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

Jun 3, 2016

$y - 5 = - \frac{1}{11} \left(x - 1\right)$

#### Explanation:

We'll be using the point gradient formula since we already have a point, x=1, which means y=5; $\left(1 , 5\right)$ no we need a gradient.

In order to find the gradient at x=1. we need to differentiate the equation. The differentiated equation is $3 {x}^{2} + 8 x$. We then substitute in x=1, and we have the gradient of 11.

However, this gradient is the tangents gradient, not the normal's. In order to find the normal's gradient, we know that the gradient of the tangent times the gradient of the normal equals -1. Therefore, the gradient of the normal is $- \frac{1}{11}$

We the substitute the numbers into the formula to get a equation