How can I find the maximum and minimum of a Parabola polynom using Java?

[LSDwhat?]

I need to make to functions in java that gives the maxim and minin of the Parabola polynom ax2+bx+c for an interval of two given points.

I have no Idea how to make this algorithm , could you help ?

I have come to something like this :

if (-b/(2*a)>=x1 && -b/(2*a)<=x2)
return (-b/(2*a));

 

[Hurkyl]

I have no Idea how to make this algorithm , could you help ?

Figure out the math first, then worry about how to program it.

 

[LSDwhat?]

Figure out the math first, then worry about how to program it.

I don't know the math that's why I'm asking. I don't want the java code.

 

[Hurkyl]

Well, what do you know about finding minima and maxima?

Alternatively, what do you know about the shape of the graphs of parabolas?

 

[LSDwhat?]

Well, what do you know about finding minima and maxima?

Alternatively, what do you know about the shape of the graphs of parabolas?

well the maxima should be the value of Y which is the bigger to a value of X and the minima the same.

About the shape its sinusoidal waves.

 

[crd]

well the maxima should be the value of Y which is the bigger to a value of X and the minima the same.

About the shape its sinusoidal waves.

Careful, you are mixing apples and oranges. The max(min) will be the value of Y which is bigger(smaller) than every other value of Y for some region around your max(min) value.

Not X like you said, X is the input variable that determines your Y.

You should try graphing ax^2 + bx + c for various values of a, b, and c to verify if it has "sinusoidal waves."

that also might give you some intuition into the the max(min) of a parabola

 

[adriank]

Theorem: If E ⊂ R and f: E → R, and f has a maximum or minimum at x ∈ E, then one of the following three is true:
(1) x is a boundary point of E,
(2) f'(x) = 0, or
(3) f is not differentiable at x.

In your case, f(x) = ax² + bx + c and E is the interval [x₁, x₂]. Then the only possibilities are these: (1) x is one of the boundary points x₁ or x₂ of E, or (2) f'(x) = 2ax + b = 0, so x = -b/2a. Look at the values of f at those three points; the largest one is the maximum, and the smallest one is the minimum.