How do you find f(x) and the vertex of f(x)=x^2+6x−16

Question

How do you find f(x) and the vertex of f(x)=x^2+6x−16

Answer 1

y= x^2 + 6x -16

complete the square by adding 9 and subtracting 9

y = x^2 +6x + 9 -16 -9

y = (x+3)^2 -25

y+25 = (x+3)^2

the vertex is (h,k) = (-3,-25) = minimum value of y or f(x)

general form of the equation for a parabola is

y-h = a(x-k)^2 where (h,k) is the vertex