Question

How do you find the focus, directrix and sketch `y=1/3x^2-x`?

Answer 1

Focus `(3/2,0)`
Vertex `(3/2,-3/4)`
directrix: y= `-3/2`

Explanation:

y=`1/3x^2-x`
`3y=x^2-3x`
`3y=9/4=x^2-3x+(3/2)^2` (completing the square)
`3(y+3/4) = (x-3/2)^2`
which is in the form: `(x-h)^2 = 4a(y-k)`
so your focus is (h,k) and your vertex is directly below your focus depending on your "a" ie your vertex has the same x-value as your focus but a different y-value
and your directrix will be "a" units BELOW your vertex

Make sure you can picture your graph as a concave up graph. A concave down graph will have a negative sign included

To find a,
4a=3
a=`3/4`
so your vertex is `(3/2,-3/4)`
and your focus is `(3/2,0)` since your focus is `3/4` units above your y-value of your vertex
and finally, your directrix will be y=`-3/2` since it is `-3/4` units below your vertex

Now, all you have to do is to graph your equation
just make sure you know your x-intercepts and y-intercepts
ie x-intercepts is when y=0 and y-intercepts is when x=0
so your x-intercepts are (0,0) AND (3,0) and your y-intercept is (0,0)
graph{y = 1/3x^2-x [-10, 10, -5, 5]}