graph the exponential function.\n\n$f(x)=-\frac{1}{2}(2)^{x}$\n\nplot five points on the graph of the…

Explanation:

Step1: Find value when x = - 2

Substitute (x=-2) into (f(x)=-\frac{1}{2}(2)^{x}), we get (f(-2)=-\frac{1}{2}(2)^{-2}=-\frac{1}{2}\times\frac{1}{4}=-\frac{1}{8}).

Step2: Find value when x = - 1

Substitute (x = - 1) into (f(x)=-\frac{1}{2}(2)^{x}), we get (f(-1)=-\frac{1}{2}(2)^{-1}=-\frac{1}{2}\times\frac{1}{2}=-\frac{1}{4}).

Step3: Find value when x = 0

Substitute (x = 0) into (f(x)=-\frac{1}{2}(2)^{x}), we get (f(0)=-\frac{1}{2}(2)^{0}=-\frac{1}{2}\times1 =-\frac{1}{2}).

Step4: Find value when x = 1

Substitute (x = 1) into (f(x)=-\frac{1}{2}(2)^{x}), we get (f(1)=-\frac{1}{2}(2)^{1}=-\frac{1}{2}\times2=-1).

Step5: Find value when x = 2

Substitute (x = 2) into (f(x)=-\frac{1}{2}(2)^{x}), we get (f(2)=-\frac{1}{2}(2)^{2}=-\frac{1}{2}\times4=-2).

The five points are ((-2,-\frac{1}{8})), ((-1,-\frac{1}{4})), ((0,-\frac{1}{2})), ((1,-1)), ((2,-2)) which can be plotted on the graph.

Answer:

The five - point set is (\left{(-2,-\frac{1}{8}),(-1,-\frac{1}{4}),(0,-\frac{1}{2}),(1,-1),(2,-2)\right})