# time in minutes: 0,2.5,3.2,6.3,7.1,9.5,11.7  temp 60,35,29,14,12,7,4  find equation of exponential function round values a & b to one decimal please help with this i dont even know where to...

time in minutes: 0,2.5,3.2,6.3,7.1,9.5,11.7

temp 60,35,29,14,12,7,4

find equation of exponential function round values a & b to one decimal

please help with this i dont even know where to start they want me to make a graph out of the information provided

embizze | Certified Educator

You are given the table of values and asked to find the exponential model.

First, to graph we use the time t as the independent variable (along the horizontal or "x" axis) and the temperature as the dependent variable (along the vertical or "y" axis). You should get a series of points getting lower as you read from left to right.

For the equation, it depends on the type of class you are in:

(1) In some classes you may be allowed to use technology (a graphing calculator or spreadsheet, for example). In this case, using a graphing calculator I found the equation to be :

`y=a*b^x` where `a=60.96783965` and `b=.7939370668` with an `r^2` value of .9995226016 and an r value of -.9997612723.

Thus your solution would be `y=61.0(.8)^t` with t the time in minutes and y the temperature of the drink.

(2) If you cannot use technology, pick a pair of points from the table. (I generally try to avoid the first and last points, as well as two points that are next to each other).

Choosing (2.5,35) and (6.3,14) we can solve for a and b:

The solution is of the form `y=a*b^t` where again y is the temperature at time t given in minutes. Plugging in the values we get the following equations:

`35=a*b^(2.5)` and `14=a*b^(6.3)`

Solving `35=a*b^(2.5)` for a we get `a=35/(b^(2.5))` . Plugging this value into the second equation we get `14=35/(b^(2.5))b^(6.3)`

Then `14/35=(b^(6.3))/(b^(2.5))`

`b^(19/5)=2/5` Raising both sides to the `5/19` power we get

`b=(2/5)^(5/19)~~.7857`

Then `a=35/(b^(2.5))=35/(.7857^(2.5))~~63.9626`

So your solution would be `y=64(.8)^t` . This isn't quite as good a fit, and the numbers will be different for different choices of points.

The graphs `y=61(.8)^t` (red) and `y=64(.8)^t` (blue):

(The vertical axis is in 10's)

` `