9514 1404 393
Answer:
y = -3x^2 -9x +30
Step-by-step explanation:
When a polynomial has a zero at x=p, it has a factor of (x -p). The factors of your quadratic are ...
f(x) = (x +5)(x -2)
At the point x=3, the value of this product is ...
f(3) = (3 +5)(3 -2) = (8)(1) = 8
In order for that value to be -24, it needs to be multiplied by a scale factor of -3. The quadratic you want is ...
y = -3(x +5)(x -2)
y = -3x^2 -9x +30 . . . . . standard form
Answer:
N(1)=50 is a minimum
N(15)=4391.7 is a maximum
Step-by-step explanation:
<u>Extrema values of functions </u>
If the first and second derivative of a function f exists, then f'(a)=0 will produce values for a called critical points. If a is a critical point and f''(a) is negative, then x=a is a local maximum, if f''(a) is positive, then x=a is a local minimum.
We are given a function (corrected)
(a)
First, we take its derivative
Solve N'(t)=0
Simplifying
Solving for t
Only t=1 belongs to the valid interval
Taking the second derivative
Which is always positive, so t=1 is a minimum
(b)
N(1)=50 is a minimum
(c) Since no local maximum can be found, we test for the endpoints. t=1 was already determined as a minimum, we take t=15
(d)
N(15)=4391.7 is a maximum