SOLUTION: identify the focus of 2x^2+5x+y+14=0

Question 384777: identify the focus of 2x^2+5x+y+14=0
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
identify the focus of 2x^2+5x+y+14=0
y=-2x^2-5x-14
factor out -2
then complete the square
y = -2(x^2+(5/2)x+25/16)-14+50/16
y = -2(x+5/4)^2-87/8
this is a parabola which opens downward whose vertex is at
(-5/4,-87/8)
y = -2(x+5/4)^2-87/8
-2(x+5/4)^2=y+87/8
(x+5/4)^2=-1/2(y+87/8)
this is now in form x^2 = 4py
4p=1/2
p=1/8
line of symmetry is at x=-5/4
focus is on the line of symmetry p units below the vertex=87/8-1/8=86/8=43/4
ans: the coordinates of the focus=(-5/4,-43/4)
see the following graph of the parabola

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