# gradient at following points at the given curve?

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• October 7th 2011, 09:05 AM
andyboy179
gradient 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
• October 7th 2011, 09:13 AM
skeeter
Re: gradient at following points at the given curve?
Quote:

Originally Posted by andyboy179
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

actually, it's dy/dx = 6x-4

what does dy/dx mean?
• October 7th 2011, 09:17 AM
andyboy179
Re: gradient at following points at the given curve?
sorry that was a typo! dy/dx means the gradient
• October 7th 2011, 09:24 AM
skeeter
Re: gradient at following points at the given curve?
Quote:

Originally Posted by andyboy179
sorry that was a typo! dy/dx means the gradient

so, the gradient is dy/dx = 6x-4 for any x-value of the given curve, and you want to find the value of the gradient when x = 1 ... what do you think happens now, andy?
• October 7th 2011, 10:00 AM
andyboy179
Re: gradient at following points at the given curve?
6x1-4 so, 6-4=2
• October 7th 2011, 10:21 AM
andyboy179
Re: gradient at following points at the given curve?
• October 7th 2011, 10:23 AM
skeeter
Re: gradient at following points at the given curve?
Quote:

Originally Posted by andyboy179

bingo!
• October 7th 2011, 10:25 AM
andyboy179
Re: 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??
• October 7th 2011, 10:44 AM
e^(i*pi)
Re: gradient at following points at the given curve?
Quote:

Originally Posted by andyboy179
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??

You need a co-ordinate on the tangent line. Since you know the tangent line meets the curve at $x=1$ then you know that $(1,f(1))$ is a point on the tangent line.

Do you know how to find $f(1)\?$
• October 7th 2011, 10:53 AM
andyboy179
Re: gradient at following points at the given curve?
no im not sure how to do this.
• October 7th 2011, 11:37 AM
e^(i*pi)
Re: gradient at following points at the given curve?
Quote:

Originally Posted by andyboy179
no im not sure how to do this.

You correctly found f'(1) though (in posts 5 and 6). Use the same method but with the original equation instead of the derivative.
• October 7th 2011, 11:50 AM
andyboy179
Re: gradient at following points at the given curve?
ok so would i do:

y= 3-4+1
y= 0?
• October 7th 2011, 12:17 PM
e^(i*pi)
Re: gradient at following points at the given curve?
Quote:

Originally Posted by andyboy179
ok so would i do:

y= 3-4+1
y= 0?

Yes, therefore $(1,0)$ is a point on the tangent line.

Now you have a co-ordinate and a gradient you can use the equation of a straight line to find this particular line's equation
• October 7th 2011, 12:27 PM
andyboy179
Re: 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?
• October 7th 2011, 12:36 PM
e^(i*pi)
Re: gradient at following points at the given curve?
Quote:

Originally Posted by andyboy179
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?

The equation of a straight line is $y = mx+c$ where m is the gradient and c the y intercept. You've worked out that $m = 2$ so you have something of the form $y = 2x+c$.

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

Spoiler:
If you've learnt the equation of a straight line as $y-y_1 = m(x-x_1)$ then $m=2$ and $(x_1,y_1) = (1,0)$
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