math help

[Fire0nIce228 1]

My problem is this, broken into 4 parts.

I have to graph these on the same grid, using 4 different colors. I can't use symbols so i'll write it out in english. i'm assuming I have to get "x" by itself on one side, and then use an x-y table to find my points on the graph, but I'm having trouble figuring how I should be setting it up. The equations are

y = cube root of x

y= negative cube root of x

y= cube root of "x-1"

y= cube root of x , +1

I tried the first one, and my x y table is all weird cause the one value will be real low,

Ie.

I cubed x and y, both sides of the equation. That takes away the cube root leaving x by itself on the right side of the equals sign. However, "y" is now to the third power. So, if I use 3 for x, then y must be 27. (3*3*3 = 27) so when I go to graph that it gets crazy. You can see how weird it would be for, 5, with its other being 125.

So am I retarted and messing this all up or something?

Help would be appreciated as the assignment is due Thursday.

[TheProfessor 2]

I cubed x and y, both sides of the equation. That takes away the cube root leaving x by itself on the right side of the equals sign. However, "y" is now to the third power. So, if I use 3 for x, then y must be 27. (3*3*3 = 27) so when I go to graph that it gets crazy. You can see how weird it would be for, 5, with its other being 125.

I think you're getting it backwards. Cube Y to get X. You'll have x = ycubed. Plug in values for y, cube them, and get x. The graph is of y = cube root (x), so set up a table like this:

X ...... Y

----------

1 ...... 1

8 ...... 2

27 ...... 3

64 ...... 4

The function grows slowly, as you can see. It takes large changes in X to see a change in Y. The other three functions are just translations/rotations of this one. for negative(cuberoot(x)), it'll be a reflection about the x axis.

For y= cube root of "x-1", just shift the original function one unit to the right (on the x axis).

For y= cube root of x , +1, shift the original function one unit up (on the y axis)

Heres a few pics of what it will look like. Note the color coding of the lines for each function. The second pic (zoomed in) shows more clearly the differences inthe functions and how they are translations/mirror of the original.

[Plymkr 0]

Just re-iterating what TheProfessor was saying, in order to solve this need to set up a table of values. Basically explaining it out more to help out, plus I'd like to increase my post count here!

Equation 1: y = x^(1/3)

here I would pick known values, starting with the origin (0,0)

When X is 0, y = 0.

When X is 1, y = 1.

When X is 8, y = 2.

When X is 27, y = 3.

When X is 64, y = 4.

When X is n, y = n^(1/3)

And so on...

Points on the Graph:

(0,0)

(1,1)

(8,2)

(27,3)

(64,4)

Basically what this graph is doing is increasing on the x-axis and barely going up on the y-axis. (as shown by the picture TheProfessor made)

[Fire0nIce228 1]

Whoa, thanks very much guys. My whole concern was one number being tiny and the others becoming huge very quick. Thank you :)

[Fire0nIce228 1]

okay, you guys were too much help for your own good. Math is my most horriblest subject ever. The second part I'm having trouble with now is finding the domain and range for the graphs.

I have done a bunch of the problems, they are in all differnt forms, and I cant figure it out. I know domain is

x cannot equal # right?

and range is everything y could be, correct?

only thing is like, the one after solving it for y and such it looks like this

y = -4 - x/2

so how the heck would I find out what x cannot be to get the domain?

2x-1, again what would i do to find what x cannot be? would that answer be x cannot = .5, since 2(.5)-1=0 ?

I'm all sorts of stuck. :(

[TheProfessor 2]

Hmm, I haven't done these in a while, but as i recall

domain is all the values x can be that produce a valid y.

range is all the values y will take on.

When you find the domain, look at the function and find what values of x cause the function to produce an invalid result. One of the things you look for is what values of x cause a "divide by zero" error. You can't divide by 0, so a function like

y = (x+3) / (x + 1)

would have a domain of all real numbers but -1 because -1 causes a divide by zero

(-1 + 3) / (-1 + 1) = 2/0

For the two functions you listed,

y = -4 - x/2

y = 2x-1

Try thinking about the graphs of these functions. Both are lines, which go on forever in both directions. The line contains points for all x values, and all have a y value. So, for both functions, the domain and range will be "all real numbers".

I think thats right, but its been a while since ive done this, so someone else tell me if I missed something.