The equation of a curve is y = 3x^2 - 2x + 1

find the coordinates of the point R on the curve at which the gradient is -14.

do i put the result of dy/dx to equal -14? doing that will not find me 2 points though..

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- Dec 4th 2005, 12:27 PMx-disturbed-xtgnts n norms
The equation of a curve is y = 3x^2 - 2x + 1

find the coordinates of the point R on the curve at which the gradient is -14.

do i put the result of dy/dx to equal -14? doing that will not find me 2 points though.. - Dec 4th 2005, 12:49 PMCaptainBlackQuote:

Originally Posted by**x-disturbed-x**

$\displaystyle \frac{dy}{dx}\ =\ -14$

to find the x-coordinate of the point, then substitute that value back

into the original equation to find the corresponding y.

RonL - Dec 4th 2005, 01:01 PMx-disturbed-xy y y
i know the gradient and the x coordinate. the original equation before differentiation / simplification was

y = 3x^2 - 2x + 1

how do i find the y coordinate? - Dec 4th 2005, 01:23 PMCaptainBlackQuote:

Originally Posted by**x-disturbed-x**

-2.

Since the required point is on the curve $\displaystyle y\ =\ 3.x^2\ -\ 2.x\ +\ 1$, and we

already know the x-coordinate, we need only substitute

this into the equation for y will give the y-coordinate.

RonL - Dec 4th 2005, 01:48 PMx-disturbed-x
y = 17?

- Dec 4th 2005, 02:00 PMCaptainBlackQuote:

Originally Posted by**x-disturbed-x**

RonL - Dec 4th 2005, 02:34 PMx-disturbed-x
Thanks for all your help :)