# 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 xis 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 xis 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 xis 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. After having gone through the stuff given above, we hope that the students would have understood "Find the maximum and minimum value of quadratic function".

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