An equation of the line normal to the graph of y=(3x^2+2x)^(1/2) at (2,4) is...?

Question

help me please it is a mult choice question, the choices are

a) -4x+y+20
b) 4x+7y=20
c) -7x+4y=2
d) 7x+4y=30
e) 4x+7y=36

Answer 1

y' = (1/2)(3x^2+2x)^(-1/2) (6x+2)
= (3x+1) (3x^2+2x)^(-1/2)

so when x=2,
we have that the derivative is equal to:
y'= (3(2) +1) ( 3(4) +4)^{-1/2}
=7(16)^{-1/2}
=7/4

therefore the slope of the normal is equal to: -4/7.
and the eq. of the normal line is:
y-4 = (-4/7) (x-2)

so the correct answer is e).

Answer 2

First we chek that the point belong to the graph (yes)
we get dy/dx = (1/2)(3x^2+2x)^(-1/2)*(6x+2)
At (2,4) the slope of the tangent = 7/4
The slope of the normal = - 4/7
its equation : (y - 4)= (- 4/7)(x - 2)
Multiplying both sides by 7
7 y - 28 = - 4x +8
i.e 4x + 7 y = 36
The correct answer is (e)