Question

graph the equation by completing the table and plotting points. identify the domain.
$y = e^x + 3$
part: 0 / 3
part 1 of 3
complete the table. round to two decimal places when necessary.

x	y
-4
-3
-2
-1
0
1

Explanation:

Step1: For \( x = -4 \)

We substitute \( x = -4 \) into the equation \( y = e^{x}+3 \). So we calculate \( e^{-4}+3 \). We know that \( e^{-4}=\frac{1}{e^{4}}\approx\frac{1}{54.5982}\approx0.0183 \). Then \( y = 0.0183 + 3=3.0183\approx3.02 \) (rounded to two decimal places).

Step2: For \( x = -3 \)

Substitute \( x=-3 \) into \( y = e^{x}+3 \). \( e^{-3}=\frac{1}{e^{3}}\approx\frac{1}{20.0855}\approx0.0498 \). Then \( y = 0.0498+3 = 3.0498\approx3.05 \) (rounded to two decimal places).

Step3: For \( x=-2 \)

Substitute \( x = -2 \) into \( y=e^{x}+3 \). \( e^{-2}=\frac{1}{e^{2}}\approx\frac{1}{7.3891}\approx0.1353 \). Then \( y=0.1353 + 3=3.1353\approx3.14 \) (rounded to two decimal places).

Step4: For \( x=-1 \)

Substitute \( x=-1 \) into \( y = e^{x}+3 \). \( e^{-1}=\frac{1}{e}\approx0.3679 \). Then \( y = 0.3679+3=3.3679\approx3.37 \) (rounded to two decimal places).

Step5: For \( x = 0 \)

Substitute \( x = 0 \) into \( y=e^{x}+3 \). We know that \( e^{0}=1 \), so \( y=1 + 3=4.00 \).

Step6: For \( x = 1 \)

Substitute \( x = 1 \) into \( y=e^{x}+3 \). \( e^{1}\approx2.7183 \), so \( y=2.7183+3 = 5.7183\approx5.72 \) (rounded to two decimal places).

Answer:

For \( x=-4 \), \( y\approx3.02 \); for \( x = -3 \), \( y\approx3.05 \); for \( x=-2 \), \( y\approx3.14 \); for \( x=-1 \), \( y\approx3.37 \); for \( x = 0 \), \( y = 4.00 \); for \( x = 1 \), \( y\approx5.72 \)

(The table filled would be:

\( x \)	\( y \)
-3	3.05
-2	3.14
-1	3.37
0	4.00
1	5.72	)