The point A lies on the curve with equation y = x^(1/2). The tangent to this curve at A is parallel to the line 3y-2x=1. Find an equation of this tangent at A. (PP JUNE 2015 AQA)  

My exact explanation would depend on the students level of understanding. The following answer assumes a basic understanding of differentiation and equations of lines; a) Gradient of tangent is equal to gradient of 3y - 2x = 1 which can be found by rearranging the equation. (2/3)  b) We can find out where along the curve y=x^1/2 we get a gradient of 2/3 (since the curve is not linear)  c) This gives us the X coordinate which we can sub into the intial curve to get the y coordinate of point A.  d) We now have the coordinates of a point on the curve and the gradient of a curve. We can therfore use the general equation of a line to work out the equation of the line. (y - y₁ = M ( x - x₁) ).