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

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

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).`

m1k3y | Student, Grade 11 | (Level 1) eNoter

Posted on

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