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

A) 4x + 7y= 20

B) -7x + 4y = 2

C) 7x + 4y = 30

D) 4x + 7y = 36

Guest Apr 25, 2019

edited by
Guest
Apr 25, 2019

#2**+1 **

I used a derivative calculator to find f' = (3x+1)/(sqrt(3x^2+2x)) subbing in x = 2 finds slope = 7/4 perpindicular is - 4/7

y = -4/7 x + b yielsds b = 5 1/7 at point 2,4

y = -4/7 x + 5 1/7

Picture

ElectricPavlov Apr 25, 2019

#2**+1 **

Best Answer

I used a derivative calculator to find f' = (3x+1)/(sqrt(3x^2+2x)) subbing in x = 2 finds slope = 7/4 perpindicular is - 4/7

y = -4/7 x + b yielsds b = 5 1/7 at point 2,4

y = -4/7 x + 5 1/7

Picture

ElectricPavlov Apr 25, 2019