The graph show the behaviour of a length of wire in the region for which the substance obeys Hooke's law. P and Q represent.

We know that by Hooke's  law,

$Y=\frac{{\displaystyle \frac{F}{A}}}{{\displaystyle \frac{x}{L}}}$

$\mathrm{F}=\left(\frac{\mathrm{AY}}{\mathrm{L}}\right)\mathrm{x}$

$$\begin{gathered} \mathrm{W}={\int }_0^{\mathrm{x}}\mathrm{F}.\mathrm{dx} \\ \mathrm{W}=\left(\frac{\mathrm{AY}}{2\mathrm{L}}\right){\mathrm{x}}^2 \end{gathered}$$

The stored elastic energy,

$U=W=\left(\frac{AY}{2L}\right)x^2$

Because of the elastic range, the graph between applied force and extension will be linear, but the graph between extension and stored elastic energy will be parabolic in shape.

Hence the correct answer is extension, stored elastic energy .