Could anyone help with this question? I'd really appreciate it
Determine the equation of the tangent of y = 3-2√(x) at x = 9
y = 3-2√(x)
when x = 9, y = 3 - 2√9 = -3
so you want the tangent at the point (9,-3)
dy/dx = -2(1/2)x^(-1/2) = -1/√x
at x = 9, dy/dx = -1/3
equation of tangent:
y = mx + b
-3 = (-1/3)(9) + b
b = 0
tangent is y = (-1/3)x