Maximum and Minimum Values of the Quadratic Expression

We will learn how to find the maximum and minimum values of the quadratic Expression ax^2 + bx + c (a ≠ 0).

When we find the maximum value and the minimum value of ax^2 + bx + c then let us assume y = ax^2 + bx + c.

Or, ax^2 + bx + c - y = 0

Suppose x is real then the discriminant of equation ax^2 + bx + c - y = 0 is ≥ 0

i.e., b^2 - 4a(c - y) ≥ 0

Or, b^2 - 4ac + 4ay ≥ 0

4ay ≥ 4ac - b^2


Case I: When a > 0 

When a > 0 then from 4ay ≥ 4ac - b^2 we get, y ≥ 4ac - b^2/4a

Therefore, we clearly see that the expression y becomes minimum when a > 0

Thus, the minimum value of the expression is 4ac - b^2/4a.

Now, substitute y = 4ac - b^2/4a in equation ax^2 + bx + c - y = 0 we have,

ax^2 + bx + c - (4ac - b^2/4a) = 0

or, 4a^2x^2 + 4abx + b^2 = 0

or, (2ax + b)^2 = 0

or, x = -b/2a

Therefore, we clearly see that the expression y gives its minimum value at x = -b/2a


Case II: When a < 0

When a < 0 then from 4ay ≥ 4ac - b^2 we get,

y ≤ 4ac - b^2/4a

Therefore, we clearly see that the expression y becomes maximum when a < 0.

Thus, the maximum value of the expression is 4ac - b^2/4a.

Now substitute y = 4ac - b^2/4a in equation ax^2 + bx + c - y = 0 we have,

ax^2 + bx + c -(4ac - b^2/4a) =0

or, 4a^2x^2 + 4abx + b^2 = 0

or, (2ax + b)^2 = 0

or, x = -b/2a.

Therefore, we clearly see that the expression y gives its maximum value at x = -b/2a.

 

Solved examples to find the maximum and minimum values of the quadratic Expression ax^2 + bx + c (a ≠ 0):

1. Find the values of x where the quadratic expression 2x^2 - 3x + 5 (x ϵ R) reaches a minimum value. Also find the minimum value.

Solution:

Let us assume y = 2x^2 - 3x + 5

Or, y = 2(x^2 - 3/2x) + 5

Or, y = 2(x^2 -2 * x * ¾ + 9/16 - 9/16) + 5

Or, y = 2(x - ¾)^2 - 9/8 + 5

Or, y = 2(x - ¾)^2 + 31/8

Hence, (x - ¾)^2 ≥ 0, [Since x ϵ R]

Again, from y = 2(x - ¾)^2 + 31/8 we can clearly see that y ≥ 31/8 and y = 31/8 when (x - ¾)^2 = 0 or, x = ¾

Therefore, when x is ¾ then the expression 2x^2 - 3x + 5 reaches the minimum value and the minimum value is 31/8.


2. Find the value of a when the value of 8a - a^2 - 15 is maximum.

Solution:

Let us assume y = 8a - a^2 -15

Or, y = - 15 - (a^2 - 8a)

Or, y = -15 - (a^2 - 2 * a * 4 + 4^2 - 4^2)

Or, y = -15 - (a - 4)^2 + 16

Or, y = 1 - (a - 4)^2

Hence, we can clearly see that (a - 4)^2 ≥ 0, [Since a is real]

Therefore, from y = 1 - (a - 4)^2 we can clearly see that y ≤ 1 and y = 1 when (a - 4)^2 = 0 or, a = 4.

Therefore, when a is 4 then the expression 8a - a^2 - 15 reaches the maximum value and the maximum value is 1.




11 and 12 Grade Math 

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