 # The graph y=-2\left(\frac{3}{2}-e^{3-x}\right) by: a) Performing the n Bevan Mcdonald 2021-10-31 Answered
The graph $y=-2\left(\frac{3}{2}-{e}^{3-x}\right)$ by:
a) Performing the necessary algebra so that the function is in the proper form (i.e., the transformations are in the proper order).
b) Listing the transformations in the order that they are to be applied.
c) Marking the key point and horizontal asymptote.
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a) Rewrite the given function as
$f\left(x\right)=-3+2×{e}^{-\left(x-3\right)}$
Now, $f\left(x\right)$ is in the required form. b) $f\left(x\right)=-3+2×{e}^{-\left(x-3\right)}$,
$f\left(x\right)$ 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) $f\left(x\right)=-3+2×{e}^{-\left(x-3\right)}$,
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
4. $y=0,$ when ${e}^{3-x}=1.5⇒3-x=\mathrm{ln}\left(1.5\right)$
$⇒x=3-\mathrm{ln}\left(1.5\right)$
$5.{y}^{\prime }=-2×{e}^{3-x}<0\mathrm{\forall }x.$ Thus $y=f\left(x\right)$
is monotonically decreasing on $\left(-\mathrm{\infty },\mathrm{\infty }\right)$
from $\mathrm{\infty }$ to $-3$ (asymptotic limit)
So, no vertical asymptote and the horizontal asymptote is $y=-3$