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Math Help - gradient at following points at the given curve?

  1. #1
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    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
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  2. #2
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    Re: gradient at following points at the given curve?

    Quote Originally Posted by andyboy179 View Post
    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?
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  3. #3
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    Re: gradient at following points at the given curve?

    sorry that was a typo! dy/dx means the gradient
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  4. #4
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    Re: gradient at following points at the given curve?

    Quote Originally Posted by andyboy179 View Post
    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?
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  5. #5
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    Re: gradient at following points at the given curve?

    6x1-4 so, 6-4=2
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  6. #6
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    Re: gradient at following points at the given curve?

    gradient = 2
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  7. #7
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    Re: gradient at following points at the given curve?

    Quote Originally Posted by andyboy179 View Post
    gradient = 2
    bingo!
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  8. #8
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    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??
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  9. #9
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    Re: gradient at following points at the given curve?

    Quote Originally Posted by andyboy179 View Post
    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)\?
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  10. #10
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    Re: gradient at following points at the given curve?

    no im not sure how to do this.
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  11. #11
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    Re: gradient at following points at the given curve?

    Quote Originally Posted by andyboy179 View Post
    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.
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  12. #12
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    Re: gradient at following points at the given curve?

    ok so would i do:

    y= 3-4+1
    y= 0?
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  13. #13
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    Re: gradient at following points at the given curve?

    Quote Originally Posted by andyboy179 View Post
    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
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  14. #14
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    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?
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    Re: gradient at following points at the given curve?

    Quote Originally Posted by andyboy179 View Post
    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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