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.
Examples:
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
Approach:
Q(x)=ax2 + bx + c.
=a(x + b/(2a))2 + c-b2/(4a).
first part second partThe 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 = -InfinityBelow is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void PrintMaxMinValue(double a, double b, double c)
{
double secondPart = c * 1.0 - (b * b / (4.0 * a));
if (a > 0) {
cout << "Maxvalue = "
<< "Infinity\n";
cout << "Minvalue = " << secondPart;
}
else if (a < 0) {
cout << "Maxvalue = " << secondPart << "\n";
cout << "Minvalue = "
<< "-Infinity";
}
else {
cout << "Not a quadratic function\n";
}
}
int main()
{
double a = -1, b = 3, c = -2;
PrintMaxMinValue(a, b, c);
return 0;
}
|
Java
import java.util.*;
class GFG
{
static void PrintMaxMinValue(double a, double b, double c)
{
double secondPart = c * 1.0 - (b * b / (4.0 * a));
if (a > 0)
{
System.out.print("Maxvalue = "
+ "Infinity\n");
System.out.print("Minvalue = " + secondPart);
}
else if (a < 0)
{
System.out.print("Maxvalue = " + secondPart + "\n");
System.out.print("Minvalue = "
+ "-Infinity");
}
else
{
System.out.print("Not a quadratic function\n");
}
}
public static void main(String[] args)
{
double a = -1, b = 3, c = -2;
PrintMaxMinValue(a, b, c);
}
}
|
Python3
def PrintMaxMinValue(a, b, c) :
secondPart = c * 1.0 - (b * b / (4.0 * a));
if (a > 0) :
print("Maxvalue =", "Infinity");
print("Minvalue = ", secondPart);
elif (a < 0) :
print("Maxvalue = ", secondPart);
print("Minvalue =", "-Infinity");
else :
print("Not a quadratic function");
if __name__ == "__main__" :
a = -1; b = 3; c = -2;
PrintMaxMinValue(a, b, c);
|
C#
using System;
class GFG
{
static void PrintMaxMinValue(double a, double b, double c)
{
double secondPart = c * 1.0 - (b * b / (4.0 * a));
if (a > 0)
{
Console.Write("Maxvalue = "
+ "Infinity\n");
Console.Write("Minvalue = " + secondPart);
}
else if (a < 0)
{
Console.Write("Maxvalue = " + secondPart + "\n");
Console.Write("Minvalue = "
+ "-Infinity");
}
else
{
Console.Write("Not a quadratic function\n");
}
}
static public void Main ()
{
double a = -1, b = 3, c = -2;
PrintMaxMinValue(a, b, c);
}
}
|
Javascript
<script>
function PrintMaxMinValue(a, b, c)
{
var secondPart = c * 1.0 -
(b * b / (4.0 * a));
if (a > 0)
{
document.write("Maxvalue = " +
"Infinity" + "<br>");
document.write("Minvalue = " +
secondPart);
}
else if (a < 0)
{
document.write("Maxvalue = " +
secondPart + "<br>");
document.write("Minvalue = " +
"-Infinity");
}
else
{
document.write("Not a quadratic function\n");
}
}
var a = -1, b = 3, c = -2;
PrintMaxMinValue(a, b, c);
</script>
|
Output: Maxvalue = 0.25
Minvalue = -Infinity
Time Complexity: O(1)
Auxiliary Space: O(1)