This is not answer as requested. It is a explanation that can hopefully give you a more intuitive understanding to help you come up with your proofs.
A linear function is any function that satisfies two properties: $ \gamma f(\vec x)= f(\gamma \vec x) $ and $ f(\vec x + \vec y)= f( \vec x)+ f(\vec y)$ where $\vec x \vec y $ are vectors and $\gamma $ is a scalar.
You're instinct to think of a parabola is correct. What you want is a function that doesn't satisfy the two properties. Basically a linear function is a function that geometrically speaking can stretch and change directions of vectors but does not transform a straight line into a any graph with a curvature. This has to do with the fact that the first property $ \gamma f(\vec x)= f(\gamma \vec x) $ only stretches or contracts lines and by the second property $ f(\vec x + \vec y)= f( \vec x)+ f(\vec y)$ that all points in a vector space remain proportional distance apart after a linear transformation.
I highly recommend checking out this video that uses very elegant computer graphics to make this point intuitively which talks about the the geometric movement of linear functions:
https://www.youtube.com/watch?v=kYB8IZa5AuE&index=4&list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab
** Also a linear function always maps a $\vec 0 $ to $\vec 0$ which is why JMoravitz solution works.
For your second question it is basically the same as the first. Hint $f(x)= e^x$ which maps the x-axis, a straight line, to the curve of $e^x$. Now think of this with context of $\mathbb R^2 \rightarrow \mathbb R^2$.
Hope this helps.