
   $$

9. To select the correct graph among the options, look for one that:  
   - Has a vertex at $$(-3, 36)$$.  
   - Opens downward.  
   - Intersects the $$x$$-axis at $$-9$$ and $$3$$.  
   - Intersects the $$y$$-axis at $$27$$.

The graph matching these features is the one that represents  
$$
y = -x^{2}-6x+27.
$$
The graph of $$ y = -x^{2} - 6x + 27 $$ is a downward-opening parabola with its vertex at $$(-3, 36)$$, $$x$$-intercepts at $$-9$$ and $$3$$, and a $$y$$-intercept at $$27$$.

Select the graph that represents

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