A differentiation/derivative/calculus problem The question is as follows: 
$$y=x^2/(x+1)$$ 
The normal to this curve at $x=1$ meets the $x$-axis at point $M$.
The tangent to the curve at $x=-2$ meets the $y$-axis at point $N$. 
Find the area of triangle $MNO$, where $O$ is the origin.
PS- this is not a school h.w  so don't worry. And I  did try....for  a good 100  min...not even joking
EDIT - The drawing is NOT EXACT, it is just to give an idea.


$y(x) = \dfrac{x^2}{x + 1} = (x + 1)^{-1}x^2; \tag 1$
$y'(x) = -(x + 1)^{-2}x^2 + 2x(x + 1)^{-1} = -(x + 1)^{-2}x^2 + 2x(x + 1)(x + 1)^{-2}$
$= -(x + 1)^{-2}x^2 + (2x^2 + 2x)(x + 1)^{-2} = (x^2 + 2x)(x + 1)^{-2} = \dfrac{x^2 + 2x}{(x + 1)^2}; \tag 2$
$y(1) = \dfrac{1}{2}; \; y'(1) = \dfrac{3}{4}; \tag 3$
$y(-2) = -4; \; y'(-2) = 0; \tag 4$
the slope of the normal line through $(1, y(1)) = (1, 1/2)$ is then
$m = -\dfrac{1}{y'(1)} = -\dfrac{4}{3}; \tag 5$
the equation of the normal line through $(1, 1/2)$ is thus
$y - \dfrac{1}{2} = -\dfrac{4}{3}(x - 1), \tag 6$
which meets the $x$-axis where $y = 0$:
$-\dfrac{1}{2} = -\dfrac{4}{3}(x - 1) \Longrightarrow x = \dfrac{3}{8} + 1 = \dfrac{11}{8}; \tag 7$
thus,
$M = \left (\dfrac{11}{8}, 0 \right); \tag 8$
likewise, the tangent line through  $(-2, y(-2)) = (-2, -4)$ is
$y + 4 = 0(x + 2) = 0, \tag 9$
which intersects the $y$-axis where $x = 0$, with $y$-coordinate given by
$y + 4 = 0 \Longrightarrow y = -4, \tag{10}$
and so
$N = \left (0,  -4 \right ); \tag{11}$
the area $A$ of $\triangle MNO$ is thus
$A = \dfrac{1}{2}ON \cdot OM = \dfrac{1}{2} 4 \cdot  \dfrac{11}{8} = \dfrac{11}{4}. \tag{12}$
