\(y'=3x^2-6x-9=3\left(x^2-2x-3\right)\)
\(y'=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)
Lập BBT ta thấy \(y\) đồng biến trên \(t< -1\cup t>3\) và nghịch biến trên \(-1< t< 3\)
\(\Rightarrow y_{max}=y\left(-1\right)=\left(-1\right)^3-3.\left(-1\right)^2-9\left(-1\right)+5=10\Rightarrow A\left(-1;10\right)\)
\(y_{min}=y\left(3\right)=3^3-3.3^2-9.3+5=-22\Rightarrow B\left(3;-22\right)\)
\(\overrightarrow{AB}=\left(4;-32\right)\Rightarrow\overrightarrow{n_p}=\left(32;4\right)=4\left(8;1\right)\)
\(\Rightarrow\left(AB\right):8\left(x+1\right)+\left(y-10\right)=0\)
\(\Rightarrow\left(AB\right):8x+y-2=0\)
\(\left(AB\right)\cap\left(Ox\right)=I\left(x;0\right)\Leftrightarrow8x+0-2=0\Leftrightarrow x=\dfrac{1}{4}\Rightarrow I\left(\dfrac{1}{4};0\right)\)
\(IA=\sqrt{\left(-1-\dfrac{1}{4}\right)^2+\left(10-0\right)^2}=\dfrac{\sqrt{1625}}{4}\)
\(IB=\sqrt{\left(3-\dfrac{1}{4}\right)^2+\left(-22-0\right)^2}=\dfrac{\sqrt{7865}}{4}\)
\(\Rightarrow\dfrac{IA}{IB}=\sqrt{\dfrac{1625}{7865}}=\sqrt{\dfrac{25}{121}}=\dfrac{5}{11}=\dfrac{b}{c}\)
\(\Rightarrow\left\{{}\begin{matrix}b=5\\c=11\end{matrix}\right.\)
\(\Rightarrow T=b-c=5-11=-6\)
