Equation of a Perpendicular to a Tangent

[Trevor]

Hey everybody, this is my first post here. I kind of stumbled onto this forum while trying to find some help for a calc assignment I have, hopefully I've found the right place

My question reads as follows:

Find the equation of the line perpendicular to the tangent line at the point (1, -2) on the graph of the equation g(x)=x^6-6x^2+3.

I believe the derivative of the equation given is 5x^5-12x. I can picture what I have to do in my head, but I don't know how to go about working it out. Thanks!

 

[wjm11]

Find the equation of the line perpendicular to the tangent line at the point (1, -2) on the graph of the equation g(x)=x^6-6x^2+3.

I believe the derivative of the equation given is 5x^5-12x.

You have started correctly. Now use the x value of the indicated point, (1,-2), in the derivative to find the tangent line slope.

A line perpendicular to this will have a slope that is the negative reciprocal of the tangent line slope.

Now you have the slope and a point – sufficient info to write the equation of the line you desire.

 

[fasteddie65]

Find the equation of the line perpendicular to the tangent line at the point (1, -2) on the graph of the equation g(x)=x^6-6x^2+3.

I believe the derivative of the equation given is 5x^5-12x. I can picture what I have to do in my head, but I don't know how to go about working it out. Thanks!

Actually, the derivative is 6x^5 - 12x. Now let x = 1, and that gives the slope of the tangent line. Take the negative reciprocal of that number and that's the slope of the line perpendicular to the tangent.

By the way, we call that the "normal" to the curve.