The formula P = 0.64x^2 - 0.044x + 3 models the approximate population P, in thousands, for a species of fish in a local pond, x years after 1997. During what year will the population reach 34,052 fish?
A.2004
b.2003
c. 2006
d.2005
To find the year when the population reaches 34,052 fish, we need to solve the equation 0.64x^2 - 0.044x + 3 = 34.052.
Step 1: Rewrite the equation
0.64x^2 - 0.044x + 3 = 34.052
Step 2: Subtract 34.052 from both sides
0.64x^2 - 0.044x - 31.052 = 0
Step 3: Use the quadratic formula
The quadratic formula is x = (-b ± √(b^2 - 4ac)) / (2a).
In this case, a = 0.64, b = -0.044, and c = -31.052.
Using these values in the quadratic formula, we can find the values of x.
x = (-(-0.044) ± √((-0.044)^2 - 4*(0.64)*(-31.052))) / (2*(0.64))
Simplifying further,
x = (0.044 ± √(0.001936 + 79.2896)) / (1.28)
x = (0.044 ± √(79.291536)) / (1.28)
x = (0.044 ± 8.907) / (1.28)
Now, we have two possible values for x.
x1 = (0.044 + 8.907) / 1.28
x2 = (0.044 - 8.907) / 1.28
Calculating these values,
x1 = 8.951 / 1.28 = 6.9984375
x2 = -8.863 / 1.28 = -6.91015625
Step 4: Convert x to the corresponding year
We know that x represents the number of years after 1997. Since x cannot be negative in this case, we discard the negative value (-6.91015625).
To find the corresponding year, we add x to 1997:
Year = 1997 + x1
Year = 1997 + 6.9984375
Year ≈ 2004.998
Therefore, the population will reach 34,052 fish in approximately the year 2005 (option d).