contestada

Complete the square to rewrite y = x^2 + 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)

Respuesta :

firuzer11
First of all, we convert the equation to a vertex form. 
[tex]y = x^{2}+6x+3 [/tex]
[tex]y=(x^{2}+6x)+3 [/tex]
[tex]y = (x+3)^{2}+3-9 [/tex]
[tex]y = (x+3)^{2}-6 [/tex].
The equation in vertex form is [tex](x+3)^{2}-6[/tex]. By looking at the equation we can understand that the y value of the vertex point is -6. By further calculation we can verify that the x value of the vertex point is -3. 
Now, there are two ways of finding if the vertex point is a maximum or a minimum. The first way is picking random x coordinates after and before the y vertex value. If the values of y decrease before the y vertex point and increase afterwards, the vertex is a minimum. If the values of y increase before the y vertex point and then decrease afterwards, the vertex is a maximum point. The second way is applying teh second derivative. Here, I will show you the way of doing it:
[tex]y = x^{2} + 6x +3[/tex]
[tex] \frac{dy}{dx} = 2x + 6[/tex]
[tex] \frac{d^{2}y}{dx^2} = 2 [/tex]
The result proves two facts about the function:
1. The function has only one vertex/turning point.
2. The vertex/turning point is a minimum (due to the result of the second derivative being positive).

The answer is C.

Answer:

vertex is (-3,-6)

Step-by-step explanation:

Complete the square to rewrite [tex]y = x^2 + 6x + 3[/tex] in vertex form.

In completing the square method, take the coefficient of middle term x

divide it by 2 and square it

6 divide by 2 is 3, [tex]3^2=9[/tex]

Add and subtract 9

[tex]y = x^2 + 6x +9-9+ 3[/tex]

[tex]y = x^2 + 6x +9-6[/tex]

Now we factor x^2+6x+9

[tex]y = (x+3)^2-6[/tex]

The equation is in the form of [tex]y=a(x-h)^2+k[/tex], where (h,k) is the vertex

In our equation, h=-3 and k=-6

So vertex is (-3,-6)