Question 295222
A new farm pond was stocked with 2500 crappies in 2003. The crappie population in 2006 was found to be 4320. 
a) Let t be the number of years after 2003 (in other words t = 0 corresponds to 2003). Write down the initial crappie population at t = 3. 
b) Find the growth function of the form f(t) = y0 bt that gives the crappie population t years after 2003. The function you have found models _________________________. 
c) Predict the crappie population in 2010. 
d) In what year will the crappie population reach 13,000? Give the exact value for t and then use your calculator to approximate t to get the year. 
e) Are your answers for (c) and (d) consistent with the given population data in your answer to (b)? 

ASSUMIMG the population growth is LINEAR the growth function is a line that passes through points (0,2500) and (3,4320). If the growth is exponential the solution is much different.

The standard form of a line is f(t) = m*t + b where m is the slope and b is the value of f(t) where the line crosses the y axis.

Since the line passes through (0,2500) and (3,4320) the slope is:

m = (4320-2500)/(3-0) 
m = 1820/3
m = 606.66

So we have y = 606.66*t + b

When t = 0, y = 2500 so substituting in the equation above we have:

2500 = 606.66*0 + b
b = 2500

The growth function then is y = 606.66*t + 2500

You can use the equation above to answer c and d:

c.) for year 2010 t = 7 so:

y = 606.66*7 + 2500

d.) 13000 = 606.66*t + 2500
solve the above for t.