Question

What is the focus of the parabola? y=−1/4x^2−x+3

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Answer 1

we know that
the equation of the parabola is of the form
y=ax²+bx+c

in this problem
y=−1/4x²−x+3
where
a=-1/4
b=-1
c=3

the coordinates of the focus are

(-b/2a,(1-D)/4a)
where
D is the discriminant b²-4ac
D=(-1)²-4*(-1/4)*3-----> D=1+3---> D=4
therefore
x coordinate of the focus
-b/2a----> 1/[2*(-1/4)]----> -2

y coordinate of the focus
(1-D)/4a------> (1-4)/(-4/4)---> 3

the coordinates of the focus are
(-2,3)