# Finding the equation of the normal line at

## Homework Statement

Find the equation, in standard form, of the normal line drawn to the graph of y=3x4 - 1/x at x = 1

## The Attempt at a Solution

-I figured out the slope of the tangent to be 12x3 + 1, then at x = 1 the slope is 13
-SO the slope of the normal being the negative reciprocal of 13 led me to get -1/13
-Plug that into the y - y1 = m(x - x1) equation and I get x +13y - 27 = 0 for the equation of the normal.

Im a little gun shy since the numbers seem abnormally large, so I would love if someone could verify this, or if im wrong, point me in the right direction of being right!

Cheers.

berkeman
Mentor

## Homework Statement

Find the equation, in standard form, of the normal line drawn to the graph of y=3x4 - 1/x at x = 1

## The Attempt at a Solution

-I figured out the slope of the tangent to be 12x3 + 1, then at x = 1 the slope is 13
-SO the slope of the normal being the negative reciprocal of 13 led me to get -1/13
-Plug that into the y - y1 = m(x - x1) equation and I get x +13y - 27 = 0 for the equation of the normal.

Im a little gun shy since the numbers seem abnormally large, so I would love if someone could verify this, or if im wrong, point me in the right direction of being right!

Cheers.

You are right that the derivative of $$3 {x^4}$$ is $$12{x^3}$$

But what is the derivative ot 1/x again?

You made a minor mistake in your derivative, it should be 12x³ + 1/x², not 12x³ + 1. However, this doesn't affect the slope at x=1, and the rest of your derivation (including the answer) is correct.

YUP i see where I made the mistake. Instead of subtracting the 1 from the x-1 I added. But your right it doesnt affect the outcome. But thanks guys!