Parabolas...?

Question

What is the vertex, focus and directrix of the equation

y=1/9 x^2 -2/9 x + 28/9


???

Answer 1

y=1/9 x^2 -2/9 x + 28/9

1/9^x2 -2/9x -y +28/9 = 0 (9)
1/9x^2(9) -2/9x(9) -y(9) + 28/9(9) = 0 (9)
x^2 - 2x - 9y + 28 = 0
(x^2 - 4x + 4) = 9y -28 completing the squares
(x^2 - 4x + 4) = 9y -28 +4
(x-2)^2 = 9(y + 24)

parabola opens upward with vertex
(2, -24)

focus
4a = 9
a = 9/4
the coordinates of the focus can be solve by -24 + 9/4 and that is (2 , 21 and 3/4)

Equation of the directrix can be solve by
-24 - 9/4
X = -26 and 1/4

Gud luck!

Answer 2

first complete the square and put the equation in standard parabolic form which is :

y= (ax+b)^2+c

look at the -2/9x term and divide it by 2 then by the root of the first term to get 1/3 now square this to get 1/9. This is the term you need to add and subtract from your eqaution to complete the square.

y=1/9 x^2 -2/9 x + 1/9 + 28/9 -1/9
y=1/9 x^2 -2/9 x + 1/9 + 27/9
y= (1/3x-1/3)^2+3

V=(1,3)

Parabola opens upward so its type isy = ax^2 and for this type:

Focus: (0, 1/4a)
Directrix: y= -1/4a

Answer 3

vertex = (1,3)