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