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
+36916 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
+36916 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
