The graph y=-2\left(\frac{3}{2}-e^{3-x}\right) by: a) Performing the n

a) Rewrite the given function as
${\displaystyle f\left(x\right)=-3+2\times e^{-(x-3)}}$
Now, ${\displaystyle f\left(x\right)}$ is in the required form. b) ${\displaystyle f\left(x\right)=-3+2\times e^{-(x-3)}}$,
${\displaystyle f\left(x\right)}$ is obtained from the basic exponential function, ${\displaystyle g\left(x\right)=e^{-x}}$, by the following sequence of transformations, in the given order:
1) shifting along x-axis: ${\displaystyle x\to x-3}$
2) Scaling the y-coordinate: ${\displaystyle y\to 2y}$
3) Translation vertically (below) by 2 units: ${\displaystyle y\to y-2}$
c) ${\displaystyle f\left(x\right)=-3+2\times e^{-(x-3)}}$,
Key points:
1. ${\displaystyle \underset{x\to -\mathrm{\infty }}{lim}f\left(x\right)=\mathrm{\infty };}$
2. ${\displaystyle \underset{x\to \mathrm{\infty }}{lim}f\left(x\right)=-3;}$ (horizontal asymptote)
3. When ${\displaystyle x=0,y=-3+2e^3}$
4. ${\displaystyle y=0,}$ when ${\displaystyle e^{3-x}=1.5\Rightarrow 3-x=\ln \left(1.5\right)}$
${\displaystyle \Rightarrow x=3-\ln \left(1.5\right)}$
${\displaystyle 5.y^{\prime }=-2\times e^{3-x}<0\mathrm{\forall }x.}$ Thus ${\displaystyle y=f\left(x\right)}$
is monotonically decreasing on ${\displaystyle (-\mathrm{\infty },\mathrm{\infty })}$
from ${\displaystyle \mathrm{\infty }}$ to ${\displaystyle -3}$ (asymptotic limit)
So, no vertical asymptote and the horizontal asymptote is ${\displaystyle y=-3}$