To find the vertex of a quadratic function of the form [tex]\( y = ax^2 + bx + c \)[/tex], we can follow these steps:
1. Identify the coefficients [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] from the given quadratic function [tex]\( y = 3x^2 + 6x + 1 \)[/tex]. Here, [tex]\( a = 3 \)[/tex], [tex]\( b = 6 \)[/tex], and [tex]\( c = 1 \)[/tex].
2. Use the formula for the x-coordinate of the vertex:
[tex]\[
x = -\frac{b}{2a}
\][/tex]
Substitute the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
[tex]\[
x = -\frac{6}{2 \cdot 3} = -\frac{6}{6} = -1
\][/tex]
3. Next, substitute the x-coordinate back into the original quadratic function to find the y-coordinate of the vertex:
[tex]\[
y = 3(-1)^2 + 6(-1) + 1
\][/tex]
Calculate each term:
[tex]\[
y = 3(1) + 6(-1) + 1 \\
y = 3 - 6 + 1 \\
y = -2
\][/tex]
Therefore, the vertex of the function is [tex]\((-1, -2)\)[/tex]. The correct answer is:
B. [tex]\((-1, -2)\)[/tex]
