y = 3+2x-x^2
dy/dx = 2-2x.
Therefore the gradient of point P with x-coordinate 1 is 2-2(1) = 0
the coordinate of point P is (1,y) <-- since y coordinate is unknown.
(1,y) satisfy the equation y = 3+2x-x^2
y = 3+2x-x^2
dy/dx = 2-2x.
Therefore the gradient of point P with x-coordinate 1 is 2-2(1) = 0
the coordinate of point P is (1,y) <-- since y coordinate is unknown.
(1,y) satisfy the equation y = 3+2x-x^2
I understand how you have got the Gradient but not how you tell which is actually the Y coordinate.
As for what i know of a derivative of a function - not much, had one lesson with my teacher on it and understood about 20% of what she said.