$x^{2}-16 x+64$

Step 1 ：Given the quadratic equation: \(x^2 - 16x + 64\)

Step 2 ：Identify the coefficients: \(a = 1, b = -16, c = 64\)

Step 3 ：Calculate the vertex: \(vertex_x = \frac{-b}{2a} = \frac{16}{2} = 8\), \(vertex_y = a(vertex_x)^2 + b(vertex_x) + c = 1(8)^2 - 16(8) + 64 = 0\)

Step 4 ：Vertex: \((8, 0)\)

Step 5 ：Calculate the discriminant: \(D = b^2 - 4ac = (-16)^2 - 4(1)(64) = 0\)

Step 6 ：\(\boxed{\text{The quadratic equation } x^2 - 16x + 64 \text{ has a vertex at } (8, 0) \text{ and has one real root since the discriminant is 0}}\)