Course:MATH110/Archive/2010-2011/003/Teams/Thurgau/Homework 12

This is where Homework 12 will be posted for the Thurgau Group.

Given that (1)/(1+e^(-x))

As we have covered in class and in other topics, if we are given f(x)=(2x^2)+2, the constant serve as a tool that shift the graph vertically upwards by a factor of 2 units. As a result, the vertex of this particular function would be (0,2) instead of (0,0). Similar to this concept and theory, this is also applicable to the given function. The following is a graph of P(t)=(1)/(1+e^(-x)) Following the concept was mentioned previously, lets analyze the difference when a constant "2" is added: P(t)=(1)/(1+e^(-x)+(2)) As shown, when a constant value is added (in this case it's 2), this changes two factors and they are: the y-intercept and the horizontal asymptote. The constant,K can be any number, and by changing this number we are able to alter the horizontal asymptote of the graph, and the y-intercept of the graph. From these two graphs, the horizontal asymptote had shifted from 1.0 to a smaller unit in the approximately 0.35. Also due to the change in the constant,K the y-intercept had shifted from 0.5 to 0.25, the value where x=0 had decreased by a factor of 50%. Due to this, the value of K can be altered in order to fit data in regards to many real life problems such as economic growth, wild life habitats and crimes.