Algebra

f(x)=(x3-27)/(x2+x-12) find the symptopes

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any one help please
al3xis503
15 October 2012
Asymptotes, I believe you mean? These describe the behaviour of the function as x becomes very large, ie when x is infinity or x is minus infinity. The way to deal with this is just to get rid of as many terms that "won't matter" when x is large as possible. In this case, the numbers stop making a difference for a start, so your function becomes 'about' f(x) = x^3(x^2+x) = x^5 + x^4. Also, when x is really really big, the x^5 will be so much large than x^4 that you can drop the last term. So the asymptotes as described by f(x) tends to x^5 as x tends to (plus or minus) infinity. Hope this is clear. Also as a note, in maths you write an arrow ---> for "tends to".
jim360
15 October 2012
Oh, darn, I misread your equation. Let me fix that: f(x) = (x^3-27)/(x^2+x-12) The first thing you have to do is factorise the bottom part of this. Using the standard techniques you'll find that: x^2+x-12 = (x+4)(x-3) So now our function reads: f(x) = (x^3-27)/[(x+4)(x-3)] Now there's a rule in maths that whenenever you divide by zero, you get infinity. So here, if x = -4 the function will "blow up" and try to go to infinity, but it will never reach it. So the first asymptote, which does indeed describe the behaviour of the function "at infinity", is x = -4. Also at x = 3, you might expect the function to blow up since that would also mean a division by zero, but not so! At x = 3 the top half of the function would become 3^3 - 27 = 27 - 27 = 0 too, and in fact these two zeroes "cancel" each other out. So x = 3 is NOT an asympote. Secondly, we also look at the behaviour as x tends to infinity. As I said before, you can start to drop terms in the graph that just stop mattering. Again, the constant numbers disappear first, giving: f(x) = (x^3)/(x^2+x) Cancel an x from top and bottom: f(x) = x^2/(x+1) and drop the one, and cancel another x: f(x) = x So as x tends to infinity f(x) tends to x. So your asymptotes are x = -4 and y = x. Hope this time I've got it right and that this is clear.
jim360
15 October 2012
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