How to predict bound states in a 1D triangular well? Assume we have a (single) particle in a potential well of the following shape:
For $x \leq 0$, $V = \infty$ (Region I)
For $x \geq L$, $V = 0$ (Region III)
For the interval $x > 0$ to $x < L$, $V = -V_0\frac{L-x}{L}$ (Region II).
The potential geometry is reminiscent of the potential energy function of a diatomic molecule (with $x$ the intra-nuclear distance).
In Region II the potential energy is a field with (positive) gradient $\frac{V_0}{L}$.
A few observations:
In Region II, $V(x)$ is non-symmetric, so we can expect eigenfunctions without definite parity.
In Region II we can expect $\psi(0) = 0$.
We can also expect $\psi(\infty) = 0$, so the wave functions should be normalisable.
A quick analytic look at the Schrödinger equation in Region II using Wolfram Alpha’s DSolve facility shows the solutions involve the Airy Functions $A_i$ and $B_i$.
For $\frac{V_0}{L} = 0$, the problem is reduced to an infinite potential wall (not a well). Incoming particles from Region III would simply be reflected by the wall at $x = 0, V = \infty$. There would be no bound states.
And this raises an interesting question: for which value of $\frac{V_0}{L}$ is there at least one bound state and approximately at which value of the Hamiltonian $E$?
I have a feeling this can be related to the Uncertainty Principle because aren’t the confinement energies of bound particles in 1 D wells inversely proportional to $L^2$? If so would calculating a $\sigma_x$ not allow calculating a $\langle p^2 \rangle$ and thus a minimum $E$ for a bound state?
 A: Disclaimer: In this answer, we will just derive a rough semiclassical estimate for the threshold between the existence of zero and one bound state. Of course, one should keep in mind that the semiclassical WKB method is not reliable$^1$ for predicting the ground state. We leave it to others to perform a full numerical analysis of the problem using Airy Functions.
     ^ V
    W| 
    a|
    l|
    l|   L
-----|---|------------------> x
     |  /
     | /
     |/
-V_0 | 
     |
     

$\uparrow$ Fig.1: Potential $V(x)$ as a function of position $x$ in OP's example.
First let us include the metaplectic correction/Maslov index. The turning point at an infinitely hard wall and an inclined potential wall have Maslov index $2$ and $1$, respectively, cf. e.g. this Phys.SE post. In total $3$. We should then adjust the Bohr-Sommerfeld quantization rule with a fraction $\frac{3}{4}$.
$$ \int_{x_-}^{x_+} \! \frac{dx}{\pi} k(x)~\simeq~n+\frac{3}{4},\qquad n~\in~\mathbb{N}_0,\tag{1}  $$
where
$$  k(x)~:=~\frac{\sqrt{2m(E-V(x))}}{\hbar}, \qquad 
V(x)~:=~-V_0 \frac{L-x}{L}. \tag{2} $$
At the threshold, we can assume $n=0$ and $E=0$. The limiting values of the turning points are $x_-=0$ and $x_+=L$. Straightforward algebra yields that the
threshold between the existence of zero and one bound state is
$$V_0~\simeq~\frac{81}{128} \frac{\pi^2\hbar^2}{mL^2} \tag{3} .$$

$^1$ For comparison, the WKB approximation for the threshold of the corresponding square well problem yields
$$V_0~\simeq~\frac{\pi^2\hbar^2}{2m L^2} \tag{4} ,$$
while the exact quantum mechanical result is
$$V_0~=~\frac{\pi^2\hbar^2}{8m L^2} \tag{5} ,$$
cf. e.g. Alonso & Finn, Quantum and Statistical Physics, Vol 3, p. 77-78. Not impressive!
     ^ V
    W| 
    a|
    l|
    l|   L 
-----|---|------------------> x
     |   |
     |   |
     |   |
-V_0 |---| 
     |
     

