Finding and plotting the impedance
The capacitor and inductor are in parallel, which means that the equivalent impedance of these two elements is

$$Z_{EQ} = \frac{Z_LZ_C}{Z_L+Z_C}$$
So, if either \$Z_L\$ or \$Z_C\$ approaches zero then \$Z_{EQ}\$ must approach zero. Now, \$Z_L \propto \omega\$ and \$Z_C \propto 1/\omega\$, so as \$\omega\$ approaches 0 or \$\infty\$ we know that \$Z_{EQ}\$ will approach 0.

When \$\omega = \frac{1}{\sqrt{LC}}\$ we find that \$Z_L = -Z_C\$, so the denominator of the \$Z_{EQ}\$ equation approaches 0, and the magnitude of \$Z_{EQ}\$ approaches \$\infty\$.

I'm not sure what you want to do with your calculator, but if you can program the formula for \$Z_{EQ}\$, using \$Z_L = j\omega L\$ and \$Z_C = \frac{1}{j\omega C}\$, then sweep \$\omega\$ you will see this behavior. Note that you must do these calculations using complex arithmetic, and your calculator may have trouble at the point where \$\omega = \frac{1}{\sqrt{LC}}\$.