Question

graph the equation by completing the table and plotting points. identify the domain.
$y = e^x + 5$
part 1 of 3
complete the table. round to two decimal places when necessary.

x	y
-3	5.05
-2	5.14
-1	5.37
0	6
1	7.72
2	12.39

part: 1 / 3
part 2 of 3
graph the equation. plot the horizontal asymptote and the points from the table.

Explanation:

Step1: Analyze the function type

The function is \( y = e^x + 5 \), which is an exponential function. The parent function \( y = e^x \) has a horizontal asymptote at \( y = 0 \), and for \( y = e^x + 5 \), the horizontal asymptote is shifted up by 5 units, so the horizontal asymptote is \( y = 5 \).

Step2: Plot the points

The points from the table are \((-3, 5.05)\), \((-2, 5.14)\), \((-1, 5.37)\), \((0, 6)\), \((1, 7.72)\), \((2, 12.39)\). Plot these points on the coordinate plane.

Step3: Draw the horizontal asymptote

Draw the horizontal line \( y = 5 \) (dashed line since it's an asymptote, not part of the function's graph).

Step4: Sketch the curve

Connect the plotted points smoothly, following the shape of an exponential curve. As \( x \to -\infty \), the function approaches the horizontal asymptote \( y = 5 \), and as \( x \to \infty \), \( y \to \infty \).

(For the domain part, since \( e^x \) is defined for all real numbers \( x \), the domain of \( y = e^x + 5 \) is all real numbers, \( (-\infty, \infty) \).)

Answer:

To graph \( y = e^x + 5 \):

1. Plot the horizontal asymptote \( y = 5 \) (dashed line).
2. Plot the points from the table: \((-3, 5.05)\), \((-2, 5.14)\), \((-1, 5.37)\), \((0, 6)\), \((1, 7.72)\), \((2, 12.39)\).
3. Sketch a smooth curve through the points, approaching \( y = 5 \) as \( x \to -\infty \) and increasing without bound as \( x \to \infty \).

Domain: \( (-\infty, \infty) \)