y = x^(3/4), find y-intercept of line normal at x = 16? Need help solving not just answer, please :)

Question

accessdenied · Answer

Are you completely unsure of how to start the problem?

anonymous · Answer

No I've started it, my answer seems strange though it has decimals / doesn't simplify.

anonymous · Answer

Here's how I started:

anonymous · Answer

I found my y1 for later on when I use the y - y1 = m(x - x1) equation. The y1 gives me a value of 8.
Then I went on to find the equation of the normal:
$$y = (3/4)x ^{3/4}$$
$$mtan when x = 16 -> (3/4)\div \sqrt[4]{16}$$
$$mtan = 3/8$$

anonymous · Answer

then for the normal I took the negative reciprocal of that slope, -8/3 and subbed that and the y1 value into the equation:
$$y - 8 = -(8\div3)(x-16)$$
$$y = (-8\div3)x + 128\div3 + 8$$
$$y = -8/3x + 152/3$$

anonymous · Answer

I don't know if it's supposed to have a weird answer like that, or if I'm doing my algebra wrong, or if I simply approached this problem the wrong way. Any advice / information?

anonymous · Answer

Oh, and of course once I figure out if that's the correct equation for the normal, I would simply find y when x = 0. However I don't feel confident in what I have so far.

accessdenied · Answer

it does look like you did everything correctly.  I get the same answer.

anonymous · Answer

Oh, okay. Hmm, guess that fraction just freaked me out. Thanks.

anonymous · Answer

I have one other one that feels way too easy, mind lookin at it on this same question? It was: y = 6x - 2x^2, find the normal at x=2, y = 4. My answer is y = 2x, but that seems too simple.

accessdenied · Answer

normal at x=2 and y=4?

accessdenied · Answer

ohh, okay, i read that it was y = 6x - x^2, and (2,4) was not on that... lol

anonymous · Answer

Mhm, it's supposed to be the normal at (2,4) if I'm not mistaken.

accessdenied · Answer

This one, I think you made a mistake somewhere...

anonymous · Answer

Yeah, my friend got a different answer. What I did was derivative, so 6 - 4x, then at x = 2 that gives you a s lope of -2. I subbed that and the y=4 into the y-y1 = m(x - x1) equation to get y = 2x for the normal. :\

accessdenied · Answer

y' = -2 is the slope of the tangent, though.  You're looking for the line normal at (2,4).

accessdenied · Answer

Since -2 is the slope of the tangent, the slope of the normal is the opposite reciprocal, or 1/2.
Then we can use that slope and (2,4) in y - y1 = m(x - x1) to get the equation.

anonymous · Answer

oooh i forgot about the opposite part

anonymous · Answer

sorry, reciprocal part