Find the equation of the normal line to the curve y=x^3 at x=2.?

Answer 1

For this, one needs to use a bit of calculus, so that is the approach I'm going to use.

First, the term "normal" means "perpendicular to."

Remember that the slopes of perpendicular lines are opposite reciplicles of each other.

Therefore, if I can find the slope of the curve at the point (2,8), I can just take the opposite reciplicle of the slope to find the slope of the normal line.

Slope of a curve can be found using the first derivative. In this case y' = 3x^2. Therefore, Slope = 3 (2)^2 = 12.

So, the slope of the normal line is -1/12

We also need the y-intercept in order to give the equation in the typical y = m*x + b form

The point we can use is where x = 2, or (2,8)

so, 8 = (-1/12)*2 + b
8 = -1/6 + b

b = 49/6 or 8 1/6

Final equation for the normal line: y = (-1/12) x +49/6