1. ## Differentiation

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 $\displaystyle y=3+2x-x^2$

Any help much appreciated

2. ## Re: 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

3. ## Re: Differentiation

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 $\displaystyle y=3+2x-x^2$
$\displaystyle y(1) = 3 + 2(1) - (1)^2 = 4$

the gradient at that point is $\displaystyle y'(1)$, the derivative of $\displaystyle y$ evaluated at the point in question.

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

4. ## Re: Differentiation

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.

5. ## Re: Differentiation

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.