SOLUTION: i am stuck on the parabolas and i just can't seem to figure it out. Here is one of the problems i am working on. x=y2 + 14y + 20. (the 2 next to the y is y squared)well i get stuc

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: i am stuck on the parabolas and i just can't seem to figure it out. Here is one of the problems i am working on. x=y2 + 14y + 20. (the 2 next to the y is y squared)well i get stuc      Log On


   

Question 32494This question is from textbook
: i am stuck on the parabolas and i just can't seem to figure it out. Here is one of the problems i am working on. x=y2 + 14y + 20. (the 2 next to the y is y squared)well i get stuck on this part x= (y2 + 14y + _____)+20 ______ i just cant figure out how to find the numbers that go in those places. please help me This question is from textbook

Answer by mukhopadhyay(490) About Me  (Show Source):
You can put this solution on YOUR website!
x=y^2 + 14y + 20
This is a parabola having directrix parallel to the y-axis (sidewise, opening on the right side).
x = y^2+14y+20
=> x = y^2 + 14y + 49 - 29
=> x = (y+7)^2 - 29
The vertex of the parabola is at (-29, -7).
The directrix of the parabola is at y = -14
Let me know if you need any additional information.