Answer:
f(x) = -(x+2)(x-1)
Step-by-step explanation:
We know the zeros are at 1 and -2
f(x) = a *(x-b1)(x-b2) where b1 and b2 are the zeros and a is a constant
f(x) =a(x-1)(x- -2)
f(x) =a(x-1)(x+2)
We know that a must be negative since the parabola opens down so
The only choice is choice D where a = -1
f(x) = -(x+2)(x-1)
Answer:
Step-by-step explanation:
(A) The difference between an ordinary differential equation and an initial value problem is that an initial value problem is a differential equation which has condition(s) for optimization, such as a given value of the function at some point in the domain.
(B) The difference between a particular solution and a general solution to an equation is that a particular solution is any specific figure that can satisfy the equation while a general solution is a statement that comprises all particular solutions of the equation.
(C) Example of a second order linear ODE:
M(t)Y"(t) + N(t)Y'(t) + O(t)Y(t) = K(t)
The equation will be homogeneous if K(t)=0 and heterogeneous if 
Example of a second order nonlinear ODE:

(D) Example of a nonlinear fourth order ODE:
![K^4(x) - \beta f [x, k(x)] = 0](https://tex.z-dn.net/?f=K%5E4%28x%29%20-%20%5Cbeta%20f%20%5Bx%2C%20k%28x%29%5D%20%3D%200)
I'm not sure how to display a graph on here, so I'm afraid I can't help you with that part.
However, the answer is C. (-2,3).
To create the graph, either put it into a graphing calculator or manually graph the equations by plugging in different numbers to the equation.
Answer:
the answer should be D
Step-by-step explanation:
Answer the answer is 25
Explanation: