Let $a^x = y$. The original function is then changed to
$$g(y) = y^2 + 3y - 2,$$ which is increasing over $\left(-\frac{3}{2}, +\infty\right)$.

When $0 < a < 1$, we have $y \in [a, a^{-1}]$ and
$$g(y)_{\max} = a^{-2} + 3a^{-1} - 2 = 8 \Rightarrow a^{-1} = 2 \Rightarrow a = \frac{1}{2}.$$
Then
$$g(y)_{\min} = \left(\frac{1}{2}\right)^2 + 3 \times \frac{1}{2} - 2 = -\frac{1}{4}.$$
When $a > 1$, we have $y \in [a^{-1}, a]$ and
$$g(y)_{\max} = a^2 + 3a - 2 = 8 \Rightarrow a = 2.$$
Then
$$g(y)_{\min} = 2^{-2} + 3 \times 2^{-1} - 2 = -\frac{1}{4}.$$
In summary, the minimum value of $f(x)$ on $x \in [-1, 1]$ is $-\frac{1}{4}$.