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)
correct, see my last response to your previous post of this
Im sorry i would help but im not even there in math yet im in 6 grade what grade are you in
@Reiny,
Thanks again!
Yes, your solution using the derivative and setting it equal to 6 is correct. Let's go through each step to make sure we get the correct coordinates:
1. Start with the given function: y = x^3 - 8x^2 + 6x - 2
2. Find the derivative of the function. You correctly found the derivative as y' = 3x^2 - 16x + 6.
3. Set the derivative equal to the given slope of 6: 3x^2 - 16x + 6 = 6
4. Simplify the equation by subtracting 6 from both sides: 3x^2 - 16x = 0
5. Factor out the common factor of x: x(3x - 16) = 0
6. Set each factor equal to zero and solve for the x-values:
- x = 0
- 3x - 16 = 0
=> 3x = 16
=> x = 16/3
7. Plug each x-value back into the original function to find the corresponding y-values:
- For x = 0: y = (0)^3 - 8(0)^2 + 6(0) - 2 = -2
- For x = 16/3: y = (16/3)^3 - 8(16/3)^2 + 6(16/3) - 2 = -1238/27
8. Therefore, the two points where the slope is 6 are: (0, -2) and (16/3, -1238/27)
Great job using the derivative to find the points!