<s>[INST] Intersection between $y=\frac{x^2+1}{x+1}$ and $y=x+2$ on $y$-axis I stumbled upon this problem: Considering the function $f:(-1;\infty)\rightarrow R$,  $$f(x)=\frac{x^2+1}{x+1}$$ Find the sum of the values on positive $y$-axis where the tangent of the function is perpendicular to the line $y=x+2$ 
After  letting $(0,a)$ be the point on positive $y$-axis, with the equation of the tangent $y-{{y}_{0}}=f'({{x}_{0}})(x-{{x}_{0}})$ gives slope to be  $-1$ So the tangent  is $y-a=-1(x-0)=-x$ 
Is this correct? Can you help me furthermore with this problem?
 [/INST] $f(x)=\frac{x^2+1}{x+1}$
$\implies f'(x) = \frac{2x(x+1)-x^2-1}{(x+1)^2}$
$\implies f'(x) = \frac{2x^2+2x-x^2-1}{(x+1)^2}$
$\implies f'(x) = \frac{x^2-1+2x}{(x+1)^2}$
set this to $-1$ as tangent of $f(x)$ is perpendicular to $y=x+2$
$\frac{x^2-1+2x}{(x+1)^2}=-1$
$x^2-1+2x=-(x^2+1+2x)$
$2x^2+4x =0$
$2x(x+2)=0$
$\therefore x=0,-2$
So at $x=0,-2$ the tangents drawn will be perpendicular to the line $y-x+2$
On the OY axis lies $x=0,y=1$, So at the point $(0,1)$ the tangent of the function $f(x)$ is perpendicular to $y=x+2$
EDIT 1:
graph for reference 
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