To find the two possible values of $b$ for which the minimum value of the function $y = x^2 + bx + 16$ is $-9$, we need to follow these steps:

1. Identify the vertex form of the quadratic function: The given quadratic function is $y = x^2 + bx + 16$.

2. Find the vertex: The vertex form of a quadratic function $y = ax^2 + bx + c$ is given by: $x = -\frac{b}{2a}$ Here, $a = 1$ and $c = 16$, so: $x = -\frac{b}{2}$

3. Substitute the vertex x-coordinate back into the function: Substitute $x = -\frac{b}{2}$ back into $y = x^2 + bx + 16$: $y = \left(-\frac{b}{2}\right)^2 + b\left(-\frac{b}{2}\right) + 16$

4. Simplify the expression: $y = \frac{b^2}{4} - \frac{b^2}{2} + 16$ $y = \frac{b^2}{4} - \frac{2b^2}{4} + 16$ $y = -\frac{b^2}{4} + 16$

5. Set the minimum value of the function to -9 and solve for $b$: Given that the minimum value is $-9$: $-\frac{b^2}{4} + 16 = -9$

6. Solve for $b$: $-\frac{b^2}{4} + 16 = -9$ $-\frac{b^2}{4} = -9 - 16$ $-\frac{b^2}{4} = -25$ $\frac{b^2}{4} = 25$ $b^2 = 100$ $b = \pm 10$

So, the two possible values of $b$ are: $b = 10$ $b = -10$

If you have any questions or need further details, feel free to ask. Here are some questions you might want to explore next:

1. How do you find the maximum value of a quadratic function?
2. How does the sign of $a$ in a quadratic function $y = ax^2 + bx + c$ affect the direction of the parabola?
3. What are the roots of the quadratic equation $y = x^2 + bx + 16$ for $b = 10$?
4. How do you complete the square for the function $y = x^2 + bx + 16$?
5. What is the axis of symmetry for the quadratic function $y = x^2 + bx + 16$?

Tip: Remember that the vertex form of a quadratic function can be very helpful in identifying the maximum or minimum value of the function, especially when the function is not easily factorable.