y = x^3 - 8x^2 + 6x -2 has a slope of 6 at two points, what are the coordinates?
I used Reiny's way to help solve this answer.
Please check if I've done it correct: y'= 3x^2-16x+6
3x^2-16x+6=6
3x^2-16x=0
x(3x-16)=0
Solve for x's
x=0
3x-16=0 ~> x=16/3
Plug x values to original function
(0)^3 - (8)(0)^2 + 6(0) - 2 = y ~> -2
(16/3)^3 - (8)(16/3)^2 + 6(16/3) - 2 = y ~> -1238/27
Coordinates will be, (0,-2) & (16/3, -1238/27)
2 answers
Reiny already did this for you. Yes, right.
@Damon,
Not sure why, but my post was posted twice. But I saw her response, thank you !
Not sure why, but my post was posted twice. But I saw her response, thank you !
7 answers