Let p(x) be a quadratic polynomial, such that its graph is tangent to the lines y = -11x - 37, y = x, and y = 9. The x-coordinates of these points of tangency are x_1, x_2, and x_3. Find x_1 + x_2 + x_3.
Let $p(x) = ax^2 + bx + c$. We know that $p(x)$ is tangent to $y = -11x - 37$, $y = x$, and $y = 9$.
First, let's consider $y = -11x - 37$. Since $p(x)$ is tangent to this line, it must have the same slope as this line at the point of tangency. The slope of $p(x)$ is $p'(x) = 2ax + b$, so we want $2ax_1 + b = -11$.
Next, we consider $y = x$. We again require the slopes to match, so $2ax_2 + b = 1$.
Finally, we consider $y = 9$. Since $p(x)$ is a quadratic and is tangent to this line, we know that it achieves its maximum value at the point of tangency. The $x$-coordinate of this point is the vertex of the parabola, which is given by $-\frac b{2a}$. Therefore, $x_3 = -\frac b{2a}$, and we know that $p(x_3) = 9$. This means that $a{x_3}^2 + bx_3 + c = 9$.
We now have three linear equations in the variables $a$, $b$, and $c$:
\begin{align*}
2ax_1 + b &= -11 \\
2ax_2 + b &= 1 \\
a{x_3}^2 + bx_3 + c &= 9
\end{align*}
We want to find $x_1 + x_2 + x_3$, which is $-\frac b{2a}$ plus the roots of the quadratic $ax^2 + bx + c$. By Vieta's formulas, the sum of the roots of this quadratic is $-\frac ba$. So, we want to solve for $a$ and $b$ in terms of $x_1$, $x_2$, and $x_3$.
From the first equation, we have $b = -11 - 2ax_1$, so we substitute into the second equation to get $2ax_2 - 11 - 2ax_1 = 1$, or $a(x_2 - x_1) = 6$.
From the third equation, we have $a{x_3}^2 + bx_3 + c = 9$. Substituting $b = -2ax_3$ from the first equation gives $a{x_3}^2 - 2ax_3 + c = 9$. Since $x_3$ is the vertex of the parabola, we know that $p(x_3) = a{x_3}^2 + bx_3 + c$ is the maximum value of $p(x)$. Therefore, $p(x) \leq 9$ for all $x$. This gives us the inequality $a(x-x_3)^2 + 9 \geq 9$ for all $x$, which simplifies to $a(x-x_3)^2 \geq 0$.
Expanding the square and simplifying, we get $ax^2 - 2ax_3 x + ax_3^2 \geq 0$. This quadratic is non-negative for all $x$ if and only if its discriminant is non-positive, so we have $(-2ax_3)^2 - 4a^2x_3^2 \leq 0$. Simplifying, we get $a^2x_3^2 \leq 0$, so $a=0$.
Therefore, we have $b = -11 - 2ax_1 = -11$, and $c = p(x_3) = 9$.
Now we can find $x_1$, $x_2$, and $x_3$. From $2ax_1 + b = -11$, we have $x_1 = \frac{-11-b}{2a} = 1$. From $2ax_2 + b = 1$, we have $x_2 = \frac{1-b}{2a} = -4$. From $ax_3^2 + bx_3 + c = 9$, we have $ax_3^2 - 2ax_3 + c = 9$, or $a(x_3-1)^2 = 0$. Since we know $a=0$, this just gives $x_3 = 1$.
Therefore, $x_1 + x_2 + x_3 = 1 - 4 + 1 = \boxed{-2}$.