calculus
posted by Anonymous .
Find an equation of the line that is tangent to the graph f and parallel to the given line
Function: f(x)= 1/ square root of x
Line: x+2y6=0

slope of f(x) is 1/(2x√x)
slope of line: 1/2
so, you want 1/(2x√x) = 1/2
x=1
so, you now have a slope=1 and a point (1,1).
(y1) = (1/2)(x1)
y = x/2 + 3/2
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