Calculus
posted by Z32 .
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 = (5x2)^.5 ??? I guess from what you wrote
dy/dx = .5 (5) (5x2)^.5
at x = 1 :
dy/dx = 2.5 / sqrt(3)
that is our slope
now at x = 1, y = sqrt(52) = 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
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