Hi
Using the vertex form of a parabola,
where(h,k) is the vertex
x^2+5+2x = 0
y = x^2+ 2x + 5 |completing Square to put into vertex form for graphing purposes
y = (x+1)^2 - 1 + 5
y = (x+1)^2 + 4 |Vertex is (-1,4) and line of symmetry is x = -1 and(0,5) the y-intercept
NO real solutions, Parabola does not cross the x-axis.
0 = (x+1)^2 + 4
-1 ± 2i = x
