Select the graph that represents the fund $$ x^{2}+12 x+27 $$

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UpStudy AI · Accepted Answer

We start with the quadratic function  
$$
x^2 + 12x + 27
$$  
and complete the square to write it in vertex form.

1. Rewrite the function by grouping the $$ x $$ terms:
   $$
   x^2 + 12x + 27 = \left(x^2 + 12x\right) + 27.
   $$

2. Complete the square for $$ x^2 + 12x $$:
   - Compute half of the coefficient of $$ x $$: $$ \frac{12}{2} = 6 $$.
   - Square it: $$ 6^2 = 36 $$.
   - Add and subtract 36 inside the parentheses:
     $$
     x^2 + 12x = \left(x^2 + 12x + 36\right) - 36 = \left(x+6\right)^2 - 36.
     $$

3. Substitute back into the expression:
   $$
   x^2 + 12x + 27 = \left(x+6\right)^2 - 36 + 27 = \left(x+6\right)^2 - 9.
   $$

4. In vertex form $$\left(x+6\right)^2 - 9$$, the vertex is at  
   $$
   (-6, -9).
   $$

5. Since the coefficient of $$\left(x+6\right)^2$$ is positive, the parabola opens upward.

6. Additionally, we can find the $$ x $$-intercepts by solving  
   $$
   \left(x+6\right)^2 - 9 = 0.
   $$
   Add 9 to both sides:
   $$
   \left(x+6\right)^2 = 9.
   $$
   Taking the square root:
   $$
   x+6 = \pm 3.
   $$
   Thus:
   $$
   x = -6+3 = -3 \quad \text{or} \quad x = -6-3 = -9.
   $$

Therefore, the graph representing the function is a parabola that:
- Has a **vertex** at $$(-6, -9)$$.
- Opens **upward**.
- Crosses the $$ x $$-axis at $$(-9, 0)$$ and $$(-3, 0)$$.

Select the graph that shows a parabola with these characteristics.
The graph is a parabola opening upward with its vertex at (-6, -9) and x-intercepts at (-9, 0) and (-3, 0).

Select the graph that represents the fund

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