Question

(a) graph the exponential function (g(x) = 2^{x-1}).
to do this, plot two points on the graph of the function, and also draw the asymptote. then click on the function button.

(b) give the domain and range of the function using interval notation.

Explanation:

Find points on the function

To graph the exponential function \(g(x) = 2^{x-1}\), we calculate the coordinates of two points by choosing simple integer values for \(x\).

• Let \(x = 1\):

$$g(1) = 2^{1-1} = 2^0 = 1$$

This gives the point \((1, 1)\).

• Let \(x = 2\):

$$g(2) = 2^{2-1} = 2^1 = 2$$

This gives the point \((2, 2)\).

Determine the asymptote

An exponential function of the form \(y = a \cdot b^{x-h} + k\) has a horizontal asymptote at \(y = k\).
For \(g(x) = 2^{x-1}\), we have \(k = 0\).
Therefore, the horizontal asymptote is the line:

$$y = 0$$

Determine the domain

The domain of any basic exponential function \(g(x) = b^{x-h} + k\) is all real numbers because we can substitute any real value for \(x\).
In interval notation, this is:

$$(-\infty, \infty)$$

Determine the range

Since the base \(2 > 0\) and there is no vertical shift (\(k = 0\)), the value of \(2^{x-1}\) is strictly greater than \(0\) for all real \(x\).
Thus, the range is all values greater than the horizontal asymptote \(y = 0\).
In interval notation, this is:

$$(0, \infty)$$

Answer:

Question a

• Two points to plot: \((1, 1)\) and \((2, 2)\)
• Horizontal asymptote: \(y = 0\) (the x-axis)

Question b

• Domain: \((-\infty, \infty)\)
• Range: \((0, \infty)\)