Answer
See image:
Work Step by Step
$ \begin{array}{llll}
\text{Function:} & f(x)=x^{2}+\sqrt{2}x+1 & & \\
\text{coefficients} & a =+1, b=1, c=-1 & & \\
\text{coefficient a} & \text{positive, opens up} & \\ & \\
\text{Vertex} & x_{V}=-b/(2a) & y_{V}=f(-b/(2a)) & \\
& x_{V}=-\sqrt{2}/2 & y_{V}=1/2 & \\
& \bf V(-\sqrt{2}/2,1/2) & & \\\\
\text{Line of symmetry} & x=-\sqrt{2}/2\\ & & \\
\text{y intercept} & c=1, & \bf(0,1) & \\
\text{point symmetric to } & & & \\
\text{the y-intercept} & x=2x_{V}=-\sqrt{2} & \bf(-\sqrt{2},1)\\ & \\
\text{Zeros:} & x^{2}+\sqrt{2}x+1=0 & & \\
& x=\frac{-b\pm\sqrt{\mathrm{b}^{2}-4ac}}{2a} & \text{... quadratic formula} & \\
& \mathrm{b}^{2}-4ac=-2 & & \\
& & & \\
& \bf \text{No zeros} & &
\end{array}$
Additional points: $(\sqrt{2},5), (-2\sqrt{2},5).$
$\sqrt{2}\approx 1.414,$
$2\sqrt{2}\approx 2.828,$
$\sqrt{2}/2\approx 0.707$
Using the above information, plot the points and join with a smooth curve (parabola).
