The point A lies on the curve with equation y=x^0.5. The tangent to this curve at A is parallel to the line 3y-2x=1 . Find an equation of this tangent at A. [5 marks]

Differentiate equation

dy/dx=0.5*x^-0.5

Gradient is the same as the second equation

2/3=0..5*x^-0.5 

Solving this will give the x coordinate

x = 9/16 

Sub into equation for y coordinate

y = (9/16)^0.5

Solve for C - constant (Y = Mx + C)

c = 3/4 - (2/3)*(9/16)

c = 3/8

Form equation

y = (2/3)x + 3/8