Find the vertex of the parabola whose equation is [tex][tex]$y = x^2 + 8x + 12$[/tex][/tex].

A. [tex](-4, -4)[/tex]
B. [tex](0, -6)[/tex]
C. [tex](-4, 12)[/tex]



Answer :

To find the vertex of the parabola given by the equation [tex]\(y = x^2 + 8x + 12\)[/tex], follow these steps:

1. Identify the coefficients:
- The coefficient [tex]\(a\)[/tex] of [tex]\(x^2\)[/tex] is 1.
- The coefficient [tex]\(b\)[/tex] of [tex]\(x\)[/tex] is 8.
- The constant term [tex]\(c\)[/tex] is 12.

2. Determine the x-coordinate of the vertex ([tex]\(h\)[/tex]):
- The x-coordinate [tex]\(h\)[/tex] of the vertex of a parabola in the form [tex]\(y = ax^2 + bx + c\)[/tex] is given by the formula:
[tex]\[ h = -\frac{b}{2a} \][/tex]
- Substituting [tex]\(a = 1\)[/tex] and [tex]\(b = 8\)[/tex] into the formula gives:
[tex]\[ h = -\frac{8}{2 \cdot 1} = -\frac{8}{2} = -4 \][/tex]

3. Determine the y-coordinate of the vertex ([tex]\(k\)[/tex]):
- The y-coordinate [tex]\(k\)[/tex] is obtained by substituting [tex]\(h\)[/tex] back into the original equation:
[tex]\[ k = a(-4)^2 + b(-4) + c \][/tex]
- Substituting [tex]\(a = 1\)[/tex], [tex]\(b = 8\)[/tex], [tex]\(c = 12\)[/tex], and [tex]\(h = -4\)[/tex] into the equation:
[tex]\[ k = 1(-4)^2 + 8(-4) + 12 \][/tex]
- Calculate each term:
[tex]\[ k = 1(16) + (-32) + 12 \][/tex]
[tex]\[ k = 16 - 32 + 12 \][/tex]
[tex]\[ k = -4 \][/tex]

So, the vertex of the parabola [tex]\(y = x^2 + 8x + 12\)[/tex] is [tex]\((-4, -4)\)[/tex].

Therefore, the correct answer is [tex]\((-4, -4)\)[/tex].