# What is the equation of the line normal to  f(x)=(x-3)^2+4x at  x=1?

Apr 19, 2018

Slope of normal is $x = 1$

#### Explanation:

The slope of tangent at a point $x = {x}_{0}$ on $f \left(x\right)$ is $f ' \left({x}_{0}\right) = {\left(\frac{\mathrm{df}}{\mathrm{dx}}\right)}_{x = {x}_{0}}$

and as normal is perpendicular to slope, its slope is $- \frac{1}{f ' \left({x}_{0}\right)}$

Here we are seeking normal at $x = 1$ on curve $f \left(x\right) = {\left(x - 3\right)}^{2} + 4 x$ i.e. at $\left(1 , {\left(1 - 3\right)}^{2} + 4 \cdot 1\right)$ or $\left(1 , 8\right)$.

As $\frac{\mathrm{df}}{\mathrm{dx}} = 2 \left(x - 3\right) + 4 = 2 x - 2$, slope of tangent at $x = 1$ is $0$ i.e. tangent at $\left(1 , 8\right)$ is $y = 8$

and normal at $\left(1 , 8\right)$ would be $x = 1$