Find the tangent to the curve y=x^2 +2x at point (1,3)

In order to find the gradient we need to differentiate d/Dx = 2x + 2using our point, the gradient is 4using y = mx+cy = 4x +cusing our pointsy= 4x - 1

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The curve C has equation: (x-y)^2 = 6x +5y -4. Use Implicit differentiation to find dy/dx in terms of x and y. The point B with coordinates (4, 2) lies on C. The normal to C at B meets the x-axis at point A. Find the x-coordinate of A.


I don't understand the point of differentiation or integration


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Relative to a fixed origin O, the point A has position vector (8i+13j-2k), the point B has position vector (10i+14j-4k). A line l passes through points A and B. Find the vector equation of this line.


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