# What is the equation of the normal line of f(x)=-x^4-2x^3-5x^2+3x+3 at x=1?

Mar 27, 2016

Equation of normal is $x - 17 y - 35 = 0$

#### Explanation:

As $f \left(x\right) = - {x}^{4} - 2 {x}^{3} - 5 {x}^{2} + 3 x + 3$ and at $x = 1$ its value is $- 1 - 2 - 5 + 3 + 3 = - 2$ and hence curve passes through $\left(1 , - 2\right)$ and we need to find equation of normal at this point.

Slope of tangent is given by function's derivative. It is $f ' \left(x\right) = - 4 {x}^{3} - 6 {x}^{2} - 10 x + 3$ and at $x = 1$ slope is $- 4 - 6 - 10 + 3 = - 17$. Hence the slope of tangent is $- 17$ and as normal is perpendicular to it, slope of normal would be $\frac{1}{17}$.

Now using point slope form, equation of normal is

$\left(y + 2\right) = \frac{1}{17} \left(x - 1\right)$ or $17 y + 34 = x - 1$ or

$x - 17 y - 35 = 0$