Posted by **Z32** on Tuesday, February 16, 2010 at 4:05am.

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=?

- Calculus -
**Damon**, Tuesday, February 16, 2010 at 5:59am
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

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