# 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.

Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.

## Recent Articles 1. ### Cardinal Numbers and Ordinal Numbers | Cardinal Numbers | Ordinal Num

Dec 07, 23 01:27 AM

Cardinal numbers and ordinal numbers are explained here with the help of colorful pictures. There are many steps in a staircase as shown in the above figure. The given staircase has nine steps,

2. ### Smallest and Greatest Number upto 10 | Greater than | Less than | Math

Dec 06, 23 11:21 PM

We will discuss about the smallest and greatest number upto 10.