What is y = 6x^2 + 12x - 10 in vertex form?

Vertex for is a (x - b) ^2 + c

first divide 6x^2 + 12x by 6 so we have

6 (x^2 + 2x) - 10

now complete the square on the expression in the brackets:-

= 6[ (x + 1) ^2 - 1] - 10

=6 (x + 1) ^2 - 6 - 10

= 6 (x + 1) ^2 - 16

Basically complete the square

so

conver tto y=a (x-h) ²+k

steps:

group x terms

factor out leading coefient

take 1/2 of ilnear coefient and square it, add negative and positive inside

factor perfect square

expand/distribute

y=6x²+12x-10

group x

y = (6x²+12x) - 10

factor out leading coefient

y=6 (x²+2x) - 10

take 1/2 of linear coefient and square it

2/2=1, 1²=1

add negativ and positivve inside

y=6 (x²+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