Answer
Vertex: $(1,2)$
Axis of Symmetry: $x=1$
Domain: all real numbers
Range: $\{y|y\ge1\}$
Graph of $f(x)=(x-1)^2+2$
Work Step by Step
Using $f(x)=a(x-h)^2+k$ or the standard form of quadratic functions, the given function, $
f(x)=(x-1)^2+2
,$ has $h=
1
$ and $k=
2
$.
Since the vertex is given by $(h,k)$, then the vertex of the given quadratic function is $
(1,2)
$.
With the axis of symmetry given by $x=h$, then the equation of the axis of symmetry is $
x=1
$.
Let $y=f(x)$. Then $y=
(x-1)^2+2
$. Substituting values of $x$ and solving for $y$ results to
\begin{array}{l|r}
\text{If }x=-1: & \text{If }x=0:
\\\\
y=(x-1)^2+2 & y=(x-1)^2+2
\\
y=(-1-1)^2+2 & y=(0-1)^2+2
\\
y=(-2)^2+2 & y=(-1)^2+2
\\
y=4+2 & y=1+2
\\
y=6 & y=3
.\end{array}
Hence, the points $
(-1,6)
$ and $
(0,3)
$ are on the graph of the parabola. Reflecting these points about the axis of symmetry gives the points $
(2,3)
$ and $
(3,6)
$.
The graph (shown above) is determined using the points $\{
(-1,6),(0,3),(1,2),(2,3),(3,6)
\}$.
Using the graph, the domain (all $x$ values used in the graph) is the set of all real numbers. The range (all $y$ values used in the graph) is $
\{y|y\ge1\}
.$
