The population of a town P(t)is modelled by the function P(t)=6t+110t+3000, where t is time in years. Note t=0 represents the years 2000.
Suspect you mean
P(t) = 6 t^2 + 110 t + 3000
oh i'm sorry. i left out that part.. the question is
When will the population reach 6000?
P(t) = 6 t^2 + 110 t + 3000
oh, hang on
The question doesn't state wheater it's a max or min
it'as grade 11 math
no I have not done calculus
oh ok
6000 = 6 t^2 + 110 t + 3000
6 t^2 + 110 t = 3000
t^2 + 18.3 t - 500 = 0
t = [ -18.3 +/-sqrt(336+2000)]/2
t = [-18.3 + sqrt(2336)]2
t = [-18.3 + 48.3 ]/2
t = 15
so 2015
Thank you
Much Appreciated !! (Y):)
Honestly, although i appreciate this help
Is there any other way to do this problem. I think this is a little to advanced because i really don't understand it.
You can go Damon's way, or the graphing way/ comnpleting the square binomial way or you can go quadratic
6 t^2 + 110 t + 3000
if you're starting to study quadratics, you have surely used the quadratic formula to find the roots.
As Damon showed,
t = [-18.3 + 48.3 ]/2
or
t = -9.15 ± √24.14
You know the parabola is symmetric, and the vertex is midway between the roots. Note that the roots are equally spaced around t = -9.15
so, the vertex is at t = -9.15. That will be the min or max.
If none of this makes any sense, you just need to study the quadratic formula some more. There are lots of examples online, and surely some in your class materials. If all that still makes no sense, no written explanation will be any clearer; some in-person help will help clear things up.
Also, a visit to
http://rechneronline.de/function-graphs
will let you play around with various formulas and see what happens when you change stuff.
where is it max and min or something?
If you do not know calculus, find vertex of parabola by completing the square
t^2 + 18.33 t + 500 = P/6
t^2 + 18.33 = P/6 - 500
t^2 + 18.33 + 84 = P/6 -500 +84
(t+9.17)^2 = P/6 - 416
(t+9.17)^2 = (1/6)(p-2496)
minimum at t = -9.17 and p = 2496
that is the year 2000 - 9.17 = 1991