a) Rewrite the given function as

$f\left(x\right)=-3+2\times {e}^{-(x-3)}$

Now, $f\left(x\right)$ is in the required form. b) $f\left(x\right)=-3+2\times {e}^{-(x-3)}$,

$f\left(x\right)$ is obtained from the basic exponential function, $g\left(x\right)={e}^{-x}$, by the following sequence of transformations, in the given order:

1) shifting along x-axis: $x\to x-3$

2) Scaling the y-coordinate: $y\to 2y$

3) Translation vertically (below) by 2 units: $y\to y-2$

c) $f\left(x\right)=-3+2\times {e}^{-(x-3)}$,

Key points:

1. $\underset{x\to -\mathrm{\infty}}{lim}f\left(x\right)=\mathrm{\infty};$

2. $\underset{x\to \mathrm{\infty}}{lim}f\left(x\right)=-3;$ (horizontal asymptote)

3. When $x=0,\text{}y=-3+2{e}^{3}$

4. $y=0,$ when ${e}^{3-x}=1.5\Rightarrow 3-x=\mathrm{ln}\left(1.5\right)$

$\Rightarrow x=3-\mathrm{ln}\left(1.5\right)$

$5.{y}^{\prime}=-2\times {e}^{3-x}<0\mathrm{\forall}x.$ Thus $y=f\left(x\right)$

is monotonically decreasing on $(-\mathrm{\infty},\mathrm{\infty})$

from $\mathrm{\infty}$ to $-3$ (asymptotic limit)

So, no vertical asymptote and the horizontal asymptote is $y=-3$