\begin{tabular}{l|l} N list & $$ \begin{array}{l} ext { For the quadratic function } \mathrm{f}(\mathrm{x})=\mathrm{x}^{2}+6 \mathrm{x}+9 ext {, answer parts (a) through (c). } \ ext { (a) Graph the quadratic function by determining whether its graph opens up or down } \ ext { and by finding its vertex, axis of symmetry, } \mathrm{y} ext {-intercept, and } \mathrm{x} ext {-intercepts, if any. } \ ext { Does the graph of } \mathrm{f} ext { open up or down? } \ ext { tion } 10\end{array} $$ \ $$ \begin{array}{ll} ext { What are the coordinates of the vertex? }\end{array} $$ \ The vertex of the parabola is $$ \square $$.\end{tabular}

Question

\begin{tabular}{l|l} N list & $$ \begin{array}{l}	ext { For the quadratic function } \mathrm{f}(\mathrm{x})=\mathrm{x}^{2}+6 \mathrm{x}+9 	ext {, answer parts (a) through (c). } \ 	ext { (a) Graph the quadratic function by determining whether its graph opens up or down } \ 	ext { and by finding its vertex, axis of symmetry, } \mathrm{y} 	ext {-intercept, and } \mathrm{x} 	ext {-intercepts, if any. } \ 	ext { Does the graph of } \mathrm{f} 	ext { open up or down? } \ 	ext { tion } 10\end{array} $$ \ $$ \begin{array}{ll}	ext { What are the coordinates of the vertex? }\end{array} $$ \ The vertex of the parabola is $$ \square $$.\end{tabular}

UpStudy AI · Accepted Answer

1. **Determine the direction the parabola opens:**

The quadratic function is given by  
   $$
   f(x) = x^2 + 6x + 9.
   $$
   Since the coefficient of $$ x^2 $$ is positive ($$1 > 0$$), the parabola opens upward.

2. **Find the vertex:**

The vertex of a parabola given by  
   $$
   f(x) = ax^2 + bx + c
   $$
   can be found using the formula  
   $$
   x = -\frac{b}{2a}.
   $$
   Here, $$ a = 1 $$ and $$ b = 6 $$, so:
   $$
   x = -\frac{6}{2 \cdot 1} = -3.
   $$
   
   To find the $$ y $$-coordinate, substitute $$ x = -3 $$ into $$ f(x) $$:
   $$
   f(-3) = (-3)^2 + 6(-3) + 9 = 9 - 18 + 9 = 0.
   $$
   Therefore, the vertex is at $$ \boxed{(-3, 0)} $$.

3. **Determine the axis of symmetry:**

The axis of symmetry is vertical through the vertex, i.e.,  
   $$
   x = -3.
   $$

4. **Find the $$ y $$-intercept:**

The $$ y $$-intercept occurs when $$ x = 0 $$:
   $$
   f(0) = 0^2 + 6(0) + 9 = 9.
   $$
   Thus, the $$ y $$-intercept is $$ (0, 9) $$.

5. **Find the $$ x $$-intercepts:**

The $$ x $$-intercepts occur when $$ f(x) = 0 $$:
   $$
   x^2 + 6x + 9 = 0.
   $$
   Factor the quadratic:
   $$
   (x+3)^2 = 0.
   $$
   Hence,
   $$
   x + 3 = 0 \quad \Longrightarrow \quad x = -3.
   $$
   This means there is one $$ x $$-intercept at $$ (-3, 0) $$ (a repeated root).

**Summary of Findings:**

- **Direction:** Opens upward.
- **Vertex:** $$ (-3, 0) $$.
- **Axis of symmetry:** $$ x = -3 $$.
- **$$ y $$-intercept:** $$ (0, 9) $$.
- **$$ x $$-intercept:** $$ (-3, 0) $$.

Thus, the coordinates of the vertex are $$\boxed{(-3, 0)}$$.
The vertex of the parabola is at $$(-3, 0)$$.

N list

The vertex of the parabola is .

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N list The vertex of the parabola is .