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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