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).
