Sketching the graph of a cubic function

Question

Consider $f: \mathbb R \rightarrow \mathbb R$ be the function $f(x) = x^3 + ax^2 + bx + c$. I have to sketch the graph of this function. I can find the solutions of $f'(x) = 0$ but it is difficult to find the intervals in which the function is increasing or decreasing since nothing is mentioned about $a$, $b$ and $c $, i.e. whether they are positive or negative and their relative positions in the number line.

Answer 1

WLOG, $a=c=0$ and $b\in\{-1,0,1\}$ ! Why ? Because

1. translating the curve horizontally, i.e. turning the equation to $$(x+t)^3+a(x+t)^2+b(x+t)+c\\=x^3+(a+3t)x^2+(2at+3t^2+b)x+at^2+bt+t^3+c$$ allows you to choose $t$ such that the quadratic coefficient vanishes.

2. varying the constant term in the general cubic equation $ax^3+bx^2+cx+d=0$ merely translates the curve vertically, which does not alter its shape. So the resulting constant term after the above manipulation, with the substituted value of t, can be ignored and the shape of the graph remains the same.

3. Next, by scaling $x$ with $x=\sqrt{|b|}y$ you have $$x^3+bx=|b|\sqrt{|b|}y^3+b\sqrt{|b|}y=|b|\sqrt{|b|}(y^3\pm y)$$ and when $b=0$, just $x^3$.

So we can show the three canonical shapes of any cubic, obtained by linear transformations of the coordinates.

The curve has a central symmetry at a single inflection point, and two extrema or none.