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Find an equation of the line that is tangent tothe graph of f and parallel tothe given line.
Function: 1/square root of x
Line: x+2y-6=0

  • Calculus-math -

    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

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