Right so I'm doing Asymptotes in maths at the moment and I missed most of the classes so I'm wondering if &T can help me with one question. This is it exactly

Let f(x)=_x_ , x not equal to 3 and is of R
x-3


i) Show that the curve f(x) has no points of inflection -> DONE
ii) Fine the equations of the asymptotes of the curve y = f(x) -> DONE
iii) Draw a sketch of the curve y = f(x) -> DONE

This is the part i'm stuck on,
iv) If the tangents to the curve at (x1, y1) and (x2, y2) are parallel, show that x1 + x2 = 6, x1 is not equal to x2

So I'm presuming that the slopes are the same so:

dy= _-3_
dx (x - 3)[ to be squared]


That's as far as I've gotten, Any ideas ? Wrong forum ? Thanks

What is "R"?

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It's X.

"IR" is the set of real numbers (denoted by the letter R with the left part of it in bold).

2OP: it's really simple.

tangent to the curve = y' = (-3)/(x-3)^2 - you figured this part out.

Now, you know that the tangents are equal in points x1 and x2. Let's write it down:

(-3)/(x1-3)^2 = (-3)/(x2-3)^2 <=>
<=> (x1-3)^2 = (x2-3)^2 <=> |x1-3| = |x2-3|

Now we know that x1<>x2. That means that x1-3 = 3-x2 <=> x1+x2 = 6.

Job done.

The only math i care about...































+1

Yes, wrong forum.

Moving to MS

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Thanks that's 80% on my maths summer exam

What level of maths is this for?

Looks pretty simple to me, I'd say some basic school maths.

I hate you. I have an IB HL calculus BC exam monday and tuesday.

*slits wrists*