Like exponential and logarithmic growth, logistic growth increases over time. One of the most notable differences with logistic growth models is that, at a certain point, growth steadily slows and the function approaches an upper bound, or limiting value. Because of this, logistic regression is best for modeling phenomena where there are limits in expansion, such as availability of living space or nutrients.
It is worth pointing out that logistic functions actually model resourcelimited exponential growth. There are many examples of this type of growth in realworld situations, including population growth and spread of disease, rumors, and even stains in fabric. When performing logistic regression analysis, we use the form most commonly used on graphing utilities:
Recall that:
 [latex]\frac{c}{1+a}[/latex] is the initial value of the model.
 when b > 0, the model increases rapidly at first until it reaches its point of maximum growth rate, [latex]\left(\frac{\mathrm{ln}\left(a\right)}{b},\frac{c}{2}\right)[/latex]. At that point, growth steadily slows and the function becomes asymptotic to the upper bound y = c.
 c is the limiting value, sometimes called the carrying capacity, of the model.
A General Note: Logistic Regression
Logistic regression is used to model situations where growth accelerates rapidly at first and then steadily slows to an upper limit. We use the command “Logistic” on a graphing utility to fit a logistic function to a set of data points. This returns an equation of the form
Note that
 The initial value of the model is [latex]\frac{c}{1+a}[/latex].
 Output values for the model grow closer and closer to y = c as time increases.
How To: Given a set of data, perform logistic regression using a graphing utility.
 Use the STAT then EDIT menu to enter given data.
 Clear any existing data from the lists.
 List the input values in the L1 column.
 List the output values in the L2 column.
 Graph and observe a scatter plot of the data using the STATPLOT feature.
 Use ZOOM [9] to adjust axes to fit the data.
 Verify the data follow a logistic pattern.
 Find the equation that models the data.
 Select “Logistic” from the STAT then CALC menu.
 Use the values returned for a, b, and c to record the model, [latex]y=\frac{c}{1+a{e}^{bx}}[/latex].
 Graph the model in the same window as the scatterplot to verify it is a good fit for the data.
Example 3: Using Logistic Regression to Fit a Model to Data
Mobile telephone service has increased rapidly in America since the mid 1990s. Today, almost all residents have cellular service. The table below shows the percentage of Americans with cellular service between the years 1995 and 2012.^{[1]}
Year  Americans with Cellular Service (%)  Year  Americans with Cellular Service (%) 

1995  12.69  2004  62.852 
1996  16.35  2005  68.63 
1997  20.29  2006  76.64 
1998  25.08  2007  82.47 
1999  30.81  2008  85.68 
2000  38.75  2009  89.14 
2001  45.00  2010  91.86 
2002  49.16  2011  95.28 
2003  55.15  2012  98.17 
 Let x represent time in years starting with x = 0 for the year 1995. Let y represent the corresponding percentage of residents with cellular service. Use logistic regression to fit a model to these data.
 Use the model to calculate the percentage of Americans with cell service in the year 2013. Round to the nearest tenth of a percent.
 Discuss the value returned for the upper limit, c. What does this tell you about the model? What would the limiting value be if the model were exact?
Solution
 Using the STAT then EDIT menu on a graphing utility, list the years using values 0–15 in L1 and the corresponding percentage in L2. Then use the STATPLOT feature to verify that the scatterplot follows a logistic pattern as shown in Figure 5:
Use the “Logistic” command from the STAT then CALC menu to obtain the logistic model,
[latex]y=\frac{105.7379526}{1+6.88328979{e}^{0.2595440013x}}[/latex]Next, graph the model in the same window as shown in Figure 6 to verify it is a good fit:

To approximate the percentage of Americans with cellular service in the year 2013, substitute x = 18 for the in the model and solve for y:
[latex]\begin{cases}y\hfill & =\frac{105.7379526}{1+6.88328979{e}^{0.2595440013x}}\hfill & \text{Use the regression model found in part (a)}.\hfill \\ \hfill & =\frac{105.7379526}{1+6.88328979{e}^{0.2595440013\left(18\right)}}\hfill & \text{Substitute 18 for }x.\hfill \\ \hfill & \approx \text{99}\text{.3 }\hfill & \text{Round to the nearest tenth}\hfill \end{cases}[/latex]According to the model, about 98.8% of Americans had cellular service in 2013.

The model gives a limiting value of about 105. This means that the maximum possible percentage of Americans with cellular service would be 105%, which is impossible. (How could over 100% of a population have cellular service?) If the model were exact, the limiting value would be c = 100 and the model’s outputs would get very close to, but never actually reach 100%. After all, there will always be someone out there without cellular service!
Try It 3
The table below shows the population, in thousands, of harbor seals in the Wadden Sea over the years 1997 to 2012.
Year  Seal Population (Thousands)  Year  Seal Population (Thousands) 

1997  3.493  2005  19.590 
1998  5.282  2006  21.955 
1999  6.357  2007  22.862 
2000  9.201  2008  23.869 
2001  11.224  2009  24.243 
2002  12.964  2010  24.344 
2003  16.226  2011  24.919 
2004  18.137  2012  25.108 
a. Let x represent time in years starting with x = 0 for the year 1997. Let y represent the number of seals in thousands. Use logistic regression to fit a model to these data.
b. Use the model to predict the seal population for the year 2020.
c. To the nearest whole number, what is the limiting value of this model?
 Source: The World Bank, 2013 ↵