$\uparrow$ Fig.2: Corresponding square well potential as a function of position $x$. Each of the 2 infinitely hard walls has Maslov index 2.
A: The wavefunction $\psi(x)$ satisfies
$$
-\frac{\hbar^2}{2m}\psi'' + V_0\left(\frac{x}{L} - 1\right) \psi = E\psi, \quad 0 \leq x \leq L\\
-\frac{\hbar^2}{2m}\psi'' = E\psi, \quad x > L
$$
Since the bound states have $E < 0$ let's introduce
$$
k = \frac{\sqrt{-2mE}}{\hbar}\\
\varkappa = \frac{\sqrt{2mV_0}}{\hbar}
$$
Then 
$$
\psi'' - \varkappa^2\frac{x}{L}\psi = (k^2 - \varkappa^2) \psi, \quad 0 \leq x \leq L\\
\psi'' = k^2 \psi, \quad x > L
$$
Introducing new dimensionless coordinate $\xi$ by
$$
x = \sqrt[3]{\frac{L}{\varkappa^2}} \xi + L - L \frac{k^2}{\varkappa^2}\\
x = L + \frac{\xi}{\gamma} -\frac{k^2}{\gamma^3}, \quad \gamma \equiv \sqrt[3]{\frac{\varkappa^2}{L}}
$$
the equation can be reduces to Airy equation
$$
\psi''(\xi) - \xi \psi(\xi) = 0\\
\psi(\xi) = \cos \alpha \operatorname{Ai}(\xi) + \sin \alpha \operatorname{Bi}(\xi)\\
$$
Since we're solving Airy equation in a limited domain $x \in [0, L]$ we cannot throw away the $\operatorname{Bi}(\xi)$ part. At $x = L$ the solution should satisfy
$$
\psi'(L) = -k \psi(L)
$$
since 
$$
\psi(x) = C_3 e^{-kx}, \quad x > L.
$$
We have following conditions to determine $k$:
$$
\text{For } \xi_1 = \frac{k^2}{\gamma^2} - L\gamma\implies \cos \alpha \operatorname{Ai}(\xi_1) +  \sin \alpha \operatorname{Bi}(\xi_1) = 0\\
\text{For } \xi_2 = \frac{k^2}{\gamma^2} \implies \frac{\cos \alpha \operatorname{Ai}'(\xi_2) +  \sin \alpha \operatorname{Bi}'(\xi_2)}{\cos \alpha \operatorname{Ai}(\xi_2) + \sin \alpha \operatorname{Bi}(\xi_2)} = -\frac{k}{\gamma}.
$$
Eliminating $\alpha$ one gets
$$
\frac{\operatorname{Bi}(\xi_1) \operatorname{Ai}'(\xi_2) - \operatorname{Ai}(\xi_1) \operatorname{Bi}'(\xi_2)}{\operatorname{Bi}(\xi_1) \operatorname{Ai}(\xi_2) - \operatorname{Ai}(\xi_1) \operatorname{Bi}(\xi_2)} = -\frac{k}{\gamma}.
$$
To simplify further let's introduce dimensionless $z = \frac{k}{\gamma}$ and parameter $q = \gamma L = \sqrt[3]{L^2 \varkappa^2} = \sqrt[3]{\frac{2mV_0L^2 }{\hbar^2}}$.
Thus we need to study the following equation for $z \geq 0$:
$$
\frac{\operatorname{Bi}(\xi_1) \operatorname{Ai}'(\xi_2) - \operatorname{Ai}(\xi_1) \operatorname{Bi}'(\xi_2)}{\operatorname{Bi}(\xi_1) \operatorname{Ai}(\xi_2) - \operatorname{Ai}(\xi_1) \operatorname{Bi}(\xi_2)} = -z, \quad \xi_1 = z^2 - q, \;\xi_2 = z^2.
$$
Manipulating with the plot of the function one can see that for $q \leq q_\text{cr}$ there are no solutions and when $q > q_\text{cr}$ there are. While $z = 0$ is not a solution ($\psi(+\infty) \neq 0$), it is useful to determine $q_\text{cr}$. That would be the least solution to the following system (plugged $z = 0$):
$$
\operatorname{Bi}(-q_\text{cr}) \operatorname{Ai}'(0) - \operatorname{Ai}(-q_\text{cr}) \operatorname{Bi}'(0) = 0\\
\frac{\operatorname{Ai}(-q_\text{cr})}{\operatorname{Bi}(-q_\text{cr})} = 
\frac{\operatorname{Ai}'(0)}{\operatorname{Bi}'(0)}\\
q_\text{cr} \approx 1.9863527074304728\\
V_0 = \frac{\hbar^2}{2mL^2} q_\text{cr}^3 \approx 7.837347 \frac{\hbar^2}{2mL^2}.
$$
