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