Question

Graph each polynomial

f(x)=(x-2)(x+1)(x+3)

Answer 1

It cuts x-axis at x=-3,-1,2
It cuts y-axis at f(x)=(0-2)(0+1)(0+3)=-6
f(x)=(x-2)(x^2+4x+3)=x^3+2x^2-5x-6
\[f \prime \left( x \right)=3x ^{2}+4x-5\]
Tangent is parallel to x-axis where \[3x ^{2}+4x-5=0,x=\frac{ -4\pm \sqrt{16+60} }{ 6 }\]
\[x=\frac{ -4\pm2\sqrt{19} }{6}=\frac{ -2\pm \sqrt{19} }{3 }\]
f(x) increases where f(x) >0
\[x<\frac{ -2-\sqrt{19} }{ 3 } and x>\frac{ -2+\sqrt{19} }{3}\]
f(x) decreases where f(x)<0,
or \[\frac{ -2-\sqrt{19} }{3 }<x<\frac{ -2+\sqrt{19} }{3 }\]
\[f \prime \prime \left( x \right)=6x+4\]
\[at x=\frac{ -2-\sqrt{19} }{3 },f \prime \prime \left( x \right)<0\]
therefore there is maxima at this point.
\[at x=\frac{ -2+\sqrt{19} }{ 3 },f \prime \prime \left( x \right)>0\]
hence there is minima at this point.
find maximum and minimum f(x) at these points.
and then draw keeping in mind above points.