Why include a constant in the delta potential $\alpha\delta(x)$? Why do we have to multiply a proportionality constant in the delta potential $V(x)=-\alpha \delta(x)$ where $\alpha$ is some positive constant? 
Isn't that $V(x)=\pm \delta(x)$ already enough to represent a very "high" potential?
 A: If $x$ is the argument of the delta function $\delta(x)$ with dimension $[x]$, then the dimensions of the delta function are,
$$[\delta(x)] = \frac{1}{[x]}.$$
As such, we must have that $V(x) = \alpha\delta(x)$ in order for dimensions to be consistent. To see this is the case, notice that,
$$\int dx \, \delta(\alpha x) f(x) = \frac{1}{|\alpha|}\int dx \, \delta(x)f(x)$$
and this is only dimensionally consistent if $\delta$ has dimensions being the inverse of its argument. The value of $\alpha$ itself matters; if we perturb a Hamiltonian by $V(x)$, the first correction to the energy is,
$$\langle \psi | V | \psi \rangle = \alpha \int |\psi(x)|^2 \delta(x)\,  dx = \alpha|\psi(0)|^2$$
and thus while conceptually we may think of $\alpha \delta(x)$ as always being an infinitely large spike regardless of the value of $\alpha$, the actual value itself does affect results.
A: 
Isn't that V(x)=±δ(x) already enough to represent a very "high"
  potential?

Consider instead, the canonical finite potential well problem.

with the additional constraint that width $L = 2a$ and depth $V_0$ are related rather than being independent parameters:
$$L V_0 = \alpha \Rightarrow V(x) = -\alpha\,\frac{\Theta(x + \frac{L}{2}) - \Theta(x - \frac{L}{2})}{L}$$
where $\alpha$ has units $\mathrm{J\cdot m}$ and $\Theta(x)$ is the Heaviside step function.
Note that as $L \rightarrow 0$, $V_0 \rightarrow \infty$ and, in the limit, we have
$$V(x) = -\alpha\; \delta(x) $$
A: The discontinuity in the wave function across a $\delta$-potential $V(x)=-\alpha \delta(x)$ is proportional to $\alpha$, and the energy $E=-\frac{\alpha^2 m}{2\hbar^2}$ for this potential also depends on $\alpha$. Thus, $\alpha$ functions as an tunable parameter for your potential.  
Likewise there is an $\alpha$-dependence on other $\delta$-type potentials, such as the double delta $-\alpha (\delta (x-a)+\delta(x+a)$ so one can adjust the energies in the problem to the particular situation at hand.
