Determine Vt and k from given transfer characteristic
Here's perhaps a quick approach to use if these come up on a test. Just use your basic knowledge about square-root function behavior and take note where the curve reaches \$I_\text{D}=0\:\text{A}\$. It might be easier for some to see the curve flipped around a bit, like this:



It's easy to see where to draw the line for \$V_\text{T}\$. Right at \$V_\text{T}=2\:\text{V}\$. Once you have that baseline to work with, just find an easy to read data point. You can see the one I picked.

With that in hand, it's easy to work out that \$k=\frac{I_\text{D}=15\:\text{mA}}{\left[\left(V_\text{GS}=7.5\:\text{V}\right)-\left(V_\text{T}=2\:\text{V}\right)\right]^2}\approx 500\:\frac{\mu\text{A}}{\text{V}^2}\$.

To test the idea, let's compute \$I_\text{D}\$ for \$V_\text{GS}=6\:\text{V}\$. \$I_\text{D}=500\:\frac{\mu\text{A}}{\text{V}^2}\cdot \left(6\:\text{V}-2\:\text{V}\right)^2=8\:\text{mA}\$. Looking back at the chart, that seems about right.

Of course, you can be more picky about it. Maybe I was too quick to pick \$V_\text{T}=2\:\text{V}\$? Or perhaps you might select different points for computing \$k\$. (You could use a least-squares fitting method that would derive both points for you. But the solution mathematics for that requires solving partial differentials and deciding how you want to handle error estimates along each axis.)