Question

What is the equation of the line that is normal to `f(x)= x-sqrt( 3x+2)` at `x=4`?

Answer 1

`1.669x+y-6.935=0`

Explanation:

`f(x)=x-sqrt(3x+2)`
Let y=f(x)
`y=x-sqrt(3x+2)`
At x=4
`y=4-sqrt(3xx4+2)=0.258`
`dy/dx=1-1/(2sqrt(3x+2))(3)`
`|dy/dx|_(x=4)=1-3/(2sqrt(3xx4+2))=0.599`

The tangent and normal are orthogonal to each other
Slope of the tangent is 0.599
m1m2=-1gives
Slope of the normal is -1/0.599=-1.669
The normal passes through the point (4,0.258)
Equation of the line passing through a point having a slope is given by

`(y-0.258)/(x-4)=-1.669`
`(y-0.258)/(x-4)+1.669=0`
Simplifying

`y-0.258+1.669(x-4)=0`

`y-0.258+1.669x-6.676=0`
Expressing in the standard form ax+by+c=0
we have
`1.669x+y-6.935=0`