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    The normal to the curve y=ax^1/2 +bx at the point where x=1 has gradient 1 and intercepts the y axis at (0,-4). Find a and b.

    Could someone tell me the steps to solve this and also any tips on how to solve these types of questions as I really struggle with figuring out the way to solve them.

    Thanks
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    (Original post by hit94)
    The normal to the curve y=ax^1/2 +bx at the point where x=1 has gradient 1 and intercepts the y axis at (0,-4). Find a and b.

    Could someone tell me the steps to solve this and also any tips on how to solve these types of questions as I really struggle with figuring out the way to solve them.

    Thanks
    You should remember that a and b are just numbers, so treat them no differently and find the normal in the usual way (it will include a and b).

    Then, note that, when a curve (or any function) intercepts the y axis, x=0.
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    (Original post by Indeterminate)
    You should remember that a and b are just numbers, so treat them no differently and find the normal in the usual way (it will include a and b).

    Then, note that, when a curve (or any function) intercepts the y axis, x=0.
    I still don't get it. I've got y' as 1/2ax^-1/2+b =-1. How do I find the y coordinate and how does the normal include a and b?
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    (Original post by hit94)
    I still don't get it. I've got y' as 1/2ax^-1/2+b =-1. How do I find the y coordinate and how does the normal include a and b?
    From \dfrac{dy}{dx}

    notice that the gradient at x=1 will involve a and b (when you substitute in) so the gradient of the normal (negative reciprocal) will also involve a and b.
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    (Original post by TenOfThem)
    This is not possible

    Can you check the question
    Why is this not possible?
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    Sorry stupid misread on my part
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    (Original post by notnek)
    I think it is possible.

    Spoiler:
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    a=-4,b=1
    Thats what I got
 
 
 
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