Answer
See image:
Work Step by Step
$ \begin{array}{llll}
\text{Function:} & f(x)=x^{2}+1 & & \\
\text{coefficients} & a =+1, b=0, c=1 & & \\
\text{coefficient a} & \text{positive, opens up} & \\ & \\
\text{Vertex} & x_{V}=-b/(2a) & y_{V}=f(-b/(2a)) & \\
& x_{V}=0 & y_{V}=1 & \\
& \bf V(0,1) & & \\\\
\text{Line of symmetry} & x=\\ & & \\
\text{y intercept} & c=1, & \bf \text{... the vertex.} & \\
\text{point symmetric to } & & & \\
\text{the y-intercept} & - & \bf \\ & \\
\text{Zeros:} & x^{2}+1=0 & & \\
& x^{2}=-1 & \text{... no real solutions} & \\
& & \bf \text{No zeros} &
\end{array}$
Additional points:$ \left[\begin{array}{l|l|l|l|l}
x & -1 & 1 & -2 & 2\\
\hline f(x) & 2 & 2 & 5 & 5
\end{array}\right]$
Using the above information, plot the points and join with a smooth curve (parabola).