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

Jan 31, 2016

5y + x - 63 = 0

#### Explanation:

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

f(x) $= {x}^{2} + 4 x + 4 - 5 x + 2 = {x}^{2} - x + 6$

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 $= - \frac{1}{5}$

equation of normal: y - b = m(x - a) , m$= - \frac{1}{5}$
and (a,b ) = ( 3 , 12)

y - 12 $= - \frac{1}{5} \left(x - 3\right)$

( multiply both sides by 5 to eliminate fraction )

5y - 60 = -x + 3 → 5y + x - 63 = 0