Write as y=(3x+2)^(1/2)
Using calculus
Let u=3x+2 => (du)/(dx) = 3....(1)
By substitution
y=u^(1/2) => (dy)/(du)=1/2u^(-1/2) ....(2)
By comparing equations (1) and (2)
(dy)/(dx)=(dy)/(cancel(du))xxcancel(du)/(dx)
=>(dy)/(dx)=1/2u^(-1/2)xx3
=>(dy)/(dx)=3/(2sqrt(3x+2))
Multiply by 1 but in the form of 1=sqrt(3x+2)/sqrt(3x+2)
dy/(dx) =(3sqrt(3x+2))/(3x+2)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
color(blue)("Determine gradient of normal")
Condition for x=1
Gradient of given curve -> (3sqrt(5))/5
"So gradient of normal is "-> -5/(3sqrt(5))
Multiply by 1 but in the form of 1=(sqrt(5))/(sqrt(5)) giving:
" "-(cancel(5)sqrt(5))/(3xxcancel(5))
color(brown)("gradient "->-(sqrt(5))/3
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
color(blue)("Determine point through which the normal passes")
For x=1
y=sqrt(3x+2) " "vec("becomes")" " y=sqrt(5)
=> "Point "->(x,y)->(1,sqrt(5))
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
color(blue)("Determine equation of normal")
y=-1/mx+c" "vec("becomes")" " sqrt(5)=(-sqrt(5)/3xx1)+c
add" "sqrt(5)/3 to both sides
sqrt(5)+sqrt(5)/3=0+c
=> c= (4sqrt(5))/3
color(blue)(y=-sqrt(5)/3 x + (4sqrt(5))/3color(white)(2/2)
