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!