Question

How do you find the vertex form of y=3x^25x3 by completing the quare? can you explain ALL the tep?

Answer 1

Below

Explanation:

To complete the square, the leading x^2 coefficient needs to be = 1 , so factor out a 3 to get

y = 3 ( x^2 + 5/3x ) +3 (I assumed it was a + sign between the terms)

Then take 1/2 of the 5/3 ( 5/6 ) , square it (25/36) , add it to the parentheses.... then subtract the amount you added (3 * 25/36) by doing this..... to have this :

y = 3 ( x^2 + 5/3 x + 25/36) - 3 * 25/36 +3 then simplify to

y = 3 ( x + 5/6)^2 + 11/12 Done.