Question

Graph y=x^2+2. Identify the vertex of the graph. Tell whether it is a minimum or maximum.

Answer 1

This is an upwards-opening parabola with a vertex of (2, 0). That +2 moves the vertex up 2 along the y axis. If this is a positive parabola, which it is, then the vertex is a minimum.

Answer 2

we know that

The equation of a vertical parabola in vertex form is equal to

`y=a(x-h)^(2)+k`

where

(h,k) is the vertex of the parabola

if
`a>0` -----> the parabola open upward (vertex is a minimum)

if
`a<0` -----> the parabola open downward (vertex is a maximum)

In this problem we have

`y=x^(2)+2`

The vertex is the point
`(0,2)`

`a=1`

so

`a>0` -----> the parabola open upward (vertex is a minimum)

Using a graphing tool

see the attached figure

The answer is

The vertex is the point
`(0,2)`

Is a minimum