# How to determine the vertex of parabola .Write the vertex of the parabola y=x^2+4x .

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For a parabola y = ax^2 + bx + c, the vertex is given by the coordinates: (-b/2a , -(b^2 - 4ac)/4a)

The equation of the parabola given is : y=x^2+4x

substituting the values, the vertex is

(-2/2 , -(16 - 0)/4)

=> (-1 , -4)

**The vertex of y=x^2+4x lies at (-1 , -4)**

The coordinates of the vertex of the parabola are:

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

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

y = ax^2 + bx + c

Comparing, we'll get the values of the coefficients:

a = 1, b = 4 , c = 0

We'll determine the x coordinate of the vertex:

xV = -b/2a

xV = -4/2

xV = -2

We'll determine 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).**