Question

graph: $f(x) = 2^{x - 1}$
$f(1) = 1$
step 2: plot the value of the function at (1, 1).
step 3: evaluate the function at two more points.
$f(0) = 1/2$ $f(2) = 2$
step 4: plot the points (0, 1/2) and (2, 2).
step 5: identify the horizontal asymptote of the function.
the asymptote is the line
options: y = 0, y = 1, y = 2

Explanation:

Step1: Recall the form of exponential functions

Exponential functions of the form \( f(x) = a^{x - h}+k \) have a horizontal asymptote at \( y = k \). For the function \( f(x)=2^{x - 1} \), we can rewrite it as \( f(x)=2^{x - 1}+0 \), so \( k = 0 \).

Step2: Analyze the behavior of \( 2^{x-1} \) as \( x\to-\infty \)

As \( x \) approaches negative infinity, \( x - 1 \) also approaches negative infinity. The function \( 2^{u} \) where \( u=x - 1 \) will approach 0 because for exponential functions with base \( a>1 \), \( a^{u}\to0 \) as \( u\to-\infty \). So the horizontal asymptote is \( y = 0 \).

Answer:

\( y = 0 \)