Question 681141
What is the vertex and the axis of symmetry of the parabola 
y=x^2+5x-6
complete the square:
y=(x^2+5x+25/4)-6-25/4
y=(x+5/2)^2-24/4-25/4
y=(x+5/2)^2-49/4
This is an equation of a parabola that opens upwards.
Its standard form: y=A(x-h)^2+k, (h,k)=(x,y) coordinates of vertex
For given equation: y=(x+5/2)^2-49/4
vertex: (-5/2,-49/4)
axis of symmetry: x=-5/2
see graph below:
{{{ graph( 300, 300, -10, 10, -20, 10,(x+5/2)^2-49/4) }}}