Given that y= (x+2) ²√(2x-3),
find the coordinates of the point at which the normal to the curve at x = 2 cuts the x-axis.
is that
y = (x+2)^2 √(2x-3)
or
y = (x+2) √(2x-3)
???
If the former, then
y' = (x+2)(5x-4)/√(2x-3)
y'(2) = 24
so the normal has slope -1/24
y(2) = 16, so the normal has equation
y-16 = -1/24 (x-2)
The x-intercept is at (386,0)