Posted by **Natalia** on Sunday, November 30, 2008 at 1:02pm.

(I posted this yesterday, if it looks familar, but no one was on who could help me). I am supposed to graph y+3=-1/12(x-1)^2. How do I find the equation of the directrix, and the coordinates of the vertex and the focus?

- Algebra -
**drwls**, Sunday, November 30, 2008 at 1:17pm
It is a downward-aimed parabola with vertex (highest value) at x = 1, y = -3.

The (1/12) coefficient in front of the (x-1)^2 term tells you about the shape of the parabola. The minus sign tells you tahat it points downward.

If you review your analytic geometry text section about parabolas, you will find that the coefficient (1/12 in this case) is called 1/(4p), where p is the distance from the vertex to the focal point and (in the other direction) to the directrix line. In your case 1/4p = 1/12, so 4p = 12 and p = 3.

The focus is therefore at y=-6 (three units below the vertex) @ x = 1, and the directrix is (y = 0), the x-axis.

- Algebra -
**Damon**, Sunday, November 30, 2008 at 1:22pm
y+3=-(1/12)(x-1)^2 I assume

At x = 1, the right hand side is zero.

a little to the right of x = 1, the right hand side is negative

a little to the left of x = 1, the right hand side is negative by the same amount.

In other words, the line is symmetric about the vertical line x = 1 and opens down.

SO

The vertex lies on x = 1

When x = 1, y = -3

SO

vertex at (1,-3)

Now you want the focus and directrix

The focus must also lie on x = 1

The form is

(x-h)^2 = 4 a (y-k)

here

(x-1)^2 = -12 (y+3)

so 4 a = -12

a = -3

vertex to focus = a = -3

so focus at -3-3 = -6 so (1,-6)

directrix is horizontal line a units above vertex so it is the line y = 0

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