 Taylor expansion of a translated Gaussian

Multiply the two exponential series in $e^{-(x-a{)}^2}=e^{-a^2}{e}^{2ax}{e}^{-x^2}$ to obtain the displayed Taylor expansion. Its first three nonzero terms depend on the real parameter $a$. For $a\ne 0,\pm 1/\sqrt{2}$, they end at degree two. At $a=0$, odd coefficients vanish and the first three terms are $1-x^2+x^4/2$. At $a=\pm 1/\sqrt{2}$, the quadratic coefficient vanishes, leaving $e^{-1/2}(1+2ax-4a{x}^3/3)$ as the first three nonzero terms. Parameter-dependent vanishing coefficients must be checked when a question counts nonzero terms rather than prescribing an expansion order.