find the co-ordinates
y=x(2-x) where the gradient is 2
y = 2x-x^2
y' = 2-2x
so, where is 2-2x = 2?
the gradient is the slope of the tangent to the curve. y' (x) gives the slope of the tangent for any value of x.
So, at x=0, y'(x) = 2-0 = 2, so the gradient is 2 at the point (0,2)
See the graphs at
http://www.wolframalpha.com/input/?i=plot+y%3D2x-x^2%2Cy%3D2x
To find the coordinates where the gradient (or slope) of the equation y = x(2 - x) is equal to 2, we need to find the first derivative of the equation and solve it for x.
Step 1: Find the derivative of y = x(2 - x) with respect to x.
To do this, we can use the product rule of differentiation.
Applying the product rule, the derivative of y with respect to x is:
dy/dx = (2 - x) + x(-1)
= 2 - x - x
= 2 - 2x
Step 2: Set the derivative equal to 2 and solve for x.
Setting dy/dx = 2 gives us the equation:
2 - 2x = 2
Simplifying the equation:
-2x = 0
Dividing both sides of the equation by -2:
x = 0
So, the value of x is 0.
Step 3: Substitute the value of x into the original equation to find the y-coordinate.
Substituting x = 0 into the equation y = x(2 - x), we get:
y = 0(2 - 0)
= 0
So, the coordinates where the gradient is 2 are (0, 0).