Suppose that f(x) = (5x + -2)^(1/2)
Find an equation for the tangent line to the graph of f(x) at x=1.
Tangent line: y=?
y = (5x-2)^.5 ??? I guess from what you wrote
dy/dx = .5 (5) (5x-2)^-.5
at x = 1 :
dy/dx = 2.5 / sqrt(3)
that is our slope
now at x = 1, y = sqrt(5-2) = sqrt 3
y = (2.5/sqrt 3) x + b
sqrt 3 = (2.5/sqrt 3)(1) + b
b = sqrt 3 - (2.5/sqrt 3)
so
y = (2.5/sqrt 3) x + sqrt 3 -2.5/sqrt 3
y sqrt 3 = 2.5 x +3 -2.5
y sqrt 3 = 2.5 x + 1/2
(2 sqrt 3) y = 5 x + 1