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]

Iwould say 48 bc the line of x is the line of 6 so 6×6=48