graph the exponential function.\n\n$f(x)=(\\frac{4}{5})^x$\n\nplot five points on the graph of the function…

graph the exponential function.\n\n$f(x)=(\\frac{4}{5})^x$\n\nplot five points on the graph of the function, and also draw the asymptote. then click on the graph - a - function button.

graph the exponential function.\n\n$f(x)=(\\frac{4}{5})^x$\n\nplot five points on the graph of the function, and also draw the asymptote. then click on the graph - a - function button.

Answer

Answer:

To plot five points and find the asymptote for (y = (\frac{4}{5})^x):

  1. When (x = 0):
    • Substitute (x = 0) into (y=(\frac{4}{5})^x). According to the rule (a^0 = 1) for (a\neq0), we have (y = (\frac{4}{5})^0=1). So the point is ((0,1)).
  2. When (x = 1):
    • Substitute (x = 1) into (y = (\frac{4}{5})^x), then (y=\frac{4}{5}=0.8). The point is ((1,0.8)).
  3. When (x = 2):
    • Substitute (x = 2) into (y = (\frac{4}{5})^x), (y = (\frac{4}{5})^2=\frac{16}{25}=0.64). The point is ((2,0.64)).
  4. When (x=- 1):
    • Substitute (x=-1) into (y = (\frac{4}{5})^x), using the rule (a^{-n}=\frac{1}{a^{n}}), we get (y=\frac{5}{4} = 1.25). The point is ((-1,1.25)).
  5. When (x=-2):
    • Substitute (x = - 2) into (y = (\frac{4}{5})^x), (y=(\frac{4}{5})^{-2}=\frac{25}{16}=1.5625). The point is ((-2,1.5625)).

For the exponential - function (y = a^x) where (0\lt a\lt1) (here (a=\frac{4}{5})), the horizontal asymptote is (y = 0).

Explanation:

Step1: Find the (y) - value for (x = 0)

Use (a^0 = 1), (y=(\frac{4}{5})^0 = 1).

Step2: Find the (y) - value for (x = 1)

Substitute (x = 1) into (y = (\frac{4}{5})^x), (y=\frac{4}{5}).

Step3: Find the (y) - value for (x = 2)

Calculate (y = (\frac{4}{5})^2=\frac{16}{25}).

Step4: Find the (y) - value for (x=-1)

Use (a^{-n}=\frac{1}{a^{n}}), (y=\frac{5}{4}).

Step5: Find the (y) - value for (x=-2)

Calculate (y = (\frac{4}{5})^{-2}=\frac{25}{16}).

Step6: Determine the asymptote

For (y = a^x) with (0\lt a\lt1), the asymptote is (y = 0).