Question

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

Answer 1

5y + x - 63 = 0

Explanation:

Require to differentiate so expand bracket in f(x).

f(x) `= x^2 +4x + 4 - 5x + 2 = x^2 - x + 6`

f'(x) = 2x - 1

Evaluating f'(3) gives the gradient of the tangent and f(3)
the value of the y-coord.

f'(3) = 2(3) -1 = 6 - 1 = 5 =m

f(3) `= (3+2)^2 -5(3) + 2 = 25 - 15 + 2 = 12 → (3 , 12)`

The product of gradients of perpendicular lines is -1

hence m of normal `= -1/5`

equation of normal: y - b = m(x - a) , m`= -1/5`
and (a,b ) = ( 3 , 12)

y - 12 `= -1/5 ( x - 3 )`

( multiply both sides by 5 to eliminate fraction )

5y - 60 = -x + 3 → 5y + x - 63 = 0