Scaling a wavelet in continuous wavelet transform
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:

$$
\begin{aligned}
f'(t) &amp; = \frac{1}{ \sigma \sqrt{2\pi}}\left[ e^{ -\frac{(t-\mu)^2}{2\sigma^2} } \left( -\frac{(t-\mu)}{\sigma^2} \right) \right] \\
 &amp; = - \frac{1}{ \sigma^3 \sqrt{2\pi}}\left[ e^{ -\frac{(t-\mu)^2}{2\sigma^2} } \left( t-\mu \right) \right] \\
\end{aligned}
$$

Taking the second derivative.

$$
\begin{aligned}
f''(t) &amp;= - \frac{1}{ \sigma^3 \sqrt{2\pi}}\left[  e^{ -\frac{(t-\mu)^2}{2\sigma^2} } \left( -\frac{(t-\mu)}{\sigma^2} \right) \left( t-\mu \right) +
e^{ -\frac{(t-\mu)^2}{2\sigma^2} }   \left( 1 \right)  \right] \\
f''(t) &amp;=  \frac{1}{ \sigma^3 \sqrt{2\pi}}\left[    \frac{(t-\mu)^2}{\sigma^2} - 1 \right]  e^{ -\frac{(t-\mu)^2}{2\sigma^2} } \\
f''(t) &amp;=  \frac{1}{ \sigma^3 \sqrt{2\pi}}\left[  \left( \frac{t-\mu}{\sigma} \right)^2 - 1 \right]  e^{ - \frac{1}{2} \left( \frac{t-\mu}{\sigma} \right)^2 } \\
\end{aligned}
$$

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):

$$ \mu = b $$

$$ \sigma = 0.63628 a $$