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
AM
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