Answer
Vertex: $(2,1)$
Axis of Symmetry: $x=2$
Domain: all real numbers
Range: $\{y|y\ge1\}$
Graph of $f(x)=3(x-2)^2+1$
Work Step by Step
Using $f(x)=a(x-h)^2+k$ or the standard form of quadratic functions, the given function, $
f(x)=3(x-2)^2+1
,$ has $h=
2
$ and $k=
1
$.
Since the vertex is given by $(h,k)$, then the vertex of the given quadratic function is $
(2,1)
$.
With the axis of symmetry given by $x=h$, then the equation of the axis of symmetry is $
x=2
$.
Let $y=f(x)$. Then $
y=3(x-2)^2+1
$. Substituting values of $x$ and solving for $y$ results to
\begin{array}{l|r}
\text{If }x=0: & \text{If }x=1:
\\\\
y=3(x-2)^2+1 & y=3(x-2)^2+1
\\
y=3(0-2)^2+1 & y=3(1-2)^2+1
\\
y=3(-2)^2+1 & y=3(-1)^2+1
\\
y=3(4)+1 & y=3(1)+1
\\
y=12+1 & y=3+1
\\
y=13 & y=4
.\end{array}
Hence, the points $
(0,13)
$ and $
(1,4)
$ are on the graph of the parabola. Reflecting these points about the axis of symmetry gives the points $
(3,4)
$ and $
(4,13)
$.
The graph (shown above) is determined using the points $\{
(0,13),(1,4),(2,1),(3,4),(4,13)
\}$.
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\}
.$
