The gradient of the curve $y=b{e}^{-x/a}$ where it crosses the y-axis is

$y=b{e}^{-x/a}-\left(1\right)$

Given (1) crosses y-axis$\Rightarrow$ put $x=0$

$\Rightarrow y=b$

(1) cuts the y-axis at $\left(0,b\right)$

Gradient$={\left(\frac{dy}{dx}\right)}_{(o,b)}=\frac{-b}{a}{e}^{-x/a}=\frac{-b}{a}$