graph the exponential function.\n\n$f(x)=-\\frac{3}{2}(2)^{x}$\n\nplot five points on the graph of the…

Answer:

To plot the points, we substitute different (x) - values into the function (y = f(x)=-\frac{3}{2}(2)^{x}).

Step1: Find the value when (x = - 2)

Substitute (x=-2) into (y =-\frac{3}{2}(2)^{x}). [ \begin{align*} y&=-\frac{3}{2}(2)^{-2}\ &=-\frac{3}{2}\times\frac{1}{4}\ &=-\frac{3}{8}=- 0.375 \end{align*} ] The point is ((-2,-0.375))

Step2: Find the value when (x=-1)

Substitute (x = - 1) into (y=-\frac{3}{2}(2)^{x}). [ \begin{align*} y&=-\frac{3}{2}(2)^{-1}\ &=-\frac{3}{2}\times\frac{1}{2}\ &=-\frac{3}{4}=-0.75 \end{align*} ] The point is ((-1,-0.75))

Step3: Find the value when (x = 0)

Substitute (x = 0) into (y=-\frac{3}{2}(2)^{x}). [ \begin{align*} y&=-\frac{3}{2}(2)^{0}\ &=-\frac{3}{2}\times1\ &=-\frac{3}{2}=-1.5 \end{align*} ] The point is ((0,-1.5))

Step4: Find the value when (x = 1)

Substitute (x = 1) into (y=-\frac{3}{2}(2)^{x}). [ \begin{align*} y&=-\frac{3}{2}(2)^{1}\ &=-\frac{3}{2}\times2\ &=-3 \end{align*} ] The point is ((1,-3))

Step5: Find the value when (x = 2)

Substitute (x = 2) into (y=-\frac{3}{2}(2)^{x}). [ \begin{align*} y&=-\frac{3}{2}(2)^{2}\ &=-\frac{3}{2}\times4\ &=-6 \end{align*} ] The point is ((2,-6))

For an exponential function of the form (y = a\cdot b^{x}+k), in the function (y=-\frac{3}{2}(2)^{x}+0), the horizontal asymptote is (y = 0) (since (k = 0)).

So the five points are ((-2,-0.375)), ((-1,-0.75)), ((0,-1.5)), ((1,-3)), ((2,-6)) and the horizontal asymptote is (y = 0)