FINDING MAXIMUM AND MINIMUM OF QUADRATIC FUNCTION

The graphical form of a quadratic function will be a parabola (u shpae). In which the maximum and minimum value will be there at vertex.

To find the maximum or minimum value from the quadratic equation, we have the following ways.

(i) Converting into the vertex form

(ii) Using formula

Find the maximum or minimum value of the following function.

Problem 1 :

y = x² - 4x + 1

Solution :

Method 1 :

y = x² - 4x + 1

Write the coefficient of x as a multiple of 2.

y = [x² - 2 ⋅ x ⋅ 2 + 2² - 2² + 1]

y = [(x - 2)² - 2² + 1]

y = [(x - 2)² - 4 + 1]

y = (x - 2)² - 3

It is exactly in the form of 

y = a(x - h)² + k

Here a = 1 > 0, then the parabola opens up.

Vertex is at (2, -3).

So,

minimum is at x = 2 and the minimum value = -3.

Method 2 :

y = x² - 4x + 1

a = 1, b = -4 and c = 1

x = -b/2a

x = -(-4)/2(1)

x = 4/2

x = 2

By applying the value of x in the given equation, we get

y = 2² - 4(2) + 1

y = 4 - 8 + 1

y = 5 - 8

y = -3

So, the minimum point is at (2, -3).

Problem 2 :

y = -x² - x + 1

Solution :

y = -x² - x + 1

Factoring negative

y = -[x² + x - 1]

Here a = -1 < 0, then the parabola opens down.

Vertex is at (-1/2, 5/4).

So,

minimum is at x = -1/2 and the minimum value = 5/4.

Problem 3 :

y = 5x² - 3

Solution :

y = 5x² - 3

y = 5(x - 0) ² - 3

Here a = 5 > 0, the parabola opens up.

Vertex (h, k) ==> (0, -3)

The minimum is at x = 0 and minimum value = -3.

Minimum point is (0, -3).

Problem 4 :

y = (1/2)x² - x - 4

Solution :

y = (1/2)x² - x - 4

Here a = 1/2 > 0, the parabola opens up.

Vertex (h, k) ==> (1, -4.5)

The minimum is at x = 1 and minimum value = -4.5

Minimum point is  (1, -4.5).