To find the vertex form of the parabola, we use the concept completing the square method.

**Vertex form of a quadratic function :**

y = a(x - h)^{2} + k

In order to find the maximum or minimum value of quadratic function, we have to convert the given quadratic equation in the above form.

**Minimum value of parabola :**

If the parabola is open upward, then it will have minimum value

If a > 0, then minimum value of f is f(h) = k

**Maximum value of parabola :**

If the parabola is open downward, then it will have maximum value.

If a < 0, then maximum value of f is f(h) = k

**Problem 1 :**

For the given function f(x) = x^{2} + 7x + 12

(a) Write f(x) in the form k(x + t)^{2} + r.

(b) Find the value of x where f(x) attains its minimum value or its maximum value.

(c) Find the vertex of the graph of f.

**Solution :**

Let y = x^{2} + 7x + 12

y = x^{2} + 2⋅x⋅(7/2) + (7/2)^{2} - (7/2)^{2} + 12

y = (x + (7/2))^{2} + 12

By comparing it with vertex form, we get the value of k . Since it is positive, the parabola is open upward. So it will minimum value.

(b) It has minimum value when x = -7/2

(c) Vertex of the parabola is (-7/2, 12)

**Problem 2 :**

For the given function f(x) = 5x^{2} + 2x + 1

(a) Write f(x) in the form k(x + t)^{2} + r.

(b) Find the value of x where f(x) attains its minimum value or its maximum value.

(c) Find the vertex of the graph of f.

**Solution :**

Let y = 5x^{2} + 2x + 1

y = 5(x^{2} + x) + 1

y = 5 [x^{2} + 2 ⋅ x ⋅ (1/2) + (1/2)^{2} - (1/2)^{2}] + 1

y = 5 [x + (1/2)]^{2} - (1/4)] + 1

y = 5 [x + (1/2)]^{2} - (5/4) + 1

y = 5 [x + (1/2)]^{2} - (1/4)

By comparing it with vertex form, we get the value of k . Since it is positive, the parabola is open upward. So it will minimum value.

(b) It has minimum value when x = -1/2

(c) Vertex of the parabola is (-1/2, -1/4)

**Problem 3 :**

For the given function f(x) = −2x^{2} + 5x − 2

(a) Write f(x) in the form k(x + t)^{2} + r.

(b) Find the value of x where f(x) attains its minimum value or its maximum value.

(c) Find the vertex of the graph of f.

**Solution :**

Let y = −2x^{2} + 5x − 2

y = −2[x^{2} - (5/2)x] − 2

y = −2[x^{2} - (5/2)x] − 2

y = -2 [x^{2} - 2 ⋅ x ⋅ (5/4) + (5/4)^{2} - (5/4)^{2}] - 2

y = -2 [x - (5/4)]^{2} - (25/16)] - 2

y = -2 [x - (5/4)]^{2} + (25/8) - 2

y = -2 [x - (5/4)]^{2} + (9/8)

By comparing it with vertex form, we get the value of k . Since it is negative, the parabola is open downward. So it will maximum value.

(b) It has maximum value when x = 5/4

(c) Vertex of the parabola is (5/4, 9/8)

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