Question

Find the vertex of the parabola.
y=-x^2+2x-5

Answer 1

option 1:
get line of symmetry
x = -b/2a, where a=-1 b=2, c=-5
x = -2/-2 = 1
substitute that in for x to find y
y = -(1)^2 +2(1) -5 = -4
vertex: (1,-4)

option 2:
use completing the square to put it in vertex form
y=a(x-h)^2 +k
vertex: (h,k)
factor out neg
y = -(x^2 -2x +5)
take half of b, square it and subtract from c
y = -((x-1)^2 +5 - (-1)^2)
y = -(x-1)^2 -4
vertex: (1,-4)