Answer:
(-1,5) and [tex](3, \frac{1}{125})[/tex] are points on the graph
Step-by-step explanation:
Given
[tex]g(x) = \frac{1}{5}^x[/tex]
Required
Determine which point in on the graph
To get which of point A to D is on the graph, we have to plug in their values in the given expression using the format; (x,g(x))
A. (-1,5)
x = -1
Substitute -1 for x in [tex]g(x) = \frac{1}{5}^x[/tex]
[tex]g(x) = \frac{1}{5}^{-1}[/tex]
Convert to index form
[tex]g(x) = 1/(\frac{1}{5})[/tex]
Change / to *
[tex]g(x) = 1*(\frac{5}{1})[/tex]
[tex]g(x) = 5[/tex]
This satisfies (-1,5)
Hence, (-1,5) is on the graph
B. (1,0)
x = 1
Substitute 1 for x
[tex]g(x) = \frac{1}{5}^x[/tex]
[tex]g(x) = \frac{1}{5}^1[/tex]
[tex]g(x) = \frac{1}{5}[/tex]
(1,0) is not on the graph because g(x) is not equal to 0
C. [tex](3, \frac{1}{125})[/tex]
x = 3
Substitute 3 for x
[tex]g(x) = \frac{1}{5}^x[/tex]
[tex]g(x) = \frac{1}{5}^3[/tex]
Apply law of indices
[tex]g(x) = \frac{1}{5} * \frac{1}{5} * \frac{1}{5}[/tex]
[tex]g(x) = \frac{1}{125}[/tex]
This satisfies [tex](3, \frac{1}{125})[/tex]
Hence, [tex](3, \frac{1}{125})[/tex] is on the graph
D. [tex](-2, \frac{1}{25})[/tex]
x = -2
Substitute -2 for x
[tex]g(x) = \frac{1}{5}^x[/tex]
[tex]g(x) = \frac{1}{5}^{-2}[/tex]
Convert to index form
[tex]g(x) = 1/(\frac{1}{5}^2)[/tex]
[tex]g(x) = 1/(\frac{1}{5}*\frac{1}{5})[/tex]
[tex]g(x) = 1/(\frac{1}{25})[/tex]
Change / to *
[tex]g(x) = 1*(\frac{25}{1})[/tex]
[tex]g(x) = 25[/tex]
This does not satisfy [tex](-2, \frac{1}{25})[/tex]
Hence, [tex](-2, \frac{1}{25})[/tex] is not on the graph
