Question

Find the points on the curve y = 2x^3 - 12 x^2 + 18 x + (-3) at which the tangent line is horizontal.

The points that have a horizontal tangent line are .
(You are to give the points, not only the x-values. If there are more than one point, separate the points with a comma in the answer space.)

Answer 1

(1,0),(3,0) thats what i got for the answer but this is incorrect

Answer 2

please help

Answer 3

y=2x^3 - 12x^2 + 18x - 3
dy/dx= 6x^2 -24x +18

tangent line is horizontal, dy/dx=0
6x^2 -24x +18=0
(x-3)(x-1)=0
x=1,3

when x=1, y= 2(1^3)-12(1^2)+18(1)-3= 5
when x=3, y=-3

(1,5) and (3,-3)
This is correct.

Answer 4

thanks

Answer 5

thanks

Answer 6

[y=2x^3−12x^2+18x+(−3)\]\[y′=6x^2−24x+18=6(x^2−4x+3)=6(x−1)(x−3)=0→x=1,x=3\]\

Answer 7

y(1)=5
y(3)=-3

Answer 8

Same as above.