Question

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

Answer 1

Equation of normal is `y-49 = -1/73(x-2)` This is in Slope-Point form.

Explanation:

To find the equation of normal line we need to find the slope of the normal line and a point through which the line passes through.

We are given

`f(x)=5x^3+3x^2+x-5` and we are asked to find equation at `x=2`
First let us find `y` when `x=2`

`y=f(2)=5(2)^3+3(2)^2+(2)-5`
`y=f(2) = 5(8)+3(4)+2-5`
`y=f(2)=40+12-3`
`y=49`

Now we have a point `(2,49)`

The slope of the tangent can be found by finding the derivative of `f(x)` and evaluated at `x=2`

`f(x)=5x^3+3x^2+x-5`
`f'(x) = 15x^2+6x+1`
`m=f'(2)=15(2)^2+6(2)+1`
`m=15(4)+12+1`
`m=60+12+1`
`m=73`

Slope of normal `m_"normal" = -1/m`
Slope of normal `m_"normal" = -1/73`
Equation of a line passing through a point `(x_1,y_1)` and slope `m` is given by

`y-y_1 = m(x-x_1)`

Here we have the point `(2,49)` and slope `-1/73`

Equation of normal is `y-49 = -1/73(x-2)`