Answer
See image:
Work Step by Step
$ \begin{array}{llll}
\text{Function:} & f(x)=-x^{2}+5 & & \\
& a =-1, b=0, c=+5 & & \\
\text{coefficient a} & \text{negative, opens down} & \\ & \\
\text{Vertex} & x_{V}=-b/(2a) & y_{V}=f(-b/(2a) & \\
& x_{V}=0 & y_{V}=5 & \\
& \bf V(0,5) & & \\\\
\text{Line of symmetry} & x=0\\ & & \\
\text{y intercept} & c=5, & \bf(vertex) & \\
\text{point symmetric to } & & & \\
\text{the y-intercept} & - & \\ & \\
\text{Zeros:} & -x^{2}+5=0 & & \\
& x^{2}=-5 & & \\
& x=\pm\sqrt{5} & & \\
& \bf\approx(\pm 2.236,0) & &
\end{array}$
Additional points:$ \left[\begin{array}{l|l|l|l|l}
x & -1 & 1 & -2 & 2\\
\hline f(x) & 4 & 4 & 1 & 1
\end{array}\right]$
Using the above information, plot the points and join with a smooth curve (parabola).