Question

find the vertex of the parabola: x = 3y^2 + 6y + 1

Answer 1

to find the x value of the vertex, use the formula x=-b/2a
plug in the x value into the equation to get the y value
then you have your vertex

Answer 2

can you help me with which ones to plug in

Answer 3

sorry but i have to go

Answer 4

oh alright :/

Answer 5

@SithsAndGiggles

Answer 6

Complete the square:
\[\begin{align*}x&=3y^2+6y+1\\
&=3\left(y^2+2y\right)+1\\
&=3\left(y^2+2y+1-1\right)+1\\
&=3\left((y+1)^2-1\right)+1\\
&=3(y+1)^2-3+1\\
&=3(y+1)^2-2\end{align*}\]
Now the parabola is in vertex form. The general form would be \(x=a(y-h)^2+k\), where \((h,k)\) is the vertex.

Answer 7

so the center would be (-1, 2) ?

Answer 8

vertex* @SithsAndGiggles

Answer 9

Close, it's (-1,-2).