Question #cd403

1 Answer
Jul 16, 2017

tangent line:
y=2x-2

normal line:
2y=-x+1

Explanation:

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)