Question

If y=6x^4, and x is halved, what happens to y?

Answer 1

I think y is multiplied by 16

Answer 2

So what, if x is doubled, wouldn't it still be multiplied by 16?

Answer 3

6 (x/2)^4 = 6 (1/2)^4 x^4. So y is multiplied by 1/16

Answer 4

Now I'm very confused.

Answer 5

yeah I think sourwing is right, confusing question:)

Answer 6

so if x is halved?

Answer 7

\[y=6x^4 \implies \frac{1}{16}y=\frac{1}{16}6x^4=6(\frac{x}{2})^4\]

Answer 8

Hey guise, turns out that if x is doubled, y is multiplied by 16.
If x is halved, y is multiplied by 1/16

Answer 9

Yes, in these problems where you have a cylinder whose radius is doubled, and the height is tripled, and so on, you can find the factor by just plugged in the ratios in the formula, and leaving everything else alone.

For example, a cube has its edges doubled, so l->2l, w->2w, h->2h:

just plug in the scale factors in the equation l*w*h = 2*2*2 = 8 — the new volume is 8x the old one. Double the radius of a circle: \(A = \pi r^2\) but we can just do \(r^2 = 2^2 = 4\): area is quadrupled when the radius is doubled.

By the same technique here: x is doubled, so plug 2 into \(x^4\) and you get \(2^4=16\), so \(y\) is multiplied by 16. Clear as mud?

Answer 10

Makes as much sense as an ice sculpture in the Sahara

Answer 11

Okay, well, here's a more formal description:

The cube example: let the original dimensions be \(l_1, w_1, h_1\) and the doubled dimensions be \(2*l_1, 2*w_1, 2*h_1\)

Original volume = \(V_1 = l*w*h = l_1*w_1*h_1 \)
Volume with doubled sides = \(V_2 = (2*l_1)*(2*w_1)*(2*h_1) = 2*2*2*l_1*w_1*h_1\)
Ratio of new one to old one:\[\frac{V_2}{V_1} = \frac{2*2*2\cancel{*l_1*w_1*h_1}}{\cancel{l_1*w_1*h_1}} = 8\]

Answer 12

I had the concept, I just reversed it for every question lml

Answer 13

Thank you for putting it in perspective though

Answer 14

Ah, bummer!

Answer 15

It is. It all started with having y=8x^2. I thought that if you halved it, then you would multiply y by 4. Someone here incorrectly confirmed that and I carried that over everywhere else. Oh well.

Answer 16

Ah, that would have been true if it was \(y = 8/x^2\) or indirect variation...that's why I prefer to always check my answers, no matter how simple the problem!