A quadratic equation of type #y=ax^2+bx+c#, which is equation of a parabola, can be written in vertex form i.e.
#y=a(x-h)^2+k#
For example, let us have the equation #y=-3x^2-6x+9#, them we can write it as
#y=-3(x^2+2x+1)+3+9=-3(x+1)^2+12#
Similarly #y=2x^2+3x-7=2(x^2+3/2x)-7#
= #2(x^2+2xx3/4x+(3/4)^2)-2xx(3/4)^2-7#
= #2(x+3/4)^2-65/8=2(x+3/4)^2-8 1/8#
Observe that #(x-h)^2>=0# and hence
if #a>0#, #a(x-h)^2# is positive and the minimum value of #y#, when #x=h# is #k# i.e. the parabola rises on either side and has lowest point as #(h,k)#. For example, #y=2x^2+3x-7=2(x+3/4)^2-8 1/8# and graph of #y=2x^2+3x-7# has minima at #(-3/4,8 1/8)# and observe that #(-3/4,8 1/8)# appears as a vertex.
graph{2x^2+3x-7 [-4, 4, -11.36, 8.64]}
but if #a<0#, #a(x-h)^2# is negative and the maximum value of #y#, when #x=h# is #k# i.e. the parabola falls on either side and has highest point as #(h,k)#. For example, #y=-3x^2-6x+9=-3(x+1)^2+12# and graph of #y=-3x^2-6x+9# has maxima at #(-1,12)# and observe that #(-1,12)# appears as a vertex.
graph{-3x^2-6x+9 [-5.433, 2.567, -6, 14]}
