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# Shape of curve watch

1. Sorry to be a pain again (afraid you'll have to get use to me pestering with questions while I attempt to teach myself some of P4-P6!).

Could anyone tell me what the graph of y=(x^2 +10 - 7x)/x would look like? I found the roots of the equation and also the position of the assymptotes however from just looking at the equation I still wouldn't have a clue of the general shape of the curve unless I plotted a few points to help. Is there an easier way?

Also, I'm trying to solve the following inequality:

[(x+3)/(x-1)] < [(x-3)/(x+1)]

and am trying to draw a graph of the curve of each of the separate sides to find the solution sets. However, I'm stuck as to how I go about drawing the curve. I know it's probably really easy but when I see (x+3)/(x-1) I don't have a clue about the shape! Is such a function the same shape as a 1/x graph and if so how do I work out the positioning of the assymptotes etc.

Thanks for all your guys' help! I'm sure you don't really mind as it'll keep your brains ticking over the summer
2. (Original post by Hoofbeat)
Sorry to be a pain again (afraid you'll have to get use to me pestering with questions while I attempt to teach myself some of P4-P6!).

Could anyone tell me what the graph of y=(x^2 +10 - 7x)/x would look like? I found the roots of the equation and also the position of the assymptotes however from just looking at the equation I still wouldn't have a clue of the general shape of the curve unless I plotted a few points to help. Is there an easier way?

Also, I'm trying to solve the following inequality:

[(x+3)/(x-1)] < [(x-3)/(x+1)]

and am trying to draw a graph of the curve of each of the separate sides to find the solution sets. However, I'm stuck as to how I go about drawing the curve. I know it's probably really easy but when I see (x+3)/(x-1) I don't have a clue about the shape! Is such a function the same shape as a 1/x graph and if so how do I work out the positioning of the assymptotes etc.

Thanks for all your guys' help! I'm sure you don't really mind as it'll keep your brains ticking over the summer
The best way to do the inequality is as follows:

(x+3)/(x-1) = 1 + 4/(x-1),

(x-3)/(x+1) = 1 - 4/(x+1),

so you want 4/(x-1) < -4/(x+1), i.e 1/(x+1) + 1/(x-1) < 0, i.e 2x/(x+1)(x-1) < 0. Now try to solve...
3. y=(x^2 +10 - 7x)/x

Factorise the top line into (x-2)(x-5)

y = (x-2)(x-5)/x

You now know that the graph will have an asymptote at 0, because y is undefined when you divide by 0. It will also be cross the x-axis at 2 and 5.

As x tends to infinity y tends to infinity, and as x tends to -infinity y tends to -infinity.

Some more ways to think about it include writing it out as

y= x +10/x -7

Now the -7 term is a constant and can be ignored for the point of sketching. We shall look at how x and 10/x look for small and big x. When x is small obviously the x term is small, but 10/x becomes large, so for small values of x i.e. |x|< 10 the graph will resemble y=10/x. But for larger x the graph resembles y=x.

Equipped with this knowledge you should have enough to sketch it. I have included a graph drawn in mathematica to illustrate.
Attached Images

4. (Original post by Hoofbeat)
...
and am trying to draw a graph of the curve of each of the separate sides to find the solution sets. However, I'm stuck as to how I go about drawing the curve. I know it's probably really easy but when I see (x+3)/(x-1) I don't have a clue about the shape! Is such a function the same shape as a 1/x graph and if so how do I work out the positioning of the assymptotes etc.
...:
Ok, you can see here that there is a denominator that gets a zero value viz. x = 1.
So the line x=1 is an asymptote.
Looking at (x+3)/(x-1), then as x-> ∞, (x+3)/(x-1) -> x/x which = 1, i.e. y ->1.
So the line y=1 is an asymtote.
Rearranging the eqn,
y = (x-1 + 4)/(x-1)
y = 1 + 4/(x-1)
Again you can see that as (x-1) -> ∞, y ->1.

And, as you said, you can now see why it looks a bit like the 1/x function.
You know that 1/x has asymptotes at x=0 and y=0, so just draw your function like 1/x, but using x=1 and y=1 as the asymptotes.

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