MAXIMUM AND MINIMUM VALUES OF QUADRATIC FUNCTIONS WORKSHEET

Problem 1 :

Find the minimum or maximum value of the quadratic function given below. 

f(x)  =  2x² + 7x + 5

Problem 2 :

Find the minimum or maximum value of the quadratic function given below. 

f(x)  =  -2x² + 6x + 12

Problem 3 :

Find the minimum or maximum value of the quadratic function given below. 

f(x)  =  -5x² + 30x + 200

Problem 4 :

Find the minimum or maximum value of the quadratic function given below. 

f(x)  =  3x² + 4x + 3

Solutions

Problem 1 :

Find the minimum or maximum value of the quadratic function given below. 

f(x)  =  2x² + 7x + 5

Solution :

Because the coefficient of x² is positive, the parabola is open upward.

So, the function will have only the minimum value and the minimum value is y-coordinate of the vertex.  

To find the y-coordinate of the vertex, first we have to find the x-coordinate of the vertex. 

Formula to find x-coordinate of the vertex is

=  -b/2a

Substitute a  =  2 and b  =  7. 

=  -7 / 2(2)

=  -7/4

To find the y-coordinate of the vertex, substitute -7/4 for x in the given function.

y-coordinate is

=  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

So, the minimum value is -9/8.

Problem 2 :

Find the minimum or maximum value of the quadratic function given below. 

f(x)  =  -2x² + 6x + 12

Solution :

Because the coefficient of x² is negative, the parabola is open downward.

So, the function will have only the maximum value and the maximum value is y-coordinate of the vertex.  

To find the y-coordinate of the vertex, first we have to find the x-coordinate of the vertex. 

Formula to find x-coordinate of the vertex is

=  -b/2a

Substitute a  =  -2 and b  =  6. 

=  -6 / 2(-2)

=  -6 / (-4)

=  3/2

To find the y-coordinate of the vertex, substitute 3/2 for x in the given function.

y-coordinate is

=  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

So, the maximum value is 33/2.

Problem 3 :

Find the minimum or maximum value of the quadratic function given below. 

f(x)  =  -5x² + 30x + 200

Solution :

Because the coefficient of x² is negative, the parabola is open downward.

So, the function will have only the maximum value and the maximum value is y-coordinate of the vertex.  

To find the y-coordinate of the vertex, first we have to find the x-coordinate of the vertex. 

Formula to find x-coordinate of the vertex is

=  -b/2a

Substitute a  =  -5 and b  =  30. 

=  -30 / 2(-5)

=  -30 / (-10)

=  3

To find the y-coordinate of the vertex, substitute 3 for x in the given function.

y-coordinate is

=  f(3)

=  -5(3)² + 30(3) + 200

  =  -5(9) + 90 + 200

=  -45 + 290

  =  245

So, the maximum value is 245.

Problem 4 :

Find the minimum or maximum value of the quadratic function given below. 

f(x)  =  3x² + 4x + 3

Solution :

Because the coefficient of x² is positive, the parabola is open upward.

So, the function will have only the minimum value and the minimum value is y-coordinate of the vertex.  

To find the y-coordinate of the vertex, first we have to find the x-coordinate of the vertex. 

Formula to find x-coordinate of the vertex is

=  -b/2a

Substitute a  =  3 and b  =  4. 

=  -4 / 2(3)

=  -2/3

To find the y-coordinate of the vertex, substitute -2/3 for x in the given function.

y-coordinate is

=  f(-2/3)

=  3(-2/3)² + 4(-2/3) + 3

  =  3(4/9) - 8/3 + 3

  =  4/3 - 8/3 + 3

  =  (4 - 8)/3 + 3  

=  -4/3 + 3

=  -4/3 + 9/3

  =  (-4 + 9) / 8

=  5/8

So, the minimum value is 5/8.

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