Question

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

Answer 1

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 ).

Answer 2

Answer:

(-4, -4)

Step-by-step 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;

$y = x^2 + 8x + 12+(\frac{8}{2} )^2-(\frac{8}{2} )^2$

⇒$y =x^2+8x+12+4^2-4^2$

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)