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

$x + y + 7 = 0$

#### Explanation:

Substituting $x = 3$ in the function $f \left(x\right) = 3 {\left(x - 2\right)}^{2} - 5 x + 2$, we get y-coordinate of point as follows

$y = f \left(3\right)$
$= 3 {\left(3 - 2\right)}^{2} - 5 \setminus \cdot 3 + 2$

$= - 10$

The slope $\frac{\mathrm{dy}}{\mathrm{dx}}$ of tangent to the curve: $f \left(x\right) = 3 {\left(x - 2\right)}^{2} - 5 x + 2$ is given by differentiating the function $f \left(x\right)$ w.r.t. $x$ as follows

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(3 {\left(x - 2\right)}^{2} - 5 x + 2\right)$

$= 6 \left(x - 2\right) - 5$

Now, the slope of tangent at $x = 3$,

$f ' \left(3\right) = 6 \left(3 - 2\right) - 5$

$= 1$

hence, the slope $m$ of normal at $\left(3 , - 10\right)$ is given as

$m = - \frac{1}{f ' \left(3\right)} = - \frac{1}{1} = - 1$

Now, equation of the normal at the point $\left({x}_{1} , {y}_{1}\right) \setminus \equiv \left(3 , - 10\right)$ & having sloe $m = - 1$ is given by following formula

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

$y - \left(- 10\right) = - 1 \left(x - 3\right)$

$x + y + 7 = 0$