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, $$ f(x)=(x+2)^{2}-3 $$ The vertex is at

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1. Start with the function:
   $$
   f(x)=x^2+4x+1
   $$
   
2. To rewrite in standard form, complete the square. Group the $$ x $$-terms:
   $$
   f(x)=\left(x^2+4x\right)+1
   $$
   
3. Complete the square inside the parentheses. Take half of the coefficient of $$ x $$ (which is 4), square it, and add and subtract that value:
   $$
   \text{Half of } 4 \text{ is } 2, \text{ and } 2^2=4.
   $$
   $$
   f(x)=\left(x^2+4x+4-4\right)+1
   $$
   
4. Rewrite the quadratic expression as a perfect square and simplify:
   $$
   f(x)=\left((x+2)^2-4\right)+1 = (x+2)^2 - 3
   $$
   
5. The function in standard form is:
   $$
   f(x)=(x+2)^2-3
   $$
   
6. The vertex of the graph is given by the values that make the squared term zero:
   $$
   x+2=0 \implies x=-2, \quad y=-3
   $$
   
Thus, the vertex is at $$\boxed{(-2,-3)}$$.
The vertex of the graph 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, \( f(x)=(x+2)^{2}-3 \) The vertex is at

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