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How To Find The Maximum Of A Quadratic

FIND THE MAXIMUM AND MINIMUM VALUE OF QUADRATIC FUNCTION

The graph of a quadratic function volition always be a parabola which is either open upward or downward.

General course of quadratic function is

f(10) = ax2 + bx + c

Maximum Value of a 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 office will exist a parabola which opens down.

The maximum value is 'y' coordinate at the vertex of the parabola.

Note :

At that place is no minimum value for the parabola which opens downward.

Minimum Value of a Quadratic Function

The quadratic function f(10) = axtwo + bx + c will accept 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 part volition exist 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 upward.

Solved Bug

Problem 1 :

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

f(10)   =  2xii + 7x + five

Solution :

Because the coefficient of ten2 is positive, the parabola is open up upwardly.

So, the function will accept 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 notice the x-coordinate of the vertex.

Formula to find x-coordinate of the vertex is

=  -b/2a

Substitute a  =  2 and b  =  seven.

=  -seven / 2(2)

=  -7/iv

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

y-coordinate is

=  f(-7/4)

=  two(-seven/4)2  + 7(-7/4) + five

  =  2(49/16) - (49/4) + 5

  =  (49/eight) - (49/4) + 5

  =  (49 - 98 + 40)/8

  =  -9/8

The minimum value is -nine/8.

Problem two :

Find the minimum or maximum value of the quadratic office 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 merely the maximum value and the maximum value is y-coordinate of the vertex.

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

Formula to detect x-coordinate of the vertex is

=  -b/2a

Substitute a = -2 and b = 6.

=  -vi/ii(-2)

=  -6/(-4)

=  3/2

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

y-coordinate is

=  f(3/2)

= -2(3/ii)2 + half dozen(three/2) + 12

  =  -2(9/4) + iii(iii) + 12

=  -9/2 + nine + 12

  =  -9/2 + 21

=  (-ix + 42)/2

=  33/ii

The maximum value is 33/2.

Trouble 3 :

Notice the minimum or maximum value of the quadratic part given below.

f(10)  = -5x2 + 30x + 200

Solution :

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

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

To observe the y-coordinate of the vertex, starting time we have to find the x-coordinate of the vertex.

Formula to find x-coordinate of the vertex is

=  -b/2a

Substitute a = -five and b = 30.

=  -30/2(-5)

=  -thirty/(-10)

=  3

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

y-coordinate is

=  f(3)

= -5(3)2 + 30(iii) + 200

  =  -5(9) + 90 + 200

=  -45 + 290

  =  245

The maximum value is 245.

Problem iv :

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

f(ten)  = 3x2 + 4x + 3

Solution :

Because the coefficient of xii is positive, the parabola is open up upwards.

So, the role 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 notice the x-coordinate of the vertex.

Formula to discover x-coordinate of the vertex is

=  -b/2a

Substitute a = 3 and b = four.

=  -4/2(3)

=  -2/3

To find the y-coordinate of the vertex, substitute -2/3 for 10 in the given function.

y-coordinate is

=  f(-2/three)

= 3(-ii/3) 2  + 4(-2/3) + 3

  =  3(4/9) - 8/3 + 3

  =  four/3 - 8/3 + 3

  =  (four - 8)/3 + three

=  -iv/3 + 3

=  -iv/3 + 9/iii

  =  (-4 + nine)/3

=  5/3

The minimum value is 5/3.

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