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# How do I draw this curve? watch

1. I find it really difficult to draw curves other than the simple functions such as y=x^2, y=x^3 etc.

How would I go about working out what the following graph would look like:
y = x^2 - 1/x?
I rearranged it to get: y= (1-x^3)/x
But now I'm stuck! arghhhh

(I know what it does look like as used graph-drawing program, but I find it difficult to work it out for myself!)
2. (Original post by Hoofbeat)
I find it really difficult to draw curves other than the simple functions such as y=x^2, y=x^3 etc.

How would I go about working out what the following graph would look like:
y = x^2 - 1/x?
I rearranged it to get: y= (1-x^3)/x
But now I'm stuck! arghhhh

(I know what it does look like as used graph-drawing program, but I find it difficult to work it out for myself!)
Well you can draw y = x^2 and y = 1/x, so you have a good idea what y = x^2 - 1/x looks like.

Other things, dy/dx = 2x + 1/x^2, If dy/dx = 0, then 2x^3 = -1, so you can get a turning point. You've also got y = 0 at x = 1, and the function is not defined as x -> 0. Also as x -> +/- inf, then 1/x -> 0.
3. (Original post by beauford)
Well you can draw y = x^2 and y = 1/x, so you have a good idea what y = x^2 - 1/x looks like.

Other things, dy/dx = 2x + 1/x^2, If dy/dx = 0, then 2x^3 = -1, so you can get a turning point. You've also got y = 0 at x = 1, and the function is not defined as x -> 0. Also as x -> +/- inf, then 1/x -> 0.
Thanks. I usually try and find roots and turning points of equations, but loads of ppl seem to be able to see a function and no excatly what the graph looks like without doing any calculus etc! arghhh I hate curves!
4. (Original post by Hoofbeat)
Thanks. I usually try and find roots and turning points of equations, but loads of ppl seem to be able to see a function and no excatly what the graph looks like without doing any calculus etc! arghhh I hate curves!
those people are mad!!
5. (Original post by Hoofbeat)
I find it really difficult to draw curves other than the simple functions such as y=x^2, y=x^3 etc.

How would I go about working out what the following graph would look like:
y = x^2 - 1/x?
I rearranged it to get: y= (1-x^3)/x
But now I'm stuck! arghhhh

(I know what it does look like as used graph-drawing program, but I find it difficult to work it out for myself!)
f(x) = 1 - x³; f(1) = 0, so x-1 is a factor
1 - x³ = (x-1)(-x²-x-1) = (1-x)(x²+x+1)

Although this doesn't help!

I would just find where it crosses the axis, and at what values there are asymptotes (like x=0 here, as 1/0 doesnt have a value)

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