Answer to Question #172677 in Calculus for Todd Phillips

Question #172677

Determine all the points on the curve y = 2x3 + 3x2 - 18x + 3 where the slope of the tangent line is -6.


1
Expert's answer
2021-03-31T13:54:09-0400

Consider the function y=2x3+3x218x+3y=2x^3+3x^2-18x+3


Differentiate with respect to xx as,


dydx=ddx(2x3+3x218x+3)\frac{dy}{dx}=\frac{d}{dx}(2x^3+3x^2-18x+3)


=2(3x2)+3(2x)18(1)+0=2(3x^2)+3(2x)-18(1)+0


=6x2+6x18=6x^2+6x-18


Set dydx=6\frac{dy}{dx}=-6 (which is the slope of the tangent line) and solve for xx as,


6x2+6x18=66x^2+6x-18=-6


x2+x3=1x^2+x-3=-1


x2+x2=0x^2+x-2=0


x2+2xx2=0x^2+2x-x-2=0


x(x+2)1(x+2)=0x(x+2)-1(x+2)=0


(x1)(x+2)=0(x-1)(x+2)=0


x=2,1x=-2,1


At x=2,y=35x=-2,y=35 and at x=1,y=10x=1,y=-10


Therefore, the points on the curve are: (2,35),(1,10)(-2,35),(1,-10).

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