Answer
$y=-2$
Work Step by Step
For x=$-10$, the point on the graph is $(-1, f(-1))$.
The slope of the tangent at x=$-1$ is $m=f^{\prime}(-1).$
The point-slope equation of the tangent line is
$y-y_{1}=m(x-x_{1})$
$y-f(-1)=f^{\prime}(-1)(x-(-1))$
$f(x)=\displaystyle \frac{x^{2}+1}{x}=\frac{x^{2}}{x}+\frac{1}{x}=x+x^{-1}$
$f^{\prime}(x)=1-x^{-2}=1-\displaystyle \frac{1}{x^{2}}$
$f(-1)=\displaystyle \frac{1+1}{-1}=-2$
$f^{\prime}(-1)=1-1=0$
An equation for the tangent line is
$y-(-2)=0\cdot(x+1)$
$y=-2$
