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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).`
y- intercept: (0,1)
interval of decrease: The graph is decreasing for
minimum and maximum point: none
equation of horizontal asymptote: x-axis, or y=0
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