The equation provided is $y = x^2 + 4x + 1$, which is a quadratic equation. The vertex of a parabola given by $y = ax^2 + bx + c$ can be found using the formula for the x-coordinate of the vertex:

$x = -\frac{b}{2a}$

In this case, the equation is $y = x^2 + 4x + 1$, where:

• $a = 1$
• $b = 4$

Now, let's find the x-coordinate of the vertex:

$x = -\frac{4}{2(1)} = -2$

To find the y-coordinate, we substitute $x = -2$ back into the equation:

$y = (-2)^2 + 4(-2) + 1 = 4 - 8 + 1 = -3$

So, the vertex is at $(-2, -3)$.

Thus, the correct answer is $\boxed{(-2, -3)}$.

Would you like more details or have any further questions?

Here are 5 related questions you might explore:

1. How do you find the axis of symmetry for a quadratic function?
2. What is the importance of the discriminant in a quadratic equation?
3. How does the value of $a$ affect the direction of a parabola?
4. Can you convert the quadratic equation $y = x^2 + 4x + 1$ into vertex form?
5. How do you graph a quadratic function using its vertex and intercepts?

Tip: The vertex form of a quadratic equation is useful for identifying the vertex directly and is given by $y = a(x-h)^2 + k$, where $(h, k)$ is the vertex.