Question

What is the equation of the line normal to `f(x)=3x^2 -5x +1` at `x=-3`?

Answer 1

23y - x - 762 = 0

Explanation:

To find the equation of the normal , y - b = m(x - a ) , first find the gradient (m) of the tangent by differentiating f(x) and evaluating f'(x) at x = - 3. To find a point on the line (a , b ) evaluate f(x) at x = -3

f'(x) = 6x - 5 and f'(-3) = 6(-3) - 5 = -18 - 5 = -23 = m of tangent

If `m_1 color(black)(" is gradient of normal")`

then : `m xx m_1 = -1 rArr -23 xx m_1 = -1 rArr m_1 = 1/23`

f(-3) `= 3(-3)^2 -5(-3) + 1 = 27 + 15 + 1 = 33`

hence (a , b ) = (-3 , 33 ) and `m_1 = 1/23`

equation of normal : `y - 33 = 1/23 (x+ 3 )`

[ multiply through by 23 to eliminate fraction ]

hence : 23y - 759 = x + 3

`rArr 23y - x - 762 = 0`