Given the parabola y = x^2 + 4x.

Then we know that:

a = 1 b = 4 c = 0.

The vertex is given as follows.

The x-coordinate is given by:

==> Vx = -b/2a

==> vx = -4 / 2 = -2.

The y-coordinate is given by:

Vy = -(b^2 -4ac)/4a = - ( 16 - 4*1*0)/ 4 = -16/4 = -4.

==> Vy = -4.

**Then the vertex of the parabola y= x^2 + 4x is V(-2, -4).**

We'll write the coordinates of the vertex of the parabola:

V(-b/2a ; -delta/4a)

a,b,c are the coefficients of the quadratic:

y = ax^2 + bx + c

Comparing, we'll identify the coefficients:

a = 1, b = 4 , c = 0

We'll calculate the x coordinate of the vertex:

xV = -b/2a

xV = -4/2

xV = -2

We'll calculate the y coordinate of the vertex:

yV = -delta/4a

delta = b^2 - 4ac

delta = 16 - 4*1*0

delta = 16

yV = -16/4

yV = -4

**The coordinates of the vertex of the parabola are: V(-2 , -4).**