Question 919253
In general, the min of {{{y = ax^2 + bx + c}}} is when {{{x = -b/(2a)}}}. This is assuming {{{a > 0}}}. If {{{a < 0}}}, we'll have a max and not a min.


{{{f(x)=0.8x^2-12.8x+54.2}}} is in that form where {{{a = 0.8}}} and {{{b = -12.8}}}. We don't need c in this case.


Plug the values in:


{{{x = -b/(2a)}}}


{{{x = -(-12.8)/(2*0.8)}}}


{{{x = 8}}}


Recall that x = 0 represents 1989. So x = 1 represents 1989+1 = 1990, etc etc


x = 8 means 1989 + 8 = 1997


So 8 years after the starting point of 1989, ie in the year 1997, the population is at the minimum.


If you want to know this min population value, plug it back into f(x) and evaluate.


Let me know if you need more help or if you need me to explain a step in more detail.
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Thanks,


Jim