Answer
$$ y=x-1, f'(1)=1$$
Work Step by Step
Given $$f(x)=x^{2}-x, \quad a=1$$
Since $f(1)=0$ and
$$ f^{\prime}(a) =\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}$$
Then
\begin{aligned}
f^{\prime}(1) & =\lim _{h \rightarrow 0} \frac{(1+h)^{2}-(1+h)-\left(0\right)}{h} \\
&=\lim _{h \rightarrow 0} \frac{1+2 h+h^{2}-1-h}{h}\\
&=\lim _{h \rightarrow 0}(1+h)=1
\end{aligned}
Then the tangent line at $x=1$ is given by
\begin{align*}
\frac{y-y_1}{x-x_1}&=f'(a)\\
\frac{y}{x-1}&=1\\
\end{align*}
Hence $$ y=x-1$$
