Question

Complete the square to rewrite y = x2 + 6x + 3 in vertex form. Then state whether the vertex is a maximum or a minimum and give its coordinates.

A. Maximum at (3, –6)
B. Minimum at (3, –6)
C. Minimum at (–3, –6)
D. Maximum at (–3, –6)

Answer 1

C. Minimum at (–3, –6)

Explanation:

The given expression is:

`y=x^(2)+6x+3`

So, to complete the square, we first have to find the squared quotient between the second-term coefficient and 2, the add and subtract this number at the same time in the expression, like this:

`b=6`

`((b)/(2) )^(2)=((6)/(2) )^(2)=9`

Then,

`y=x^(2)+6x+3+9-9`

Now, we just have to group the terms that can be factorized:

`y=(x^(2)+6x+9)+3-9`

Then, the factorization would be made using the squared root of
`x^(2)` and
`9`, which is:

`y=(x+3)^(2) +3-9`

At the end, we just operate independent number outside the factorization:

`y=(x+3)^(2) -6`

So, according to the complete square of the expression, we can see that the vertex has coordinates
`(-3;-6)`, which is a minimum, because the squared coefficient is positive, that means the parabola is concave up.

The reason why the vertex has that coordinates is because the explicit expression of a parabola is:

`y=(x-h)^(2) +k`

Where
`(h;k)` is vertex coordinates.

Therefore, the answer is C.

Answer 2

y = x² + 6x + 3

y = x² + 6x + 9 - 6

y = (x + 3)² - 6

since the coefficient of x² is positive, then min at (-3,-6). the answer is C.