+0  
 
0
43
1
avatar+17 

My guess is the first one

 Dec 14, 2018
 #1
avatar+94526 
+1

f(x) = x^3 - 4

 

The derivative gives the slope of a tangent line at any x value

 

So

 

f ' ( x)   =   3x^2

 

At  x = 2.... the slope of the tangent line is    3(2)^2 =  12

 

The slope of a normal line  =  -1/12

 

 

So...the equation of the normal line is

 

 

y = - (1/12) ( x - 2) + 4       simplify

 

y = (-1/12)x + 1/6 + 4

 

y = (-1/12)x + 25 / 6

 

 

cool cool cool

 Dec 14, 2018

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