# MAXIMUM AND MINIMUM VALUES OF QUADRATIC FUNCTIONS WORKSHEETS

Maximum and minimum values of quadratic functions worksheets :

Here we are going to see some practice questions on finding maximum and minimum values of quadratic functions.

## Maximum and minimum values of quadratic functions worksheets - Practice questions

(1)  Find the minimum or maximum value of the quadratic equation given below.

f(x)  =  2x² + 7x + 5

(2)  Find the minimum or maximum value of the quadratic equation given below.

f(x)  =  -2x² + 6x + 12

(3)  Find the minimum or maximum value of the quadratic equation given below.

f(x)  =  -5x² + 30x + 200

(4)  Find the minimum or maximum value of the quadratic equation given below.

f(x)  =  3x² + 4x + 3

## Find the maximum and minimum value of quadratic function - Solution

Question 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.

Question 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.

Question 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.

Question 4 :

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 245.

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