Rewrite the equation in standard form, then find the vertex of the graph of $$ f(x) $$. $$ f(x)=x^{2}+4 x+1 $$. In standard form, $$ f(x)=\square $$ The vertex is at

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To rewrite the equation in standard form, we need to complete the square for the quadratic expression $$x^2 + 4x + 1$$.

The standard form of a quadratic equation is $$f(x) = a(x - h)^2 + k$$, where $$(h, k)$$ is the vertex of the parabola.

Let's complete the square for the given quadratic expression:

$$f(x) = x^2 + 4x + 1$$

To complete the square, we take half of the coefficient of $$x$$, which is $$4/2 = 2$$, and square it, which gives us $$2^2 = 4$$. We then add and subtract 4 inside the expression:

$$f(x) = (x^2 + 4x + 4) - 4 + 1$$

Simplifying, we get:

$$f(x) = (x + 2)^2 - 3$$

Therefore, the standard form of the equation is $$f(x) = (x + 2)^2 - 3$$.

The vertex of the graph of $$f(x)$$ is at $$(-2, -3)$$.
In standard form, $$ f(x) = (x + 2)^2 - 3 $$. The vertex is at $$(-2, -3)$$.

Rewrite the equation in standard form, then find the vertex of the graph of \( f(x) \). \( f(x)=x^{2}+4 x+1 \). In standard form, \( f(x)=\square \) The vertex is at

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