Graph f(x)=x^2-3x-10/x-2. Find the domain, range, VA, HA/SA

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Graph f(x)=x^2-3x-10/x-2. Find the domain, range, VA, HA/SA

Mathematics
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HI!!
Hey:)
is it \[\frac{x^2-3x-10}{x-2}\]

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Other answers:

Yes
ok lets do them one at a time
domain is easy enough, set the denominator equal to zero and solve then say "all real numbers except that one"
what do you get for that? (you can do it in your head)
2
yeah so all numbers except 2
now before we continue, lets make sure this is not a trick question can you factor the numerator and cancel?
Let me see...
hmmi think it does factor, but not cancel !
(x-5)(x+2)/x-2
yeah but nothing cancels, so lets continue but you should always check that first
once we have the domain, we also have the vertical asymptote too
clear or no?
So it is also 2
yes!
careful here
the vertical asymptote is a vertical line, not a number
so you want to return and answer of \(x=2\) the vertical line, not just the number 2
clear?
Oh! Yes! Clear
now for the horizontal asympote the degree of the numerator is ?
2
and the denominator?
there is no horizontal then since it's greater than the denom.
bingo
but the degree of the numerator is one more than the degree of the denominator that means there will be a slant asymptote do you know how to find it?
I never learned about a slant asymptote so I dont know :(
ok it is easy to say what to do , but not easy for me to write in one word "divide"
when you divide a polynomial of degree 2 by a polynomial of degree 1 you get a polynomial of degree 1 ( a line) and a remainder ignore the remainder, the line is the slant asymptote
you can use either long division or (much easier) synthetic division do you know how to do that?
Yes :)
whew like i said it is very hard for me to write division here what do you get?
take your time
I did it in a rush so I got x^2-1x-6.
hmm no you should have a polynomial of degree 1, not a quadratic
1 -3 -10 2 ______________ 1
that is the iniital set up then 1 -3 -10 2 2 ______________ 1 -1
the next line is unimportant since we only need the line, not the remainder the line has the coefficients from the final row so \(x-1\) slant asymptote is \(y=x-1\)
Oh! Now I know what I did wrong! When did the fak divison, i fogot to add the "0^2" Lol but yes, I know understand
ok then we are done!
oh except for the range
I'm so bad at finding that.
in this case it is real easy, since it has a vertical asympote and no horizontal asymptote that means it goes from \(-\infty\) to \(\infty\)
you can write your answer as an interval \((-\infty, \infty)\) or just say "all real numbers"
it is harder for some other ones, but not for this one
Yay! Thanks so much:)))
\[\color\magenta\heartsuit\]

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