Question #cd403

Original equation: #y=x^3-x#
slope of the tangent to y, or #dy/dx=2# when #x=1#

In order to form a straight line equation, #Y=Mx+C#
we would need both the value of #x "and" y# at the point we would like the line to pass through.

In this case, we need to find #y# when #x=1#

Substitute #x=1# into #y=x^3-x#
#y=1^3-1#
#y=0#

#:. x=1, y=0#

Now, we can form equation of the tangent line

I will be using the gradient formula:
#m=(y_1-y_2)/(x_1-x_2)#

We know that #m=2# at #(1,0)#
#2=(y-0)/(x-1)#
#2=(y)/(x-1)#

#y=2x-2# is the tangent line equation at #(1,0)#

Normal line?
It is a line perpendicular to tangent line, in this case at #(1,0)#

Find #m# of the normal line

#m_1m_2=-1#

#m_1=-1/2#

form the equation using of the normal line.

Now, i will use another method
(of course, you still can use the method i showed above. Both yield the same answer)

We know that
#Y=Mx+C#

We have #m# but not #C#

To find #C# , we will need to fill the information we have:
#m=-1/2, x=1, y=0#

#0=-1/2+C#
#C=1/2#

Substitute #C=1/2# and #m=-1/2# into #Y=Mx+C#

#y=-1/2x+1/2#
#2y=-x+1# is the equation of normal at #(1,0)#