How do I graph the logistic function $P (t) = \frac{11.5}{1 + 12.8 e^{- 0.0266 t}}$ $P(t)=11.5/(1+12.8e^(-0.0266t))$ on a TI-83?

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Answer 1 · 2016-12-25T05:27:54

See the Socratic graph and the explanation.
graph{x-1/.266ln((12.8y)/(11.5-y))=0 [-10, 10, -5 5]}

graph{x-1/.266ln((12.8y)/(11.5-y))=0 [-100, 100, -20, 20]}
P>0.

The inverse t-sxplicit formula

$t = \frac{1}{0.266} ln (\frac{12.8 P}{11.5 - P})$ $t=1/0.266ln((12.8P)/(11.5-P))$ is used

The P-intercept ( t = 0 ) is 11.5/13.8=0.8333, nearly.

As $t \to - \infty, P \to 0$ $t to -oo, P to 0$ .

As $t \to \infty, P \to 11.5$ $t to oo, P to 11.5$ .

So, P = 11.5 is aother horizontal asymptote, besides P = 0.

The second graph is on a befitting scale to reveal this important

characteristic if this logistic function.

I hope that these data and graph will be useful.

How do I graph the logistic function P(t)=11.51+12.8e−0.0266tP(t)=11.5/(1+12.8e^(-0.0266t)) on a TI-83?