Final answer:
The normal line parallel to y = -3x + 6 for the function f(x) = √(2x+1) can be found by taking the derivative to find the slope of the tangent and locating the point where the slope of the tangent is the negative reciprocal of -3.
Step-by-step explanation:
The normal line to the function f(x) = √(2x+1) at a certain point is parallel to the line y = -3x + 6 if its slope is the negative reciprocal of the slope of the tangent to f(x) at that point. Since the slope of the given line is -3, the slope of the normal line would also be -3. To find the slope of the tangent at a point on f(x), we differentiate to find f'(x), which gives us 1/√(2x+1) * (2) by the chain rule. Once we know the value of x that makes this derivative equal to 1/3 (the reciprocal of -3), we can then calculate the y-coordinate by plugging this x-value back into the original function f(x). With the coordinates of the point on f(x), we can also find the equation of the normal line which would be y = -3x + b, where b is determined using the point we found.