To find the vertex of the quadratic function [tex]\( y = 2x^2 + 4x + 1 \)[/tex], we follow these steps:
1. Identify the coefficients: The quadratic function is given in the form [tex]\( y = ax^2 + bx + c \)[/tex]. For our function:
- [tex]\( a = 2 \)[/tex]
- [tex]\( b = 4 \)[/tex]
- [tex]\( c = 1 \)[/tex]
2. Find the x-coordinate of the vertex: The x-coordinate of the vertex for a parabola [tex]\( y = ax^2 + bx + c \)[/tex] can be found using the formula:
[tex]\[
x = -\frac{b}{2a}
\][/tex]
Plugging in the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
[tex]\[
x = -\frac{4}{2 \cdot 2} = -\frac{4}{4} = -1
\][/tex]
3. Find the y-coordinate of the vertex: Substitute the x-coordinate back into the original equation [tex]\( y = 2x^2 + 4x + 1 \)[/tex]. This gives us:
[tex]\[
y = 2(-1)^2 + 4(-1) + 1
\][/tex]
Simplify each term:
[tex]\[
y = 2(1) + (-4) + 1 = 2 - 4 + 1 = -1
\][/tex]
Therefore, the vertex of the quadratic function [tex]\( y = 2x^2 + 4x + 1 \)[/tex] is at the point [tex]\((-1, -1)\)[/tex].
The correct answer is:
D. [tex]\((-1, -1)\)[/tex]
