Question

quiz: exponential functions
hs: algebra 2b m (sequential) / 3:exponential
graph of an exponential function. then four options:
y = 0.5(2)^{x + 1} + 3

y = 0.5(2)^{x - 1} + 3

y = 0.5(2)^{x + 1} - 3

y = 0.5(2)^{x - 1} - 3

Explanation:

Step1: Identify the horizontal asymptote

The graph has a horizontal asymptote at \( y=-3 \). For an exponential function of the form \( y = a(b)^{x - h}+k \), the horizontal asymptote is \( y = k \). So \( k=-3 \), which eliminates the first two options (they have \( k = 3 \)).

Step2: Find a point on the graph

The graph passes through the point \( (1, -2.5) \) (or we can check \( x = 1 \) in the remaining options). Let's check \( x = 1 \) in the third option \( y=0.5(2)^{x + 1}-3 \):
\( y=0.5(2)^{1 + 1}-3=0.5\times4 - 3=2 - 3=-1
eq - 2.5 \)

Now check the fourth option \( y = 0.5(2)^{x - 1}-3 \):
\( y=0.5(2)^{1-1}-3=0.5\times1 - 3=0.5 - 3=-2.5 \), which matches the point on the graph.

Answer:

D. \( y = 0.5(2)^{x - 1}- 3 \)