# Question #cd906

Equation of tangent: $y = 5 x - 2$, equation of normal: $y = - \setminus \frac{1}{5} x + \setminus \frac{16}{5}$
Tangents and normals are lines, which means to find their equations we need their gradients and a fixed point, which is $\left(1 , 3\right)$ in this case.
We are given $y = 3 {x}^{2} - x + 1$. Differentiating, we get $\setminus \frac{\mathrm{dy}}{\mathrm{dx}} = 6 x - 1$. Thus the gradient of the tangent is given by $6 \left(1\right) - 1 = 5$ and the gradient of the normal, perpendicular to the tangent, is given by $- \setminus \frac{1}{5}$. Thus the equation of the tangent is $y - 3 = 5 \left(x - 1\right) \setminus \Leftrightarrow y = 5 x - 2$ and the equation of the normal is $y - 3 = - \setminus \frac{1}{5} \left(x - 1\right) \setminus \Leftrightarrow y = - \setminus \frac{1}{5} x + \setminus \frac{16}{5}$.