Where is found the vertex of the function f(x) = x^2 + 5x + 1 ?

We have to establish the quadrant where the vertex of the
parable is located.

We know that the coordinates of the
parabola vertex are:

 V(-b/2a;-delta/4a), where a,b,c are
the coefficients of the  function and delta=b^2
-4*a*c.

y=f(x)=x^2 + 5x +
1

We'll identify the
coefficients:

a=1, b=5, c=1, 2a=2,
4a=4

delta=5^2
-4*1*1=25-4=21

V(-b/2a;-delta/4a)=V(-5/2;-21/4)

Because
the coordinates are both negative, the vertex is located in the third quadrant:
V(-5/2;-21/4).