Question

graph: $f(x) = \frac{3}{2}(2)^x$
step 1: (partially visible, likely evaluate initial value)
$f(0) = 1.5$
step 2: plot the initial value of the function at (0, 1.5).
step 3: evaluate the function at two more points.
$f(1) = 3$ $f(-1) = 0.75$
step 4: plot the points (1, 3) and (-1, 0.75).
step 5: identify the horizontal asymptote of the function.
the asymptote is the line dropdown with y=0, y=1, y=4.
(graph with grid, points plotted; table with x=-1,0,1)

Explanation:

Step1: Recall exponential function asymptote

For an exponential function \( f(x) = a(b)^x \) (where \( a
eq0 \), \( b > 0 \), \( b
eq1 \)), the horizontal asymptote is \( y = 0 \) because as \( x\to-\infty \), \( b^x\to0 \) (if \( b > 1 \)) or \( x\to\infty \), \( b^x\to0 \) (if \( 0 < b < 1 \)), so \( f(x)\to0 \). Here, \( f(x)=\frac{3}{2}(2)^x \), \( a=\frac{3}{2} \), \( b = 2>1 \). As \( x\to-\infty \), \( 2^x\to0 \), so \( f(x)\to0 \). Thus, the horizontal asymptote is \( y = 0 \).

Step2: Verify with options

The options are \( y = 0 \), \( y = 1 \), \( y = 4 \). From the function's behavior, the asymptote is \( y = 0 \).

Answer:

\( y = 0 \)