SOLUTION: Give the vertex, focus, directrix, and the distance across parabola at focus, and then graph the parabola. x^2+6x+2y+15=0

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Give the vertex, focus, directrix, and the distance across parabola at focus, and then graph the parabola. x^2+6x+2y+15=0      Log On


   

Question 188602: Give the vertex, focus, directrix, and the distance across parabola at focus, and then graph the parabola.
x^2+6x+2y+15=0
Answer by Mathtut(3670) About Me  (Show Source):
You can put this solution on YOUR website!
2y=-x%5E2-6x-15
:
y=-1%2F2%28x%5E2%2B6x%2B9%29-15%2F2%2B9%2F2
:
y=-1%2F2%28x%2B3%29%5E2-3
:
vertex is at (-3,-3)
:
1/4p=-1/2
:
p=-1/2
:
directrix is at y=-5/2
focus is at (-3,-7/2)
:
distance across parabola is abs value of 4p which equals abs value of 4(-1/2)= 2:
graph%28300%2C300%2C-10%2C10%2C-10%2C10%2C%28-1%2F2%29%28x%2B3%29%5E2-3%29