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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

 Apr 25, 2019
edited by Guest  Apr 25, 2019

Best Answer 

 #2
avatar+19792 
+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

 Apr 25, 2019
 #1
avatar+106539 
+1

*******

 

See EP's answer....I made a mistake......

 

 

cool cool cool

 Apr 25, 2019
edited by CPhill  Apr 25, 2019
 #2
avatar+19792 
+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
 #3
avatar+106539 
+1

Taking EP's  correct answer...we have

 

y  = - (4/7)x  + 5 1/7

 

y = -(4/7)x + 36/7       multiply through by 7

 

7y  = -4x + 36

 

4x + 7y  = 36

 

cool cool cool

 Apr 25, 2019

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