Find equation of a normal to a graph, but using what?

[jlynh]

The question is,

Find the equation in standard form, of the normal line drawn to the graph of y=(3x^4)-(1/x) at x = 1.

So my thought process was to take the derivative of the tangent using first principles... that was a whole run around... then I thought since x=1 and y=2 that I would just do trace value on the calc, and use y2-y1/x2-x1 ... that also gave me a weird fraction answer that I didn't like and it also wasn't right

So my question is, what do I use to find whatever in order to give me more information to start/finish the question!!! Be it slope, or another point, I am not sure what to try anymore.

Please help me! I don't know what to do!!

Thank you in advance!

 

[jlynh]

I am not sure if this is right, but would I just take the derivative of 3x^4 to find the slope of the tangent and then take the reciprocal for the normal??

The only reason I am unsure is because there is also -1/x.

 

[wjm11]

Find the equation in standard form, of the normal line drawn to the graph of y=(3x^4)-(1/x) at x = 1.

So my thought process was to take the derivative of the tangent using first principles... that was a whole run around... then I thought since x=1 and y=2...

...would I just take the derivative of 3x^4 to find the slope of the tangent and then take the reciprocal for the normal??

The only reason I am unsure is because there is also -1/x.

You are doing fine; you've got all the right ideas. You do need to find the derivative of the entire function (not just of 3x^4). For the 1/x part, think of it as x^(-1). Does that help you in finding the derivative?

Once you have the derivative, plug in x = 1 to find its value at x = 1. The normal line will be the negative reciprocal (not just reciprocal).

You now know the value of the slope and the point it must pass through, (1,2). Proceed.