Thread: 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

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?

sorry that was a typo! dy/dx means the gradient

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?

6x1-4 so, 6-4=2

gradient = 2

Originally Posted by andyboy179

gradient = 2

bingo!

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??

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 then you know that is a point on the tangent line.

Do you know how to find

no im not sure how to do this.

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.

ok so would i do:

y= 3-4+1
y= 0?

Originally Posted by andyboy179

ok so would i do:

y= 3-4+1
y= 0?

Yes, therefore 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

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?

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 where m is the gradient and c the y intercept. You've worked out that so you have something of the form .

To find c (the y intercept) sub in your point into and solve for c.

Spoiler:

If you've learnt the equation of a straight line as then and