SOLUTION: Currently, we are working on horizontal parabolas. I am supposed to find the vertex, focus, and directrix of this formula: {{{ y^2-y-x+6=0 }}} I've been working the equation, an

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Currently, we are working on horizontal parabolas. I am supposed to find the vertex, focus, and directrix of this formula: {{{ y^2-y-x+6=0 }}} I've been working the equation, an      Log On


   

Question 854090: Currently, we are working on horizontal parabolas. I am supposed to find the vertex, focus, and directrix of this formula: +y%5E2-y-x%2B6=0+
I've been working the equation, and my work currently looks like this:
+y%5E2-y=x-6+
+%28y-%281%2F2%29%29%5E2=x-%2823%2F4%29+
I don't know what to do with the 23/4. Any help would be appreciated :) Thank you!
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Find the vertex, focus, and directrix of this formula: +y%5E2-y-x%2B6=0+
y^2-y-x+6=0
complete the square:
(y^2-y+(1/4))-x+6=0+1/4
(y-(1/2)^2=x+(1/4)-6
%28y-%281%2F2%29%29%5E2=%28x-%2823%2F4%29%29
This is an equation of a parabola that opens right.
Its basic form of equation: %28y-k%29%5E2=4p%28x-h%29, (h,k)=coordinates of the vertex
For given problem:
vertex:((23/4),(1/2))
axis of symmetry: y=(1/2)
4p=1
p=1/4
focus: ((24/4),(1/2))=(6,(1/2)) (p-distance to the right of vertex on the axis of symmetry)
directrix: x=(22/4)=11/2 (p-distance to the left of vertex on the axis of symmetry)