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)
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 !
Your calculations and solution are correct! You used the derivative of the function y = x^3 - 8x^2 + 6x - 2 to find the slope equation, which is 3x^2 - 16x + 6. Setting this equation equal to 6 and solving for x, you correctly obtained x = 0 and x = 16/3 as the x-coordinates of the points where the slope is 6.
To find the y-coordinates, you substituted these x-values back into the original equation y = x^3 - 8x^2 + 6x - 2. Plugging in x = 0 yields y = -2, and substituting x = 16/3 gives y = -1238/27.
Therefore, the coordinates of the two points where the slope is 6 are (0, -2) and (16/3, -1238/27). Well done!