How to plot $\frac{2x-4}{1+x^2}$

Question

Can someone suggest me how to plot this function (without any software)?

$$\frac{2x-4}{1+x^2}$$

I have found out such points as:

1. $(0,-4)$ and $(2,0)$
2. I know that $y = 0$ is horizontal asymptote (when $x \rightarrow\pm\infty$)
3. I have found out also extrema points and those are $(x=2\pm\sqrt(5))$
4. And some other stuff, but not sure how to draw a plot

Answer 1

Here I gave a method without Calculus.

Let $$y=\frac{2(x-2)}{(x^2+1)}.$$ Note that $$\lim_{x\to\pm\infty}y=0$$ $y$ has only one zero $x=2$ and $$y\ge 0\iff x\ge 2.$$ Also $$yx^2-2x+(y+4)=0$$ Since we consider only real $x$ values discriminant of this equation is non-negative. $$\Delta_x=2^2-4y(y+4)\ge 0$$ $$y^2+4y-1\le 0$$ $$(y+2)^2-5\le 0.$$ Therefore $y$ is always in the range $$-2-\sqrt5\le y\le -2+\sqrt5.$$ Find $x$ values corresponding to $y=-2\pm\sqrt5.$
Now you can draw the graph of your function using these details. your graph will looks like

Answer 2

Find inflection points $ y^{''} =0 $ with tangent slopes $y^{'}$ , maximum/minimum points and the asymptotes by standard procedure ( divide by $ x^2, x\rightarrow \infty $ etc. ), find coordinates of points for $ y=0, x=0, $ joining them with a smooth curve. Connect the curve at a few points between these points calculated by hand.