1. **State the problem:** Find the vertex of the parabola given by the quadratic function $f(x) = x^2 + 10x + 9$. 2. **Formula for vertex:** The vertex of a parabola $f(x) = ax^2 + bx + c$ is given by the point $$\left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right).$$ Here, $a=1$, $b=10$, and $c=9$. 3. **Calculate the x-coordinate of the vertex:** $$x = -\frac{b}{2a} = -\frac{10}{2 \times 1} = -\frac{10}{2} = -5.$$ 4. **Calculate the y-coordinate of the vertex by substituting $x=-5$ into $f(x)$:** $$f(-5) = (-5)^2 + 10(-5) + 9 = 25 - 50 + 9 = -16.$$ 5. **Write the vertex as an ordered pair:** The vertex is $$(-5, -16).$$ This means the parabola reaches its minimum point at $x=-5$ and $y=-16$.