It sort of depends on how detailed a graph you want. Note first of all that the graph is symmetrical about the $y$-axis. This observation immediately cuts our work in half! Just do the graph for $x$ positive, and reflect the picture across the $y$-axis.
So we take $x \ge 0$. Near $x=0$, we have that $y \approx \sqrt{2}$. As $x$ increases, $\sqrt{x^2+2}$ increases.
For $x$ at all large positive, we have that $\sqrt{x^2+2}\approx x$. More precisely, it is a little bigger than $x$, but awfully close to $x$ when $x$ is large. (You may want to do a small number of calculations, using a calculator.)
Draw the half-line $y=x$, for $x \ge 0$. To say that $\sqrt{x^2+2}$ is awfully close to $x$, when $x$ is large, means that the curve $y=\sqrt{x^2+2}$ hugs the half-line $y=x$, for $x \ge 0$, when $x$ is large. So for large values, your curve will be roughly indistinguishable from $y=x$.
In technical language, the half-line is an (oblique) asymptote to the curve $y=\sqrt{x^2+2}$.
With this information, you should be able to get a good idea of the first-quadrant part of the curve. Then don't forget to reflect across the $y$-axis.
Wish I could supply a picture. There are (free) graphing calculator programs, or more generally graphing programs, available for download.
Look up also Wolfram Alpha. It will draw the curve for you. But first do it yourself.
Apparently, the problem is not that easy after all. What your teacher believes is the solution is obviously wrong. After all, she suggests that $\frac12$ is one of the solutions, but
$$ (\tfrac12)^2-(k+1)\tfrac12+k+1\ne 0$$
Best Answer
It would help you to factor $$f(x) = -x^2 + 4x - 3 = -(x - 3)(x - 1)$$
Then $$f(x) = 0 \iff x - 3 = 0 \;\text{ or } \; x - 1 = 0$$
As suggested, did you graph the function? You can then visually see where $f(x)$ intersects the $x$-axis: those are the "zeros" of the function.