Transcript 8.8 Logistic Growth Functions P. 517 Hello, my name is Super Power Hero.

8.8 Logistic Growth Functions
P. 517
Hello, my name is
Super Power Hero.
General form
Logistic Growth Functions
• a, c, r are positive real
constants
c
•y=
 rx
1  ae
Evaluating
•
100
f(x) =
2 x
1  9e
100
• f(-3) =
 23 ≈ .0275
1  9e
•
100
f(0) =
 20
1  9e
= 100/10 = 10
Graph on your calculator:
1
y
x
1 e
Graph on your calculator:
10
y
2 x
1  5e
Graph on your calculator:
5
y
2 x
1  10 e
• From these graphs you can see
that a logistic growth function has
an upper bound of y=c.
• Logistic growth functions are used
to model real-life quantities whose
growth levels off because the rate
of growth changes – from an
increasing growth rate to a
decreasing growth rate.
Decreasing growth rate
Increasing growth rate
Point of maximum
Growth where the graph
Switches from growth
To decrease.
c
The graphs of y 
 rx
1  ae
• The horizontal lines y=0 & y=c are asymptotes
c
• The y intercept is (0, 1  a )
• The Domain is all reals and the Range is
0<y<c
• The graph is increasing from left to right
• To the left of it’s point of maximum growth,
 ln a c 
, ,

 r 2
the rate of increase is increasing.
• To the right of it’s point of maximum growth,
the rate of increase is decreasing
6
Graph y 
.5 x
1  2e
•
•
•
•
•
•
•
Asy:
y=0, y=6
Y-int:
6/(1+2)=6/3=2
Max growth:
(ln2/.5 , 6/2) =
(1.4 , 3)
(0,2)
Your turn! Graph:
3
y
2 x
1  5e
•Asy: y=0 & y=3
•Y-int: (0,1/2)
•Max growth: (.8, 1.5)
Solving Logistic Growth Functions
• Solve:
•
•
•
•
•
•
50
 40
3 x
1  10e
50 = 40(1+10e-3x)
50 = 40 + 400e-3x
10 = 400e-3x
.025 = e-3x
ln.025 = -3x
1.23 ≈ x
1  10e  3 x
 40
50
Your turn!
• Solve:
• .46 ≈ x
30

10
2 x
1  5e
Lets look at Example #5 p.519
• We’ll use the calculator to model a Logistic
Growth Function.
Assignment