- #1

LSDwhat?

- 13

- 0

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));

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- Thread starter LSDwhat?
- Start date

- #1

LSDwhat?

- 13

- 0

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));

- #2

Hurkyl

Staff Emeritus

Science Advisor

Gold Member

- 14,967

- 19

Figure out the math first,I have no Idea how to make this algorithm , could you help ?

- #3

LSDwhat?

- 13

- 0

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

I dont know the math that's why I'm asking. I dont want the java code.

- #4

Hurkyl

Staff Emeritus

Science Advisor

Gold Member

- 14,967

- 19

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

- #5

LSDwhat?

- 13

- 0

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.

- #6

crd

- 14

- 0

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

- #7

adriank

- 534

- 1

(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) = ax

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