What are the horizonatal asymptotes of y=(1000)/(1+9e^(-.1656x))?
To find the horizontal asymptotes of a function, we need to determine the behavior of the function as x approaches positive infinity and negative infinity.
In the case of the function y = 1000 / (1 + 9e^(-0.1656x)), we can analyze its behavior by examining the exponential term, 9e^(-0.1656x), as x approaches infinity and negative infinity.
As x approaches infinity, the exponential term, e^(-0.1656x), will approach 0 since the negative exponent will make the term smaller and smaller. This means that the function will approach 1000 / (1 + 9 * 0) = 1000 as x tends to infinity. Therefore, the horizontal asymptote on the positive side is y = 1000.
Similarly, as x approaches negative infinity, the exponential term, e^(-0.1656x), will also approach 0 since the x value becomes increasingly negative. Thus, the function will again approach 1000 / (1 + 9 * 0) = 1000 as x tends to negative infinity. Therefore, the horizontal asymptote on the negative side is also y = 1000.
In summary, the horizontal asymptotes of the function y = 1000 / (1 + 9e^(-0.1656x)) are y = 1000 both on the positive and negative sides.