hi,

i need to work out the gradient at following points at the given curve:

y=3x^2-4x+1 at x=1

i know how to start it but after that im stuck and require some help, please!

i can do this:

dy/dy= 6x-4

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- Oct 7th 2011, 09:05 AMandyboy179gradient at following points at the given curve?
hi,

i need to work out the gradient at following points at the given curve:

y=3x^2-4x+1 at x=1

i know how to start it but after that im stuck and require some help, please!

i can do this:

dy/dy= 6x-4 - Oct 7th 2011, 09:13 AMskeeterRe: gradient at following points at the given curve?
- Oct 7th 2011, 09:17 AMandyboy179Re: gradient at following points at the given curve?
sorry that was a typo! dy/dx means the gradient

- Oct 7th 2011, 09:24 AMskeeterRe: gradient at following points at the given curve?
- Oct 7th 2011, 10:00 AMandyboy179Re: gradient at following points at the given curve?
6x1-4 so, 6-4=2

- Oct 7th 2011, 10:21 AMandyboy179Re: gradient at following points at the given curve?
gradient = 2

- Oct 7th 2011, 10:23 AMskeeterRe: gradient at following points at the given curve?
- Oct 7th 2011, 10:25 AMandyboy179Re: gradient at following points at the given curve?
but then what else do i do to find the equation of the tangent to y=3x^2-4x+1 parallel to y=2x+7??

- Oct 7th 2011, 10:44 AMe^(i*pi)Re: gradient at following points at the given curve?
- Oct 7th 2011, 10:53 AMandyboy179Re: gradient at following points at the given curve?
no im not sure how to do this.

- Oct 7th 2011, 11:37 AMe^(i*pi)Re: gradient at following points at the given curve?
- Oct 7th 2011, 11:50 AMandyboy179Re: gradient at following points at the given curve?
ok so would i do:

y= 3-4+1

y= 0? - Oct 7th 2011, 12:17 PMe^(i*pi)Re: gradient at following points at the given curve?
- Oct 7th 2011, 12:27 PMandyboy179Re: gradient at following points at the given curve?
im not sure how to do this next step. i can do the equation of a straight line but i only have one set of brackets which is (1,0) what do i do?

- Oct 7th 2011, 12:36 PMe^(i*pi)Re: gradient at following points at the given curve?
The equation of a straight line is $\displaystyle y = mx+c$ where m is the gradient and c the y intercept. You've worked out that $\displaystyle m = 2$ so you have something of the form $\displaystyle y = 2x+c$.

To find c (the y intercept) sub in your point $\displaystyle (1,0)$ into $\displaystyle y = 2x+c$ and solve for c.

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