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

Jan 11, 2016

$29 y + x - 88 = 0$

#### Explanation:

Differentiate the function to find its gradient.
$f ' \left(x\right) = 20 {x}^{3} + 9 {x}^{2} + 2 x - 2$
Use the point $x = 1$ to find the gradient of the curve at that point.
$f ' \left(1\right) = 20 + 9 + 2 - 2 = 29$
The gradient of the normal is the negative reciprocal of the gradient of the tangent at a point.
So the gradient of the normal at $x = 1$ is $m = - \frac{1}{29}$
To find the coordinates of the point the normal passes through, substitute $x = 1$ into $f \left(x\right)$
$f \left(1\right) = 5 + 3 + 1 - 2 - 4 = 3$
So coordinates are $\left({x}_{1} , {y}_{1}\right) \implies \left(1 , 3\right)$
Equation of a line is: $y - {y}_{1} = m \left(x - {x}_{1}\right)$
$y - 3 = - \frac{1}{29} \left(x - 1\right)$

$29 y - 87 = - \left(x - 1\right)$

$29 y + x - 88 = 0$