What is h(x) = –3x2 – 6x + 5 written in vertex form?
h(x) = –3(x + 1)2 + 2
h(x) = –3(x + 1)2 + 8
h(x) = –3(x – 3)2 – 4
h(x) = –3(x – 3)2 + 32

H=-3x^2-6x+5  subtract 5 from both sides

h-5=-3x^2-6x  divide both sides by -3

(5-h)/3=x^2+2x  halve the linear coefficient, square it, add it to both sides

(2/2)^2=1 add 1 to both sides...

(5-h+3)/3=x^2+2x+1, now the right side is a perfect square...

(8-h)/3=(x+1)^2  multiply both sides by 3

8-h=3(x+1)^2  divide both sides by -1

h-8=-3(x+1)^2  add 8 to both sides

h=-3(x+1)^2+8

h(x)= -3(x+1)^2+8

Second answer down from the top :)

The answer is option c.