**OBJECTIVES**

1. Use a Graphing Utility to Fit an Exponential Function to Data

2. Use a Graphing Utility to Fit a Logarithmic FUnction to Data

3. Use a Graphing Utility to Fit a Logistic Function to Data

**Objective 1:** Use a Graphing Utility to Fit an Exponential Function to Data

The following data represents the number of bacteria after hours of being infected:

Hours Infected | # of Bacteria |
---|---|

0 | 500 |

1 | 597 |

3 | 817 |

5 | 1153 |

7 | 1620 |

9 | 2240 |

15 | 6160 |

Using a graphing utility, find the exponential function that best models the data.

1. Input the data into a table in the graphing utility

a. This will plot the points for you

2. Enter y_{1}~a*b^{x}_{1} into the graphing utility in a separate box

a. read y sub-1 tilde “a” times “b” to the power of “x” sub-1,

3. The graphing will fit an exponential regression

a. It gives the R^{2} value: a statistical measure of how close the data are to the fitted regression line

i. Between 0 and 1 (0 being no correlation and 1 being a perfect fit)

b. It gives the value for “a” and “b”

c. Use the give “a” and “b” values to write a function that best models the data

Example from the table above:

R^{2} value: 1

“a” value: 497.928

“b” value: 1.18255

The exponential regression that best fits the data is y=497.928(1.18255)^{x}

**Objective 2:** Use a Graphing Utility to Fit a Logarithmic Function to Data

The following data represent the population of a small town months after incorporation:

Months After Incorporation | Population of the Town |
---|---|

6 | 3,000 |

18 | 4,000 |

42 | 5,000 |

90 | 6,000 |

150 | 7,000 |

Using a graphing utility, find the logarithmic function that best models the data.

**1.** Input the data into a table in the graphing utility

a. This will plot the points for you

**2.** Enter y_{1}~a+b*lnx_{1} in a separate box

a. read: y sub-1 tilde “a” plus “b” times natural logarithm of “x” sub-1,

**3.** The graphing will fit a logarithmic regression

a. It gives the R^{2} value: a statistical measure of how close the data are to the fitted regression line

i. Between 0 and 1 (0 being no correlation and 1 being a perfect fit)

b. It gives the value for “a” and “b”

c. Use the give values to write a function that best models the data

Example from the table above:

R^{2} value: .983

“a” value: 621.55

“b” value: 1221

The logarithmic regression that best fits the data is y=621.55+1221*ln(x)

**Objective 3:**Use a Graphing Utility to Fit a Logistic Function to Data

The following data represents the height of a sunflower, in cm, after days of growing.

Day | Height (cm) |
---|---|

0 | 0.00 |

7 | 17.93 |

14 | 36.36 |

21 | 67.76 |

28 | 98.10 |

35 | 131.00 |

42 | 169.50 |

49 | 205.50 |

56 | 228.30 |

63 | 247.10 |

70 | 250.50 |

77 | 253.80 |

84 | 254.50 |

Using a graphing utility, find the logistic function that best models the data.

1. Input the data into a table in the graphing utility

a. This will plot the points for you

2. Enter y_{1}~a/(1+b*e^{t*x}_{1}) in a separate box

a. read: y sub-1 tilde “a” divided by 1 plus “b” times euler’s number to the “t” times “x” sub-1 power

3. The graphing will fit a logistic regression

a. It gives the R^{2} value: a statistical measure of how close the data are to the fitted regression line

i. Between 0 and 1 (0 being no correlation and 1 being a perfect fit)

b. It gives the value for “a”, “b”, and “t” values

c. Use the give values to write a function that best models the data

Example from the table above:

R^{2} value: 0.9976

“a” value: 259.963

“b” value: 21.8277

“t” value: -0.0900829

The logistic regression that best fits the data is y=259.963/(1+21.8277e^{-0.0900829x})