The Political Noise Problem. The amount of background noise is important to television news reporters. One station developed the formula showing the noise level in decibels (N) as it relates to the time after the speaker stops talking in seconds (t). Use the equation below to compute how many seconds after the speaker stops will the noise level be the greatest? Write and tell how you decided
N = t^2 + 12t + 54
Something is not right about your equation.
The way it is, it would be a parabola opening upwards, and of course it would not have a maximum, but rather a minimum value of N
Are you sure the first term wasn't -t^2 ?
I assume you don't know calculus, so you will have to use the "completing the square" technique (or graphing) to find the t value for which N is greatest.
N = t^2 + 12t + 54
= t^2 + 12t + 36 + 18
= (t+6)^2 + 18
That is SMALLEST when t = -6. I assume you are only interested in t>0. There is no maximum. Are you sure you didn't mean to write
N = t^2 - 12t + 54 ?
Sorry, Yes the correct equation is
-t^2 + 12t + 54
To find the number of seconds after the speaker stops when the noise level is at its maximum, we need to determine the maximum value of the equation N = t^2 + 12t + 54.
To find the maximum value of a quadratic equation in the form of N = at^2 + bt + c (where a, b, and c are constants), we can use the vertex formula:
t = -b / (2a).
In this case, a = 1, b = 12, and c = 54.
Substituting these values into the formula, we have:
t = -12 / (2 * 1) = -12 / 2 = -6.
Since time cannot be negative in this context, we can conclude that the value of t that maximizes the noise level is t = -6.
However, this negative value doesn't make sense in the context of the problem. So, we can disregard it.
Now, let's consider a different approach to find the maximum value.
To determine the maximum value, we can complete the square to rewrite the equation in vertex form:
N = t^2 + 12t + 54
We complete the square by adding and subtracting the square of half the coefficient of t:
N = (t^2 + 12t + 36) + 54 - 36
N = (t + 6)^2 + 18
Now, we can see that N takes its minimum value of 18 when t = -6. Since the quadratic term (t + 6)^2 is always non-negative, the value of N will increase indefinitely as t approaches infinity.
Therefore, there is no maximum value for N, and the noise level will continue to increase as time progresses.