Find an equation of the tangent line to the curve at the given point.
\(f(x)=\sqrt(x)\) (9,3)
f(x) = x^1/2
f ' (x) = (1/2 (x ^-1/2)) = slope at x = 9 = 1/2 (1/3) = 1/6
y= 1/6 x + b at point 9,3
3 = 1/6 (9) + b b = 1 1/2
y = 1/6 x + 1 1/2
I did it way different than you. Idek how to do it the way you did