Answer
$f(x)=x^{2}-1.$
Work Step by Step
A function with $f^{\prime}(x)=2x$ (for all x)
has form $f(x)=x^{2}+c$,
where c is some constant.
The point $(1,0)$ is on the graph, so
$f(1)=0$
Since $f(1)=1^{2}+c,$
$0=1+c$
it follows that $c=-1$ and
$f(x)=x^{2}-1.$
