Question

How do you find the vertex, focus and sketch `x+y^2=0`?

Answer 1

Please see below.

Explanation:

As `x+y^2=0`, we have `x=-(y-0)^2+0` i.e. a parabola of the type `x=a(y-k)^2+h`, where vertex is `(h,k)` and axis of symmetry is `y-k=0`. Focus of such a parabola is `(h+1/(4a),k)`

Hence `x+y^2=0hArrx=(-1)(y-0)^2+0` is a parabola whose vertex is `(0,0)` and axis of symmetry is `y=0` i.e. `x`-axis. Focus is `(0+1/(4xx(-1)),0)` or `(-1/4,0)`.

We can sketch the curve by taking various values of `y`, say `{0,+-1,+-2,+-3,+-4,+-5}`, which gives us points

`(0,0),(-1,-1),(-1,1),(-4,-2),(-4,2),(-9,-3),(-9,3),(-16,-4),(-16,-4),(-25,-5),(-25,5)`

and graph appears as shown below:

graph{(x+y^2)((x+1/4)^2+y^2-0.02)=0 [-12.9, 7.1, -5.14, 4.86]}