# State for y=(2/7)^x: x- and y- intercept, domain, range, intervals of increase/decrease, minimum and maximum point, equation of horizontal asymptote.

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### 2 Answers

You should remember that you may find x intercepts solving the equation `y = 0` such that:

`(2/7)^x = 0` invalid

Since `(2/7)^x` can never be zero, hence, there are no x intercepts for the given function.

You should remember that you may find y intercepts considering `x= 0` such that: `y = (2/7)^0 =gt y = 1`

Hence, the graph intercepts `y` axis at the point `(0,1).`

Notice that the given function is an exponential function, hence, its domain of definition is R and the range is `(0,oo).`

You need to solve the equation `(dy)/(dx) = 0` to check if the function has extreme points such that:

`(dy)/(dx) = (2/7)^x*ln(2/7) < 0`

Since the function never cancels, hence the function has no extreme points.

**Since the first derivative is negative, hence the function strictly decreases over `(-oo,oo).` **

x-intercept: **none**

y- intercept: **(0,1)**

domain:

range:

interval of decrease: **The graph is decreasing for **

minimum and maximum point:** none**

equation of horizontal asymptote: **x-axis, or y=0**