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.
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.
The vertex lies on x = 1
When x = 1, y = -3
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)
(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
Algebra - I am supposed to write the standard equation of the parabola with the ...
Algebra - Hi again. I am supposed to graph y+3=-1/12(x-1)^2. How do I find the ...
Graph problems!!! - Ok. so i'm supposed to find the coordinates of a line using ...
Algebra 1 - how do i graph x + y = 1 put y on one side y=-x+1 im doing an online...
Algebra 1 (Reiny or Kuai) - 1. Graph each equation. Tell whether the equation ...
precalculus HELP! - I posted this question before: What is the area enclosed by...
Linear Equation - I posted a question below but forgot to put in the equation! ...
chem - The study of kinetics group 1: Temp: 12.6 C T(K): 12.6+273= 285.6 1/T(K...
algebra - Can you use algebra to find a quadratic equation if your know the ...
MATH/ALGEBRA PLEASE - I am having a real issue with this one problem the problem...