What is the equation of the line normal to f(x)=3x2−5x+1 at x=-3?

Question

Naima Silva · Accepted Answer

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}

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 {(- 3)}^{2} - 5 (- 3) + 1 = 27 + 15 + 1 = 33

hence (a , b ) = (-3 , 33 ) and

m_{1} = \frac{1}{23}

equation of normal :

y - 33 = \frac{1}{23} (x + 3)

[ multiply through by 23 to eliminate fraction ]
hence : 23y - 759 = x + 3

\Rightarrow 23 y - x - 762 = 0

What is the equation of the line normal

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