The animation below shows values of x being related to values of y by a relationship named f. Play the video, then answer the questions following the video using meanings for terms in the question. You can replay the video and pause it, or scroll the play bar to examine parts of the video. (Move your pointer away from the video to remove the scroll bar.)
The value of x appears to vary through all values from -7 to 7. Each of these values is related to a value of y. The domain of f is therefore all values of x such that $-7\le x \le 7$.
The value of y appears to range from -5 to 8 (and all values between) as the value of x varies from -7 to 7.
Yes, f is a function from x to y. Each value of x in the domain of f is related to exactly one value of y.
f is not continuous over its domain. It appears from the video that f is discontinuous at $x=0$. The value of y "jumps" from 0 to -5 as the value of x increases from 0. Put more formally, Let $\epsilon=1$. There is no interval around $x=0$ such that $|f(u)-f(v)|\lt \epsilon$ for all values of u and v in that interval. $f(x)$ will be near 0 for negative values of x near 0 and $f(x)$ will be near -5 for positive values of x near 0.
f is not 1-1 over its domain. $f(-7)=7.5$ and $f(6)=7.5$. Therefore, there exist two values of x related related by f to the same value of y.
Another counterexample to a claim that f is 1-1 is that all values of x in $[6.5,7]$ (i.e., from 6.5 to 7 inclusive) are associated with $y=8$.
It appears $f(u)\lt f(v)$ when $u\lt v$ for u and v in the domain of f and $0\le u,v\le 6.2$.
Another way to say this is the value of y increases as the value of x increases from 0 to 6.2.
It appears $f(u)\gt f(v)$ when $u\lt v$ for u and v in the domain of f and $u,v\lt 0$.
Another way to say this is the value of y decreases as the value of x increases from -7 to 0.