Question

AN EQUATION OF THE LINE NORMAL TO THE GRAPH OF
Y= SQRT[3X^2 + 2X] AT (2, 4) IS??

Answer 1

y' = (3x+1)/(3x^2+2x)

So the slope of the tangent at (2,4) is
7/4
so the slope of the normal is -4/7

let the equation of the normal be
4x + 7y = C, then at (2,4)
8 + 28 = C = 36

the equation of the normal is 4x + 7y = 36

Answer 2

y' should have bee

y' = (3x+1)/(3x^2+2x)^(1/2)

my calculations reflected the correct version

Answer 3

y = (3 x^2 + 2 x)^.5
dy/dx = .5 [(3x^2+2x)^-.5 ] (6 x +2)
at (2,4), x = 2
dy/dx = .5 [ (3*4+4)^-.5 ] (14)
= .5 [ 1/4 ] (14)
= 7/4
the slope of the normal to that is +4/7
so
y = (4/7) x + b
now at (2,4)
4 = (4/7)*2 + b
4 = 8/7 + b
28/7 = 8/7 + b
b = 20/7
so
7 y = 4 x + 20

Answer 4

CALCULUS - Damon, Friday, January 2, 2009 at 5:23pm

y = (3 x^2 + 2 x)^.5
dy/dx = .5 [(3x^2+2x)^-.5 ] (6 x +2)
at (2,4), x = 2
dy/dx = .5 [ (3*4+4)^-.5 ] (14)
= .5 [ 1/4 ] (14)
= 7/4
the slope of the normal to that is -4/7
so
y = -(4/7) x + b
now at (2,4)
4 = -(4/7)*2 + b
4 = -8/7 + b
28/7 = -8/7 + b
b = 36/7
so
7 y = -4 x + 36

Answer 5

Well, my answer goes through (2,4) now :)