This is application question.

Bacterial Control. If t days after treatment the bacteria count per cubic centimeter in a body of water is given by C(t) = 10t^2 - 120x + 800
0 <= t <= 9
Thn in how many days will the jcount be a minimum?

I got this result:

20(t^2 - 6 + 40)
20(t - 5)(t - 8)
t - 5 = 0 t - 8 = 0
t = 5 t = 8
f(5) = 700
f(8) = 1120
f(0) = 800
f(9) = 1340

20(3)^2-120(3)+800
500-600+800 = 700

20(8)^2-120(8)+800
1280-960+800 = 1120

20(0)^2-120(0)+800 = 800

20(9)^2-120(9)+800 =1340

minimum = 700

but this is incorrect. What am I doing wrong and how do I solve it?

First off, t^2 - 6t + 40 does not factor into (t-5)(t-8) That would be t^2 - 13t + 40.

A parabola ax^2 + bx + c reaches its minimum where x = -b/2a = 120/20 = 6

C(6) = 10*36 - 120*6 + 800 = 440

Thank you Steve. But i'm still a little confused. The answer in my answer list shows minimum as 620 at t=3.?

In that case, I'm also confused. If

C(t) = 10t^2 - 120x + 800

C(3) = 10*9 - 120*3 + 800
= 90 - 360 + 800
= 530

Is the function correct?

To find the minimum value of the bacteria count, we need to find the minimum point of the quadratic function C(t) = 10t^2 - 120t + 800.

First, let's rewrite the function in a standard quadratic form: C(t) = 10t^2 - 120t + 800 = 10(t^2 - 12t + 80).

To find the minimum point of the quadratic function, we can use the vertex formula. The vertex formula states that the x-coordinate of the vertex of a quadratic function in the form f(x) = ax^2 + bx + c is given by x = -b/(2a).

In our case, a = 10 and b = -120. Plugging these values into the formula, we get t = -(-120)/(2*10) = 6.

Therefore, the bacteria count will be at a minimum at t = 6 days.

To verify this, we can substitute t = 6 into the function C(t) = 10t^2 - 120t + 800:

C(6) = 10(6^2) - 120(6) + 800 = 10(36) - 720 + 800 = 360 - 720 + 800 = 1440 - 720 = 720.

Hence, the correct minimum value of the bacteria count is 720.