What are the coordinates of the points on the curve

Question

y=x^3-2x^2 where the gradjent is -1.?

 

Is it:
- (1,-1)
- (-1,1)
- (1/3,-5/27)
- (-1/3,5/27)
- (1/3,-1)

 

Answer 1

The Curve Equation is y = x³ - 2x².

Gradient or slope is -1.

Now y = x³ - 2x².

To find the slope of the Equation

Differentiate with respect to x on both sides.

y' = 3x² - 4x

m = 3x² - 4x

-1 = 3x² - 4x

3x² - 4x + 1 =0.

Roots of the Quadratic Equation 3x² - 4x + 1 =0.

3x² - 4x + 1 =0

3x² - 3x - x + 1 = 0

3x(x-1) -1(x-1) = 0

(3x-1)(x-1) = 0

x = 1/3 and 1

Therefore the the values of x is (1/3 , 1).