Answer:
[tex](0,2)[/tex]; minimum
Step-by-step explanation:
Given:
The function is, [tex]y=x^{2}+2[/tex]
The given function represent a parabola and can be expressed in vertex form as:
[tex]y=(x-0)^{2}+2[/tex]
The vertex form of a parabola is [tex]y=(x-h)^{2}+k[/tex], where, [tex](h,k)[/tex] is the vertex.
So, the vertex is [tex](0,2)[/tex].
In order to graph the given parabola, we find some points on it.
Let [tex]x=-2,y=(-2)^{2}+2=4+2=6[/tex]
[tex]x=-1,y=(-1)^{2}+2=1+2=3[/tex]
[tex]x=0,y=(0)^{2}+2=0+2=2[/tex]
[tex]x=2,y=(2)^{2}+2=4+2=6[/tex]
[tex]x=1,y=(1)^{2}+2=1+2=3[/tex]
So, the points are [tex](-2,6),(-1,3),(0,2),(1,3),(2,6)[/tex].
Mark these points on the graph and join them using a smooth curve.
The graph is shown below.
From the graph, we conclude that at the vertex [tex](0,2)[/tex], it is minimum.
