# tgnts n norms

• Dec 4th 2005, 12:27 PM
x-disturbed-x
tgnts 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 PM
CaptainBlack
Quote:

Originally Posted by x-disturbed-x
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..

There is only one point at which the gradient is -14 and yes you do solve:

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

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 PM
x-disturbed-x
y 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 PM
CaptainBlack
Quote:

Originally Posted by x-disturbed-x
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?

The x-coordinate of the point at which the derivavtive is -14 is
-2.

Since the required point is on the curve $y\ =\ 3.x^2\ -\ 2.x\ +\ 1$, 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 PM
x-disturbed-x
y = 17?
• Dec 4th 2005, 02:00 PM
CaptainBlack
Quote:

Originally Posted by x-disturbed-x
y = 17?

Yep.

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