posted by Anonymous .
Find an equation of the line that is tangent tothe graph of f and parallel tothe given line.
Function: 1/square root of x
using slope-intercept, the given slope is -1/2
the slope of the tangent to the function is the 1st derivative
f(x) = 1 / sqrt(x) = x^(-1/2)
f'(x) = (-1/2) x^(-3/2)
substitute to find x and then y at the point of tangency
use point-slope to write the equation of the new parallel line