Consider the function y=2x3+3x2−18x+3
Differentiate with respect to x as,
dxdy=dxd(2x3+3x2−18x+3)
=2(3x2)+3(2x)−18(1)+0
=6x2+6x−18
Set dxdy=−6 (which is the slope of the tangent line) and solve for x as,
6x2+6x−18=−6
x2+x−3=−1
x2+x−2=0
x2+2x−x−2=0
x(x+2)−1(x+2)=0
(x−1)(x+2)=0
x=−2,1
At x=−2,y=35 and at x=1,y=−10
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