$\color{blue}{\boxed{\color{red}{(} \color{#FF5500}{っ} \color{#FFAA00}{◔} \color{#FFFF00}{◡} \color{#AAFF00}{◔} \color{#55FF00}{)} \color{#00FF00}{っ} \color{#00FF55}{ ♥}\begin{matrix}\color{#00FFAA}{t}\color{#00FFFF}{r} \color{#00AAFF}{i} \color{#0055FF}{e}\color{#0000FF}{t} \color{#5500FF}{s} \color{#AA00FF}{i} \color{#FF00FF}{g} \color{#FF00AA}{a} \color{#FF0055}{i}\\\color{#00FFAA}{♥} \color{#00FFFF}{♥} \color{#00AAFF}{♥} \color{#0055FF}{♥}\color{#0000FF}{♥} \color{#5500FF}{♥} \color{#AA00FF}{♥} \color{#FF00FF}{♥} \color{#FF00AA}{♥} \color{#FF0055}{♥}\\\color{#00FFAA}{n} \color{#00FFFF}{h} \color{#00AAFF}{i} \color{#0055FF}{x}\color{#0000FF}{i} \color{#5500FF}{n} \color{#AA00FF}{h} \color{#FF00FF}{i} \color{#FF00AA}{u}\end{matrix}}}$

$D=|\mathbb{R}$

$y'=\frac{(x^2-3x+2)'(x^2-3x+2)}{|x^2-3x+2|}$

$~~~=\frac{2x-3)(x^2-3x+2)}{|x^2-3x+2|}$

$~~~=\frac{2x^3-9x^2+13x-6}{|x^2-3x+2|}$

+) $ |x^2-3x+2|>0\forall x\in \mathbb{R}$\$\{1;2\}$

+) $2x^3-9x^2+13x-6=0$

$\Rightarrow x=1;~x=1,5;~x=2$

+) $a=2 >0 \Rightarrow$ BBT:

$\begin{array}{|c|ccccc|}\hline \text{x}&\text{$-\infty$}&\text{$1$}&\text{$1,5$}&\text{$2$}&\text{$+\infty$}\\\hline \text{y'}&\text{}&\text{$-~~~~~~0~~~~~+$}&\text{$~~~~~~0~~~~-$}&\text{0}\\\hline\end{array}$
Vậy hàm có 3 cực trị tại x=1; x=2; x=1,5

$\begin{matrix}\color{#00FF55}{ ♥}\\\color{#00FF55}{ ♥}\\\color{#00FF55}{ ♥}\end{matrix}\begin{matrix}\color{#00FFAA}{♥} \color{#00FFFF}{♥} \color{#00AAFF}{♥} \color{#0055FF}{♥}\color{#0000FF}{♥} \color{#5500FF}{♥} \color{#AA00FF}{♥} \color{#FF00FF}{♥} \color{#FF00AA}{♥} \color{#FF0055}{♥}\\\color{#00FFAA}{t}\color{#00FFFF}{r} \color{#00AAFF}{i} \color{#0055FF}{e}\color{#0000FF}{t} \color{#5500FF}{s} \color{#AA00FF}{i} \color{#FF00FF}{g} \color{#FF00AA}{a} \color{#FF0055}{i}~~~~\color{#FF0055}{♥}\\\color{#00FFAA}{♥} \color{#00FFFF}{♥} \color{#00AAFF}{♥} \color{#0055FF}{♥}\color{#0000FF}{♥} \color{#5500FF}{♥} \color{#AA00FF}{♥} \color{#FF00FF}{♥} \color{#FF00AA}{♥} \color{#FF0055}{♥}\end{matrix}$