Find an equation of the tangent line to the curve f(x) = √x at the point (9, 3)

Answer 1

f'(x)= (1/2)x^-1/2 which is the slope of the tangent line to f(x)=x^1/2

at (9,3) f'(x)= (1/2)/sqr9= 1/6

use the point slope formula

y-3 = (1/6)(x-9) is the equation of the tangent line at (9,3)

multiply by 6 to eliminate the fraction

6y-18 = x-9

6y-x=18-9=9

6y-x=9