Find the equation of the tangent and normal to the graph given function at the given point. y^2=x^3/4-x ,(2,2)

plsss anyone can answer this plsss... 5 stars for the best answer.. :D

y = ((x^3/4)-x)^1/2

dy/dx = 1/2((x^3/4)-x)^(-1/2)*((3/4x^-1/4)+(-1))
= 1/2(x^3/4-x)^2*(3/(4(x^4))-1)
at x = 2
f'(2) = 1/2(2^3/4-2)^2*(3/(4(2^4))-1)
-4.7067 = m

y-y1 = m(x-x1)
y - 2 = -4.7067(x-2)
y - 2 = -4.7067x + 9.4134


y + 4.7067x -11.4134 = 0 <----- tangent formula

Normal:
y - 2 = 1/4.7067(x-2)
4.7067y - 9.4134 = x-2
4.7067y -x - 7.4134 = -
Should be approx right