Find The Equation This is the question:
Find the equations of the tangent lines to the curve $y = x − \frac 1x + 1$ that are parallel to the line $x − 2y = 3$.
There are two answers: 1) smaller y-intercept; 2) larger y-intercept
The work:
The slope of the line is (1/2).
$y' = ((x + 1) - (x - 1))/(x + 1)^2 = 2/(x + 1)^2 = 1/2$
=> $(x + 1)^2 = 4$
=> $(x + 1) = 2$ and $ x + 1 = -2$
=> $x = 1$ and $x = -3$
At $x = 1$, $y = (x - 1)/(x + 1) = 0$
The equation of the tangent is $y/(x - 1) = (1/2)$
=> $2y = x - 1$
=> $x - 2y - 1 = 0$
At $x = -3$,$ y = (x - 1)/(x + 1) = -4/-2 = 2$
The equation of the tangent is $(y - 2)/(x + 3) = (1/2)$
=> $2y - 4 = x + 3$
=>$ x - 2y + 7 = 0$

there are no points on the graph of $y = x - \frac1x + 1$ has a tangent that is parallel to $x - 2y = 3.$
i will explain why.
you will get a better idea of the problem if you can draw the graph of $y = x - \frac1x + 1$ either on your calculator or by hand. the shape of the graph is called a hyperbola, similar looking to the rectangular hyperbola $y = \frac1x$ but here the asymptotes $x = 0$ and $y = x+1$ are not orthogonal. you will two branches one in the left half plane and the other in the right half plane. if you at the right branch, you can see that the slope starts very large positive fro $x$ very close to zero and positive, then gradually decreases to match the slope $1$ of the asymptote. that is the smallest slope on this graph. but the line you have $x - 2y = 3$ has slope $\frac12$ that is smaller than $1,$ the least slope on the graph.
here is how you see with calculus. taking the derivative of $y = x - \frac1x + 1$ you find $$\frac{dy}{dx} = 1 + \frac1{x^2} \ge 1 \text{ for all} x. $$
