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

Feb 7, 2016

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 ${m}_{1} \textcolor{b l a c k}{\text{ is gradient of normal}}$

then : $m \times {m}_{1} = - 1 \Rightarrow - 23 \times {m}_{1} = - 1 \Rightarrow {m}_{1} = \frac{1}{23}$

f(-3) $= 3 {\left(- 3\right)}^{2} - 5 \left(- 3\right) + 1 = 27 + 15 + 1 = 33$

hence (a , b ) = (-3 , 33 ) and ${m}_{1} = \frac{1}{23}$

equation of normal : $y - 33 = \frac{1}{23} \left(x + 3\right)$

[ multiply through by 23 to eliminate fraction ]

hence : 23y - 759 = x + 3

$\Rightarrow 23 y - x - 762 = 0$