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

Answer:

C. Minimum at (–3, –6)

Step-by-step 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:

$b=6$

$(\frac{b}{2} )^{2}=(\frac{6}{2} )^{2}=9$

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.