# Use a technique used to rewrite the quadratic function x^{2} + 6x + 8 from standard form to vertex form.

Quadratic equations are equations with a degree of two. They can be represented in the standard form ax^{2} + bx + c as well as in the vertex form a(x - h)^{2} + k = 0.

### Answer: The equation x^{2} + 6x + 8 in vertex form is represented as [(x + 3)^{2} - 1].

Let's understand the solution. The technique used is completing the square method.

**Explanation:**

To convert the given standard equation to vertex form, we follow the below steps:

⇒ First, we identify the coefficient of x, that is, 6.

⇒ Now, we half and square the coefficient of x, that is, (6 / 2)^{2} = 9.

⇒ After this, we add and subtract the above result from the equation.

⇒ Now, we write the equation as (x^{2} + 6x + 9 - 9 + 8).

⇒ Now, using the identity of ( a + b)^{2} = a^{2} + 2ab + b^{2}; we get (x^{2} + 6x + 9 - 9 + 8) = [(x + 3)^{2} - 1].

The above equation is the vertex form of the given equation. The vertex of the parabola representing the equation is the point (-3, -1).

### Hence, The equation x^{2} + 6x + 8 in vertex form is represented as [(x + 3)^{2} - 1].

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