Write the vertex of the parabola y=x^2+4x.
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).