Cho đồ thị hàm số bậc ba $y=f\left( x...

Ta có .
Do đồ thị hàm số có hai điểm cực trị là và nên
$\left\{ \begin{aligned}
& 3a-2b+c=0 \\
& 3a+2b+c=0 \\
& -a+b-c+d=4 \\
& a+b+c+d=0 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& a=1 \\
& b=0 \\
& c=-3 \\
& d=2 \\
\end{aligned} \right.$$\Rightarrow y={{x}^{2}}-3x+2d:y=mx+n\left( -2\text{ ; 0} \right),\left( \text{0 ; 2} \right)d:y=x+2{{S}_{1}}=\dfrac{1}{2}{{.2}^{2}}+\int\limits_{0}^{1}{\left| {{x}^{3}}-3x+2 \right|\text{d}x=}2+\int\limits_{0}^{1}{\left( {{x}^{3}}-3x+2 \right)\text{d}x=}=2+\left. \left( \dfrac{{{x}^{4}}}{4}-\dfrac{3{{x}^{2}}}{2}+2x \right) \right|_{0}^{1}=\dfrac{11}{4}{{S}_{2}}=\int\limits_{0}^{2}{\left| \left( x+2 \right)-\left( {{x}^{3}}-3x+2 \right) \right|\text{d}x=}\int\limits_{0}^{2}{\left( x+2-{{x}^{3}}+3x-2 \right)\text{d}x=}\int\limits_{0}^{2}{\left( -{{x}^{3}}+4x \right)\text{d}x=}4\Rightarrow \dfrac{{{S}_{1}}}{{{S}_{2}}}=\dfrac{p}{q}=\dfrac{11}{16}p+q+2022=2049$.