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    Find the equation of the tangent to the curve with equation y=2x^2-x-2 which is perpendicular to the straight line with equation 2x-3y+4=0

    I've managed to do it and got the answer at the back of the book, however it took me a very long amount of time and was very long winded, which I don't imagine the exam would ask you for. So I was wondering about how you guys would go about doing it, because my method took atleast 20 mins lol

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    rearrange 2x-3y+4=0 in the form y=mx + c
    use m1 x m2 = -1 to find gradient of the tangent I have to find
    find dy/dx of the quadratic function
    make dy/dx = the gradient I found above
    solve for x
    sub x in original quadratic equation to find y
    use these (x,y) and the gradient to form an equation for the tangent

    you would be expected to allocate a fair amount of time on this as it would typically be worth 8 marks or more
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    (Original post by gdunne42)
    rearrange 2x-3y+4=0 in the form y=mx + c
    use m1 x m2 = -1 to find gradient of the tangent I have to find
    find dy/dx of the quadratic function
    make dy/dx = the gradient I found above
    solve for x
    sub x in original quadratic equation to find y
    use these (x,y) and the gradient to form an equation for the tangent

    you would be expected to allocate a fair amount of time on this as it would typically be worth 8 marks or more
    Ah right, that's exactly what I did!


    hmmm no wonder it took ages, rthx!

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    You will find with practice that these questions become a lot quicker to do yourself. When you realise all the steps it becomes second nature.
 
 
 
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