Question

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

Answer 1

`y = -1/19x +44 5/19`

or

#y = -1/19 x + 841/19

Explanation:

Remember `m = f'(x)` is slope of tangent line

Slope of normal line is `m_2 = 1/m` as it is perpendicular to the tangent line

`f(x) = 2x^2 - x - 1` at `x= 5`

Step 1: Find derivative, to determine the slope of tangent line

`f'(x) = d/dx(2x^2) -d/dx(x) -d/dx(1)`

`f'(x) = 4x -1`

Step 2: Find the slope of the tangent line

`m= f'(5) = 4(5) -1 = 19`

Slope of normal line is `m_2 = (-1)/19`

Step 3 Determine the `y` coordinate, of `f(x)` , when `x= 5`

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

`f(5) = 50-5-1`

`f(5) = 44`

Ordered pair `(5, 44)`

Step 4: Write the equation of the line using point slope formula

`m_2 = -1/19 , " " " (5, 44)`

`y -44 = -1/19(x-5)`

`y-44 = -1/19 x +5/19`

`y = -1/19x + 5/19 +44`

`y = -1/19x +44 5/19`

or
`y = -1/19 x + 841/19`