What is the equation of the line that is normal to f(x)= (x+2)^2-5x+2 at x=3 ?
1 Answer
Jan 31, 2016
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