given.......(2x^2-3x+1)e^x
for the turning point, dy/dx=0
from the product rule
dy/dx=vdu/dx+udv/dx
0=e^x(4x-3)+(2x^2-3x+1)e^x
0=e^x{(4x-3)+(2x^2-3x+1)}
2x^2+x-2=0
using the general formula for quadratic eqns
x=1.28 or -0.78
from
y=(2x^2-3x+1)e^x
when x=1.28
y={2(1.28)^2-3(1.28)+1}e^1.28
y=3.6
when x=-0.78
y={2(-0.78)^2-3(-0.78)+1}e^-0.78
y=0.6
the turning point is (x,y)=(1.28,3.6) , (-0.78,0.6)