Question

What is the equation of the normal line of `f(x)=5x^4+3x^3+x^2-2x-4` at `x=1`?

Answer 1

`29y+x-88=0`

Explanation:

Differentiate the function to find its gradient.
`f'(x)=20x^3+9x^2+2x-2`
Use the point `x=1` to find the gradient of the curve at that point.
`f'(1)=20+9+2-2=29`
The gradient of the normal is the negative reciprocal of the gradient of the tangent at a point.
So the gradient of the normal at `x=1` is `m=-1/29`
To find the coordinates of the point the normal passes through, substitute `x=1` into `f(x)`
`f(1)=5+3+1-2-4=3`
So coordinates are `(x_1, y_1)=>(1, 3)`
Equation of a line is: `y-y_1=m(x-x_1)`
`y-3=-1/29(x-1)`

`29y-87=-(x-1)`

`29y+x-88=0`