Question

Find the vertex of the parabola whose equation is y = x2 + 8x + 12.

Answer 1

(-4, -4)

Explanation:

A quadratic equation
`y = ax^2+bx+c` has written in vertex form using the completing square method is:

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

where, (h, k) is the vertex

As per the statement:

`y = x^2 + 8x + 12`

Using the completing square method we have;

`y = x^2 + 8x + 12+((8)/(2) )^2-((8)/(2) )^2`

⇒
`y =x^2+8x+12+4^2-4^2`

Using identity rule:

`(a+b)^2 = a^2+2ab+b^2`

then;

`y=(x+4)^2+12-16`

⇒
`y =(x+4)^2-4`

vertex of the given equation is: (h, k) = (-4, -4)

Therefore, the vertex of the parabola whose equation
`y = x^2 + 8x + 12` is (-4, -4)

Answer 2

We will transform the equation to the vertex form:
y = x² + 8 x + 12 = x² + 8 x + 16 - 16 + 12 =
= ( x + 4 )² - 4
Vertex form is: y = a ( x - k )² + h
Vertex coordinates are: ( - 4, - 4 ).