Determine equations for the tangents to the points on the curve y = -x^4 + 8x^2 such that the tangent lines are perpendicular to the line x = 1.

Question

I've found the derivative, but not really sure where to go from here.

Answer 1

The first derivative is the slope of the tangent at a given point
hoping you had
-4x^3 + 16x for the derivative, which, when x = 1 becomes -4+16 or 12

Since the new line is perpendicular to the tangent , it's slope must be -1/12 ,the negative reciprocal.

-4x^3 + 16x = -1/12
4x^3 - 16x - 1/12 = 0
48x^3 - 192x - 1 = 0

perhaps your problem is solving this cubic.
It does not factor , so it is hard to do, and quite messy

You could use something like Wolfram to solve it, .....
It shows 3 real x values, appr. -2, +2, and 0
https://www.wolframalpha.com/input/?i=48x%5E3+-+192x+-+1+%3D+0

I was expecting some nicer numbers, since we are not done yet
pick x = 2.0026, sub this into the original to get y = 15.99989
let's go with (2,16)

so the tangent at the point (2,16) is
y - 16 = (-1/12)(x - 2)

Answer 2

thanks reiny! my problem was that it was indeed cubic, and I thought i may have perhaps been doing the question wrong. You've helped me on 2 questions today, appreciate it.

Answer 3

Of course, the maxima are at x = ±√8, so if (2,16) is one point of interest
is (√8-0.828,16) (just to the left of the peak), then the other point is at x = -√8-0.828 = -3.656

Answer 4

hmmm. and yet, the cubic works at -2, not -3.656

I'll have to ponder this...