What is the equation of the normal line of #f(x)=2x^3-x^2-3x+9# at #x=-1#?

1 Answer
Jun 19, 2016

Let's first find what point the tangent passes through, given #x = a#.

Explanation:

#f(x) = 2(-1)^3 - (-1)^2 - 3(-1) + 9#

#f(x) = -2 - 1 + 3 + 9#

#f(x) = 9#

Therefore, the tangent passes through #(-1, 9)#.

Now that we know this, we must differentiate the function.

By the power rule:

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

The slope of the tangent is given by evaluating #f(a)# inside the derivative, a being #x = a#.

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

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

#f'(-1) = 5#

The slope of the tangent is #5#. The normal is always perpendicular to the tangent, so the slope will be the negative reciprocal of that of the tangent.

This means the slope of the normal line is #-1/5#. By point slope form, we can find the equation of the normal line:

#y - y_1 = m(x - x_1)#

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

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

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

#y = -1/5x + 44/5#

#:.# The equation of the normal line is #y = -1/5x + 44/5#.

Hopefully this helps!