Question

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

Answer 1

`x+7y=-6`

Explanation:

We need to obtain first the slope m which is negative reciprocal of the slope of the tangent line at x=1 on the curve `f(x)=3x^2+x-5`

Compute the first derivative `f'(x)` and `m=f' (1)`

`f(x)=3x^2+x-5`

`f' (x)=d/dxf(x)=d/dx(3x^2+x-5)=6x+1`

`f' (x)=6x+1`

at `x=1`

`f' (1)=6(1)+1=7`

Compute `m`

`m=-1/(f' (1))=-1/7`

Determine the ordinate for the given abscissa `x=1` using the original equation `f(x)=3x^2+x-5`

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

The point is at `(1, -1)`

Determine the Normal Line now using Point-Slope Form

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

`y--1=-1/7(x-1)`

`7(y+1)=-x+1`

`7y+7=-x+1`
`x+7y=-6`

See the graph of `f(x)=3x^2+x-5` and `x+7y=-6`

graph{(y-3x^2-x+5)(x+7y+6)=0[-22,22,-10,10]}