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

$y = - \frac{1}{5} x + \frac{6}{5}$
The derivative is $f ' \left(x\right) = 8 x - 3$ so that $f ' \left(1\right) = 5$. The slope of the normal line is the opposite reciprocal of this, which is $- \frac{1}{5}$.
Since the point $\left(1 , f \left(1\right)\right) = \left(1 , 1\right)$ is on the line, the equation of the line is $y - 1 = - \frac{1}{5} \left(x - 1\right)$, which is equivalent to $y = - \frac{1}{5} \left(x - 1\right) + 1$ and $y = - \frac{1}{5} x + \frac{6}{5}$.