Okay, I am not quite understanding your question. Let's start with the definition of the Gaussian, aka the Bell Curve, in its general form.
$$ f(t) = \frac{1}{ \sigma \sqrt{2\pi}} e^{ -\frac{(t-\mu)^2}{2\sigma^2} } $$
$\mu$ is the mean, and represents where the peak occurs.
$\sigma$ is the standard deviation, and identifies where the inflection points are.
Taking the first derivative:
$$
\begin{aligned}
f'(t) & = \frac{1}{ \sigma \sqrt{2\pi}}\left[ e^{ -\frac{(t-\mu)^2}{2\sigma^2} } \left( -\frac{(t-\mu)}{\sigma^2} \right) \right] \\
& = - \frac{1}{ \sigma^3 \sqrt{2\pi}}\left[ e^{ -\frac{(t-\mu)^2}{2\sigma^2} } \left( t-\mu \right) \right] \\
\end{aligned}
$$
Taking the second derivative.
$$
\begin{aligned}
f''(t) &= - \frac{1}{ \sigma^3 \sqrt{2\pi}}\left[ e^{ -\frac{(t-\mu)^2}{2\sigma^2} } \left( -\frac{(t-\mu)}{\sigma^2} \right) \left( t-\mu \right) +
e^{ -\frac{(t-\mu)^2}{2\sigma^2} } \left( 1 \right) \right] \\
f''(t) &= \frac{1}{ \sigma^3 \sqrt{2\pi}}\left[ \frac{(t-\mu)^2}{\sigma^2} - 1 \right] e^{ -\frac{(t-\mu)^2}{2\sigma^2} } \\
f''(t) &= \frac{1}{ \sigma^3 \sqrt{2\pi}}\left[ \left( \frac{t-\mu}{\sigma} \right)^2 - 1 \right] e^{ - \frac{1}{2} \left( \frac{t-\mu}{\sigma} \right)^2 } \\
\end{aligned}
$$
In your example $\mu=0$ and $\sigma=1$. In your graph, the zero is where the peak is, and the one is the +/-1 on your horizontal scale. Notice that the second derivative is zero at these points (inflection in the original function.)
You can translate the graph by setting $\mu$. You can stretch the graph horizontally by setting $\sigma$.
I am not familiar with wavelets, so I don't know what this $a$ and $b$ are that you are talking about. You can definitely to another coordinate conversion if you want:
$$ t = a \tau + b $$
Gosh, but that is awfully similar to
$$ \frac{1}{\sigma} t - \frac{\mu}{\sigma} $$
that you already got.
Based on the comments, to retrofit the function call to MATLAB (or wherever):
$$ \mu = b $$
$$ \sigma = 0.63628 a $$
