calculus
posted by jon .
Find the equation of the tangent line to the curve y=sqrt(x) at x = c. Put the answer in slope intercept form y=mx+b. Use the definition of the slope of the tangent with no short cuts.

so the point of contact is (c, √c)
y = √x = x^(1/2)
dy/dx = (1/2)x^(1/2)
at our point
dy/dx = (1/2)c^(1/2) = 1/(2√c) = slope = m
y = (1/(2√c) x + b
at(c, √c)
√c = (1/(2√c)(c) + b
times 2√c
2c = c + 2√c b
c = 2√c b
b = c/√c
y = (1/(2√c)) x + c/√c
check my arithmetic
You could rationalize the denominator if you had to.