FIND THE MAXIMUM AND MINIMUM VALUE OF QUADRATIC FUNCTION
About "Find the maximum and minimum value of quadratic function"
Find the maximum and minimum value of quadratic function :
The graph of the quadratic equation will always be a parabola which is open upward or downward.
General form of quadratic equation is
ax2 + bx + c
Maximum value of quadratic function
The quadratic function f(x) = ax² + bx + c will have only the maximum value when the the leading coefficient or the sign of "a" is negative.
When "a" is negative the graph of the quadratic function will be a parabola which opens down.
The maximum value is "y" coordinate at the vertex of the parabola.
Note :
There is no minimum value for the parabola which opens down.
Minimum value of quadratic function
The quadratic function f(x) = ax² + bx + c will have only the minimum value when the the leading coefficient or the sign of "a" is positive.
When "a" is positive, the graph of the quadratic function will be a parabola which opens up.
The minimum value is "y" coordinate at the vertex of the parabola.
Note :
There is no maximum value for the parabola which opens up.
Find the maximum and minimum value of quadratic function - Examples
Example 1 :
Find the minimum or maximum value of the quadratic equation given below.
f(x) = 2x² + 7x + 5
Solution :
Since the coefficient of x2 is positive, the parabola is open upward.
So, the function will have only the minimum value.
x-coordinate of minimum value = -b/2a
y-coordinate of minimum value = f(-b/2a)
a = 2 b = 7 and c = 5
x-coordinate = -7/2(2) ==> -7/4
y-coordinate = f(-7/4)
f(x) = 2x² + 7x + 5
f(-7/4) = 2(-7/4)² + 7(-7/4) + 5
= 2(49/16) - (49/4) + 5
= (49/8) - (49/4) + 5
= (49 - 98 + 40) / 8
= -9/8
Hence the minimum value is -9/8.
Example 2 :
Find the minimum or maximum value of the quadratic equation given below.
f(x) = -2x² + 6x + 12
Solution :
Since the coefficient of x2 is negative, the parabola is open downward.
So, the function will have only the maximum value.
x-coordinate of maximum value = -b/2a
y-coordinate of maximum value = f(-b/2a)
a = -2 b = 6 and c = 12
x-coordinate = -6/2(-2) ==> 6/4 ==> 3/2
y-coordinate = f(3/2)
f(x) = -2x² + 6x + 12
f(3/2) = -2(3/2)² + 6(3/2) + 12
= -2(9/4) + 3(3) + 12
= -9/2 + 9 + 12
= -9/2 + 21
= (-9 + 42)/2
= 33/2
Hence the maximum value is 33/2.
Example 3 :
Find the minimum or maximum value of the quadratic equation given below.
f(x) = -5x² + 30x + 200
Solution :
Since the coefficient of x2 is negative, the parabola is open downward.
So, the function will have only the maximum value.
x-coordinate of maximum value = -b/2a
y-coordinate of maximum value = f(-b/2a)
a = -5, b = 30 and c = 200
x-coordinate = -30/2(-5) ==> 30/10 ==> 3
y-coordinate = f(3)
f(x) = -5x² + 30x + 200
f(3) = -5(3)² + 30(3) + 200
= -5(9) + 90 + 200
= -45 + 290
= 245
Hence the maximum value is 245.
Example 3 :
Find the minimum or maximum value of the quadratic equation given below.
f(x) = 3x² + 4x + 3
Solution :
Since the coefficient of x2 is positive, the parabola is open upward.
So, the function will have only the minimum value.
x-coordinate of minimum value = -b/2a
y-coordinate of minimum value = f(-b/2a)
a = 3, b = 4 and c = 3
x-coordinate = -4/2(3) ==> -4/6 ==> -2/3
y-coordinate = f(-2/3)
f(x) = -3x² + 4x + 3
f(-2/3) = 3(-2/3)² + 4(-2/3) + 3
= 3(4/9) - (8/3) + 3
= 12/9 - 8/3 + 3
= 4/3 - 8/3 + 3
= (4 - 8 + 9)/3
= (13 - 8)/3
= 5/3
Hence the minimum value is 5/3.
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