# How do you graph y=(1/5)^x and y=(1/5)^(x-2) and how do the graphs compare?

##### 2 Answers
Dec 4, 2017

Graph them by manually calculating points and plotting them, or use a spreadsheet or plotting program.

#### Explanation:

No matter where you put the range of values, the relative shape of the two curves remains the same. Both are inverse exponential curves (logarithmic curves) with an asymptote at y = 0.
y = (1/5)^x and y = (1/5)^(x−2) Dec 4, 2017

See below.

#### Explanation:

Graph 1:

$y = {\left(\frac{1}{5}\right)}^{x}$

$y = {5}^{-} x$

This has the graph of the standard negative exponential function $f \left(x\right) = {a}^{-} x$; where $a = 5$. The properties of such a graph are as follows:

• The graph passes through the point $\left(0 , 1\right)$
• The domain is $\left(- \infty , + \infty\right)$
• The range is $\left(0 , + \infty\right)$
• The graph is decreasing
• The graph is asymptotic to the x-axis as $x \to + \infty$
• The graph increases without bound as $x \to - \infty$
• The graph is smooth and continuous.

This graph is shown below.

graph{(1/5)^x [-10, 10, -5, 5]}

Graph 2:

$y = {\left(\frac{1}{5}\right)}^{x - 2}$

$y = {5}^{- \left(x - 2\right)} = {5}^{-} x \times {5}^{2}$

This is the Graph 1 above scaled by 25 as shown below..

graph{y=(1/5)^(x-2) [-10, 10, -5, 5]}

This graph has all the properties of Graph 1 above except that it passes through the point $\left(0 , 25\right)$