y^2+4y-x+1=0 what is the vertex of this equation and the axis of symmetry for this parabola?
Y^2 + 4Y -X + 1 = 0.
X = Y^2 + 4Y + 1 = 0,
K = Yv = -b/2a = -4 / 2 = -2.
h = Xv = (-2)^2 + 4*-2 + 1 = -3.
V(-3,-2).
Axis: Y = K = -2.
X = Y^2 + 4Y + 1 = 0,
K = Yv = -b/2a = -4 / 2 = -2.
h = Xv = (-2)^2 + 4*-2 + 1 = -3.
V(-3,-2).
Axis: Y = K = -2.