The standard form of the equation of a parabola is x=y^2+6y+1. What is the vertex form of the equation?

A. X = (y+3) ^2-8

B. X = (y+3) ^2-5

C. X = (y+6) ^2-35

A. x = (y+3) ^2-8 is the correct answer.

What you want to do is get the quadratic (y^2+6y+1) into it's root factor form, in other words, the quantity of (y+what) squared will equal y^2+6y+1. Well it's not a perfect square so we'll have to play around with it a little.

In the format ay^2 + by + c, completing the square first requires finding out what value would do that: c = (b/2) ^2 = (6/2) ^2 = 3^2 = 9

So we need y^2+6y+1 to have the last number (c) to equal 9 to have (y+b/2) ^2, or (y+3) ^2

How do we get 9 down to the 1 that we have for c? 9-1=8, so just subtract 8.

That means that y^2+6y+1 is the same as (y^2+6y+9) - 8, and y^2+6y+9 = (y+3) ^2, so now we have: y^2+6y+1 = (y+3) ^2-8, which is answer [A]