Question

What is the equation of the line that is normal to `f(x)=-2x^3-3x^2+5x-2` at `x=1`?

Answer 1

`" "y=1/7x-15/7`

Explanation:

The gradient of the line is found by differentiation

Let `" "y=-2x^3-3x^2+5x-2` .......................(1)

Gradient `= (dy)/(dx)= -6x^2-6x+5`

Thus the gradient of `f(x)` at `x=1` is

`(dy)/(dx)=-6(1)^2-6(1)+5 = -7`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The gradient of the line that is normal to `f(x)` is
`" "(-1)xx(dx)/(dy)= +1/7`

So we have

`" "y=1/7x+c`......................(2)

We have a known point `P_("(x,y)") ->P_("(1,y)")`

We can determine the value of y by `f(x)->f(1)`

`" "f(1)->y=-2(1)^3-3(1)^2+5(1)-2`

`" "y= -2`

We now have `P_("(x,y)") ->P_("(1,-2)")`

So by substitution in equation (2)
`" "-2=1/7(1)+c`......................(2)

`" "c= -15/7`

Giving

`" "y=1/7x-15/7......................(2_a)`