Question

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

Answer 1

`x+y+7=0`

Explanation:

Substituting `x=3` in the function `f(x)=3(x-2)^2-5x+2`, we get y-coordinate of point as follows

`y=f(3)`
`=3(3-2)^2-5\cdot 3+2`

`=-10`

The slope `dy/dx` of tangent to the curve: `f(x)=3(x-2)^2-5x+2` is given by differentiating the function `f(x)` w.r.t. `x` as follows

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

`=6(x-2)-5`

Now, the slope of tangent at `x=3`,

`f'(3)=6(3-2)-5`

`=1`

hence, the slope `m` of normal at `(3, -10)` is given as

`m=-1/{f'(3)}=-1/1=-1`

Now, equation of the normal at the point `(x_1, y_1)\equiv(3, -10)` & having sloe `m=-1` is given by following formula

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

`y-(-10)=-1(x-3)`

`x+y+7=0`