To find the vertex of the quadratic function $g(x) = x^2 + 10x + 9$, we can either complete the square or use the vertex formula. The vertex formula for a quadratic function $g(x) = ax^2 + bx + c$ is:

$x_{\text{vertex}} = \frac{-b}{2a}$

Here:

• $a = 1$ (coefficient of $x^2$),
• $b = 10$ (coefficient of $x$),
• $c = 9$ (constant).

Step 1: Find the x-coordinate of the vertex

$x_{\text{vertex}} = \frac{-10}{2(1)} = \frac{-10}{2} = -5$

Step 2: Find the y-coordinate of the vertex

Now, substitute $x = -5$ into the original function $g(x) = x^2 + 10x + 9$:

$g(-5) = (-5)^2 + 10(-5) + 9 = 25 - 50 + 9 = -16$

Vertex:

The vertex of the function $g(x) = x^2 + 10x + 9$ is at $(-5, -16)$.

Would you like further details on how this relates to the graph?

Here are 5 related questions to explore this topic:

1. What is the axis of symmetry for this quadratic function?
2. How do you determine if the vertex is a maximum or minimum point?
3. Can you rewrite $g(x) = x^2 + 10x + 9$ in vertex form?
4. How does the discriminant of a quadratic equation relate to the graph?
5. What is the domain and range of the function $g(x)$?

Tip: The vertex of a parabola can always be found using the vertex formula or by completing the square!