Can someone explain to me how to find the vertex form of this equation? y= -x^2 + 12x-4

ANSWER:  The vertex of given equation  [tex]y=-x^{2}+12 x-4[/tex] is (6, 32). SOLUTION: Given, quadratic equation is  [tex]y=-x^{2}+12 x-4[/tex] Above equation is a parabola and we need to find the vertex of that parabola.  We know that, general form of the parabola is  [tex]y=a(x-h)^{2}+k[/tex]Where, (h, k) is the vertex. So, let us convert the given equation in parabolic equation. [tex]\begin{array}{l}{y=-x^{2}+12 x-4} \\ {y=-\left(x^{2}-12 x+4\right)} \\ {y=-\left(x^{2}-(2)(x)(6)+6^{2}-6^{2}+4\right)} \\ {\left.y=-\left((x-6)^{2}-36+4\right)\right)} \\ {y=-(x-6)^{2}+32}\end{array}[/tex]Now, by comparing above equation with general form,  h = 6, k = 32 and a = -1. Hence, the vertex of given equation is (6, 32).