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    The equation of the tangent to the curve 1/(1−2x) at the point (1, -1) is given by y=ax+b, where
    a=
    b=
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    (Original post by fahd20428)
    The equation of the tangent to the curve 11−2x at the point (1, -1) is given by y=ax+b, where
    a=
    b=
    y=11-2x is a straight line. The tangent to a straight line is the line itself. The line doesn't pass (1,-1), have you written the question correctly?
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    (Original post by NotNotBatman)
    y=11-2x is a straight line. The tangent to a straight line is the line itself. The line doesn't pass (1,-1), have you written the question correctly?
    sorry it is 1/(1-2x)
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    (Original post by fahd20428)
    sorry it is 1/(1-2x)
    Find the derivative of 1/(1-2x), the derivative is the gradient function, which will allow you to find out the gradient of the tangent at some point x. sub x=1 into the gradient function and use the formula y-y1=m(x-x1)
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    (Original post by NotNotBatman)
    Find the derivative of 1/(1-2x), the derivative is the gradient function, which will allow you to find out the gradient of the tangent at some point x. sub x=1 into the gradient function and use the formula y-y1=m(x-x1)
    is the answer going to be a=-2 b=1
 
 
 
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