Question

How do you find the vertex and intercepts for `x^2=8y`?

Answer 1

`"see explanation"`

Explanation:

`"the standard form of a parabola with the y-axis as it's"`
`"principal axis and opening vertically is"`

`color(red)(bar(ul(|color(white)(2/2)color(black)(x^2=4py)color(white)(2/2)|)))`

`"this parabola has it's vertex at the origin "(0,0)`

`• " if 4p ">0" then opens vertically up "uuu`

`• " if 4p "<0" then opens vertically down "nnn`

`x^2=8y" is in this form"`

`rArr4p=8`

`"since 4p ">0" then opens up"`
graph{x^2=8y [-10, 10, -5, 5]}

Answer 2

Vertex `=(0,0)` which is also the only intercept of the `xy-`axes.

Explanation:

`x^2=8y -> y = 1/8x^2`

The graph of `y` is a parabola, with a vertex at the absolute extremum of `y`

Since the coefficient of `x^2 = 1/8>0` the vertex of `y` will be an absolute minimum.

Since `x^2>=0 forall x in RR -> y_min = y(0)`

`:. y_min = 1/8 xx 0 = 0`

Hence, the vertex of `y = (0,0)`

Since the vertex is the absolute minimum of `y` there can be no other intercepts than `(0,0)`

This result can be seen from the graph of `y` below.

graph{x^2/8 [-16.11, 15.93, -2.15, 13.87]}