Question

How do you find the vertex and intercepts for `y^2=1/3x`?

Answer 1

`(0,0)` is only intercept.
`"vertex"=(oo,oo)`

Explanation:

For `y=0`,
`color(white)xxx=0`

Therefore `(0,0)` is only intercept.

`y=sqrt(1/3x)`
Therefore `y` is positive for `AA x inRR^+`.

`y'` is also positive for `AA x inRR^+`:
`y'=1/(2sqrt3sqrtx)`

Therefore there is no real vertex:
`"vertex"=(oo,oo)`

graph{sqrt(1/3x) [-10, 10, -5, 5]}

Answer 2

The vertex and intercepts are all at `(0, 0)`

Explanation:

If `x = 0` then `y^2 = 0`, so `y = 0`.

If `y = 0` then `1/3 x = 0`, so `x = 0`.

So the only `x` and `y` intercepts are at `(0, 0)`

`y^2 >= 0` for any Real value of `y`, hence the minimum possible value of `x` is `0`. So the vertex of the parabola is at `(0, 0)`.

Both positive and negative values are possible for `y`.

`y` is not a function of `x`, but `x` is a function of `y`...

graph{y^2=x/3 [-5.22, 12.56, -4.484, 4.405]}