....sorry should have made it clearer
its y= 2/ (x^2+3)
no sorry- i have never heard of a truncus- i am aware that this is an even function if that has anything to do with it 
No problem.
Ok, well we can differentiate that function in much the same way, but we'll also need the chain rule:
^{-1}\\y&=2u^{-1}\\u&=x^2+3\\\frac{dy}{du}&=-2u^{-2}\\\frac{du}{dx}&=2x\\\therefore\frac{dy}{dx}&=-\frac{2}{u^2}\cdot 2x\\&=-\frac{4x}{\left(x^2+3\right)^2}\end {align*})
From this, it is clear that the derivative is zero at \(x=0\).
To find the second derivative, we can use the quotient rule:
^2\\\frac{dv}{dx}&=2x\left(x^2+3\right)\cdot 2\\&=4x\left(x^2+3\right)\\\frac{d^2y}{dx^2}&=\frac{\frac{du}{dx}\cdot v-\frac{dv}{dx}\cdot u}{v^2}\\&=\frac{-4\left(x^2+3\right)^2-4x\left(x^2+3\right)\left(-4x\right)}{\left(x^2+3\right)^4}\\&=-\frac{4\left(x^2+3\right)\left(x^2+3-4x^2\right)}{\left(x^2+3\right)^4}\\&=-\frac{4\left(x^2+3\right)\left(3-3x^2\right)}{\left(x^2+3\right)^4}\end{align*})
This is only zero when \(3-3x^2=0\), that is at \(x=\pm 1\).
Additionally, \(\frac{d^2y}{dx^2}>0\) for \(x<-1\) and \(x>1\), and \(\frac{d^2y}{dx^2}<0\) for \(-1<x<1\), which means that there are points of inflexion at \(x=\pm 1\).
For the stationary point, \(y\left(0\right)=\frac{2}{3}\) and for the points of inflexion, \(y\left(-1\right)=y\left(1\right)=\frac{1}{2}\). Additionally, \(\frac{d^2y}{dx^2}=-\frac{4}{9}\) at \(x=0\) so \(\left(0,\frac{2}{3}\right)\) is a local maximum.
For the asymptote, \(\lim_{x\rightarrow\pm\infty}\left(\frac{2}{x^2+3}\right)=0\), so \(y=0\) is an asymptote (and is, in fact, the only asymptote).
Finally, to sketch the graph, it is easiest to sketch the parabola, \(y=\frac{x^2+3}{2}\) first, then sketch its reciprocal on the same axes:
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