Math question (pre calculus)?

Question

y = ((sqrt of x) - 2) (3(sqrt of x) + 8) at P (4, 0)

Find the equation of the tangent.
I got 3x-16y-12=0 but this is wrong according to book. if possible show me the steps thannks

Answer 1

Product rule:

y' = (√x - 2)(3/(2√x)) + (1/(2√x))(3√x + 8)

Plug in x = 4:

y' = (√4 - 2)(3/(2√4)) + (1/(2√4))(3√4 + 8)
y' = (2 - 2)(3/4) + (1/4)(3(2) + 8)
y' = 0 + 1/4(6 + 8) = 14/4 = 7/2, if I didn't mess up.

So:

y - 0 = (7/2)(x - 4)
2y = 7(x - 4)
2y = 7x - 28
0 = 7x - 2y - 28

If I didn't mess up.