# What is the equation of the line normal to f(x)=4x^2-7x+6  at x=1?

Mar 31, 2016

Equation of normal to $f \left(x\right)$ at $x = 1$ is $x + y = 4$.

#### Explanation:

At $x = 1$, $f \left(x\right) = 4 - 7 + 6 = 3$, hence normal will be at point $\left(1 , 3\right)$ on the curve $f \left(x\right) = 4 {x}^{2} - 7 x + 6$.

The slope of tangent will be given by value of $f ' \left(x\right)$ at $x = 1$/

As $f ' \left(x\right) = 8 x - 7$, slope of tangent is $1$.

As normal is perpendicular to tangent, slope of normal will be $- \frac{1}{1} = - 1$;

Hence equation of normal given by point slope form will be

$\left(y - 3\right) = - 1 \cdot \left(x - 1\right)$ or $y - 3 = - x + 1$ or $x + y = 4$.