Question

What is the equation of the line normal to `f(x)=(3x-1)(2x+4)` at `x=0`?

Answer 1

`10y = -x +40`

Explanation:

Differentiate the formula to find the slope `m`. The line normal to the function is then `-1/m`. Given the point through which it passes, the equation of the line can then be calculated.
`f(x) = (3x-1)(2x+4)`
`f(x)=6x^2 +10x -4`
`f'(x) = 12x +10`

At `x=0`, `f'(x) = 10`
`:. -1/m = -1/10`

At `x=0`, `f(x) = 4`
`4 = -x/10 +c`
`:. c= 4`

the equation of the normal is `10y = -x +40`