Given a quadratic function ax2 + bx + c. Find the maximum and minimum value of the function possible when x is varied for all real values possible.
Input: a = 1, b = -4, c = 4 Output: Maxvalue = Infinity Minvalue = 0 Quadratic function given is x2 -4x + 4 At x = 2, value of the function is equal to zero. Input: a = -1, b = 3, c = -2 Output: Maxvalue = 0.25 Minvalue = -Infinity
Q(x)=ax2 + bx + c. =a(x + b/(2a))2 + c-b2/(4a). first part second part
The function is broken into two parts.
The first part is a perfect square function. There can be two cases:
- Case 1: If value of a is positive.
- The maximum value would be equal to Infinity.
- The minimum value of the function will come when the first part is equal to zero because the minimum value of a square function is zero.
- Case 2: If value of a is negative.
- The minimum value would be equal to -Infinity.
- Since a is negative, the task to maximize the negative square function. Again maximum value of a negative square function would be equal to zero as it would be a negative value for any other value of x.
The second part is a constant value for a given quadratic function and hence cannot change for any value of x. Hence, it will be added in both cases. Hence, the answer to the problem is:
If a > 0, Maxvalue = Infinity Minvalue = c - b2 / (4a) If a < 0, Maxvalue = c - b2 / (4a) Minvalue = -Infinity
Below is the implementation of the above approach:
Maxvalue = 0.25 Minvalue = -Infinity
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