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

Question

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Answer 1

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

Take the derivative dy/dx getting [.5(3x^2 +2x)^-.5](6x+2)
Set this = 0 getting x=-1/3
So the slope of the normal to the tangent is 3
So y=3x+b
so 4 = 3(2) + b --> b= -2
So y= 3x-2 is the equation desired

Answer 2

y' = (1/2)(1/y)(6x+2)

At (2, 4), y' = 7/4

Therefore, the equation of the normal line is,

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