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

##### 1 Answer

#y = -1/5 x - 9/5#

#### Explanation:

To build the equation of the normal line, you need a point on the normal line and its slope.

**1) Point on the normal line**

As you are searching for the normal line at

To find the

#f(1) = - 1^4 + 2 * 1^3 + 2 * 1^2 - 1 - 4 = - 1 + 2 + 2 - 1 - 4 = -2#

Thus, the point is

**2) Slope of the normal line**

To compute the slope of the normal line, first of all, you need the derivative of

#f'(x) = - 4 x^3 + 6 x^2 + 4 x - 1#

Now, you need to evaluate the derivative at

#f'(1) = - 4 * 1^3 + 6 * 1^2 + 4 * 1 - 1 = - 4 + 6 + 4 - 1 = 5#

Thus, the slope of the tangent at

We know that the slope of the normal line

#m_n = - 1 / m_t = - 1 /5#

**3) Computing the equation of the normal line**

Now that we have the point and the slope, we can build the equation.

Any line has the form

#y = m * x + n#

We already have

#-2 = - 1 /5 * 1 + n#

#-2 = - 1/5 + n#

... add

#-9 /5 = n#

Thus, the equation of the normal line is

#y = -1/5 x - 9/5#