Question

How do you find the vertex and intercepts for `y=x^2+6`?

Answer 1

vertex at `(0,6)`
y-intercept at `6`
there is no x-intercept

Explanation:

vertex
The general vertex form is
`color(white)("XXX")y=color(green)(m)(x-color(red)(a))^2+color(blue)(b)` with vertex at `(color(red)(a),color(blue)(b))`

Re-writing the given: `y=x^2+6` into explicit vertex form:
`color(white)("XXX")y=color(green)(1)(x-color(red)(0))^2+color(blue)(6)`
with vertex at `((color(red)(0),color(blue)(6))`

y-intercept
The y-intercept is the value of `y` when `x=0`
Setting `x=0` in the given equation:
`color(white)("XXX")y=(0)^2+6=6`
The y-intercept is at `y=6`

x-intercept
The x-intercept is the value of `x` which causes `y=0`
For the given equation `y=x^2+6`
`color(white)("XXX")`if `y=0` then `x^2=-6`
`color(white)("XXX")` but `x >= 0` for all Real values of `x`
Therefore
`color(white)("XXX")`there is no x-intercept.
graph{x^2+6 [-15.54, 12.94, -1.06, 13.18]}