Thread: Differentiation

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

Originally Posted by jacobjacob44

I am having trouble understanding the method for finding the y-coordinate.

My question is:
Find the y-Coordinate and the value of the dradient at the point P with x-coordinate 1 on the curve with wquation

the gradient at that point is , the derivative of evaluated at the point in question.

what do you know about finding the derivative of a function?

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.

Originally Posted by jacobjacob44

Apologies i should have put it in calculus yes. 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.

y = f(x) ... if x = 1 , then y = f(1)

it's that simple.