# What is the slope of the line normal to the tangent line of #f(x) = e^(x^2-1)+2x-2 # at # x= 1 #?

##### 1 Answer

May 16, 2017

#y=-1/82.34x+22.09#

#### Explanation:

Given

#y=e^(x^2-1)+2x-2#

At

#y=2.71828^(2^2-1)+2(2)-2#

#y=2.71828^3+4-2#

#y=20.085+2=22.085#

At Point

Look at the graph

The slope of the tangent is equal to the slope of the given curve.

The slope of the given curve is-

#dy/dx=2xe^(x^2-1)+2#

At

#dy/dx=2(2)(2.71828^3)+2#

#dy/dx=4xx 20.085+2#

#dy/dx=82.34#

The slope of the tangent

The slope of the normal is

The equation of the Normal is -

#y=m_2x+c#

#22.085=-1/82.34(2)+c#

#22.085+1/82.34=c#

#22.09=c#

#y=-1/82.34x+22.09#