Which equation is y = 6x^2 + 12x – 10 rewritten in vertex form?
a. y = 6(x + 1)^2 – 11
b. y = 6(x + 1)^2 – 10
c. y = 6(x + 1)^2 – 4
d. y = 6(x + 1)^2 – 16

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so just complete the square
isolate xterms and factor out a in ax^2+bx+c=y
y=6(x^2+2x)-10
take 1/2 of linear coefient (b) and square it
2/2=1, 1^2=1
add negative in positive inside
y=6(x^2+2x+1-1)-10
factor perfect square
y=6((x+1)^2-1)-10
distribute
y=6(x+1)^2-6-10
y=6(x+1)^2-16
D is answer


easy way is to just expand your choices
we see that the similar parts are the 6(x+1)^2, so just expand that

6(x+1)^2=6(x^2+2x+1)=6x^2+12x+6
we just need the last constant
look at constant of 6(x+1)^2 and 6x^2+12x-10 (original function)

6 and -10
what should be added to 6 to get -10
that's right, -16

y=6(x+1)^2-16
easy
D is answer
work smart not hard

D is the answer

Y = 6x^2 + 12x - 10 in vertex form gives,
y = 6(x^2 + 2x - 5/3) = 6(x^2 + 2x + 1 - 5/3 - 1) = 6(x + 1)^2 + 6(-8/3) = 6(x + 1)^2 - 16

Therefore, option D is the correct answer.