and that's the question I tried to answer in post 5.

If that isn't right then the only other interpretation I can think of is

That has x = 0 as an asymptote and doesn't cross the y axis at all.

**This shows the importance of correct placing of brackets.**

Oh wait a minute. Did you mean:

In which case, same asymptotes as post 5 and same crossing .... in fact all the answers are unchanged, except .....

As x tends to infinity the function tends towards

So now we have a horizontal asymptote.

Bob

]]>it should be f(x) = 3x^2+4 over x^2 -x -6. If you understand what I mean

]]>I'm assuming that is

when x = 0

The vertical asymptotes are at x = -2 and x = + 3.

I'm never happy with these questions that ask for the domain. I think the setter should be declaring a domain as part of the function definition rather than expecting the student to guess what domain is expected.

But, assuming the questioner means "The domain is the set of real numbers less any values that are inadmissible", then the inadmissible values are x = -2 and x = +3, because at these x values the f(x) value cannot be computed as it would involve division by zero.

As x tends to + or - infinity the graph approaches f(x) = 3x^2 so there are no horizontal asymptotes.

g(x) : the quadratic can be similarly factorised to give the asymptotes and inadmissible values.

As x tends to + or - infinity the graph approaches g(x) = 2x. As this is a line, it counts as an asymptote, but it is not horizontal.

Bob

]]>Says there you also have to graph it. Did you go here and do that?

]]>f(x) 3x^2 + 4/x^2-x-6 and

g(x) = 2x-15/x^2-8x+15

I was marked 4/8 but wasn't told what I did wrong.

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