Graphing question...

[404]

f(x)=(1/8)x^3

I think because the 1/8 is less than 1, therefore it's wider, but if it's greater than 1, it's narrower right? I tried to graph it, but looks like mine was way to wide, and not matching what the answer book had... Can someone explain and sketch it for me? Thanks!

[vsage]

A little algebraic trick can solve this. Isn't 1/8 * x^3 the same as (x/2)^3? That may be a little easier to see. f(2) on the original graph of x^3 is now at f(4), f(6) on the original graph is at f(12). see what's going on here?

[Math Is Hard]

Check carefully the scale of the solution in your book. This might vary from what you are drawing and cause it to look different.

[404]

I do see the pattern, it's doubled... but can someone just PLEASE just answer it straight forward, there's a whole bunch of those questions to do and I'm really really really confused right now...

And as for the scale, the graph was rather tiny but I know for sure mine is a lot wider than it should... :(

[vsage]

Well put two and two together. If f(6) on the old graph is f(12) now and for every f(x) on the old graph that value is at f(2x), by what factor horizantally is the graph stretched?

Edit: What threw you off was most likely that 1/8. It is tempting to just expand the entire graph by 1/8 but coefficients on graphs outside the orders (like 1/8 * x^3) actually mean a vertical scale (in this case a scaledown by a factor of 8). If the coefficient is included with x in the power (like (1/2 * x)^3) then that is a horizantal scretch. However, on linear graphs (y = mx + b), m means a vertical scale up by m or a horizantal shrink by 1/m. It's confusing yes but really the only things you need to know where the first three sentences of my edit.

[404]

Ok, thanks, I sorta see it now. :)