What is the equation of the normal line of #f(x)= x^8 + 6 e^x # at #x=0#?

1 Answer
Dec 25, 2015

#y-6=-1/6(x-0)# in the slope point form. #y=-1/6x + 6# in slope intercept form. The detailed explanation is given below.

Explanation:

We need to find the equation of the normal line.

To find the equation of the line we need two things.
1. Slope #m#.
2. Point #(x_1,y_1)#
Then the equation can be formed by using the slope point form

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

We are to find the equation of normal at #x=0# let us use that to find #f(0)# this would give our points #(x_1,y_1)#
#f(x)=x^8+6e^x#
#f(0) = 0^8 +6e^0#
#f(0) = 0+6#
#f(0)=6#

Our #(x_1,y_1) = (0,6)#

Now to find the slope of required equation. For this, we need to find the slope of the tangent. The slope of the tangent is found by finding #f'(x)#

#f'(x)=8x^7+6e^x#
Slope of tangent at #x=0#
#f'(0) = 8(0)^7+6e^0#
#f'(0)=6#

The slope of the normal is the negative reciprocal of slope of the tangent.

Slope of normal #m = -1/6#

Equation of the normal line at #x=0# is

#y-6=-1/6(x-0)# in the slope point form.

#y=-1/6x + 6# in slope intercept form.