## anonymous one year ago What is the simplified form of x minus 5 over x squared minus 3x minus 10 ⋅ x plus 2 over x squared plus x minus 12?

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1. anonymous

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2. anonymous

@Michele_Laino

3. anonymous

I got 1/(x+3)(x-2) as an answer

4. Michele_Laino

is your expression like this: $\frac{{x - 5}}{{{x^2} - 3x - 10}}:\frac{{x + 2}}{{{x^2} + x - 12}}$

5. anonymous

x-5/(x-5)(x+2) timex x+2/(x+4)(x-3)

6. Michele_Laino

ok! sorry

7. Michele_Laino

we have these steps:

8. anonymous

ok

9. Michele_Laino

$\begin{gathered} \frac{{x - 5}}{{{x^2} - 3x - 10}} \cdot \frac{{x + 2}}{{{x^2} + x - 12}} = \hfill \\ \hfill \\ = \frac{{x - 5}}{{\left( {x - 5} \right)\left( {x + 2} \right)}} \cdot \frac{{x + 2}}{{\left( {x - 3} \right)\left( {x + 4} \right)}} = ...? \hfill \\ \end{gathered}$

10. anonymous

Was I wrong?

11. anonymous

what's going on? @Michele_Laino

12. Michele_Laino

after a simplification I got this: $\frac{1}{{\left( {x - 3} \right)\left( {x + 4} \right)}}$

13. anonymous

Thanks :) Can I ask another?

14. Michele_Laino

ok!

15. anonymous

What is the simplified form of the quantity x over 4 plus y over 3 all over the quantity x over 3 minus y over 4?

16. anonymous

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17. anonymous

i got 3x+4y/ 4y-3x

18. Michele_Laino

do you mean this: $\frac{{\frac{x}{4} + \frac{y}{3}}}{{\frac{x}{4} - \frac{y}{3}}}$

19. Michele_Laino

$\Large \frac{{\frac{x}{4} + \frac{y}{3}}}{{\frac{x}{4} - \frac{y}{3}}}$

20. anonymous

no the bottom one is x/3 - y/4

21. Michele_Laino

ok! we have to compute the least common multiple at numerator and at denominator, of your main fraction, which is 12, so we get: $\Large \frac{{\frac{x}{4} + \frac{y}{3}}}{{\frac{x}{3} - \frac{y}{4}}} = \frac{{\frac{{3x + 4y}}{{12}}}}{{\frac{{4x - 3y}}{{12}}}}$

22. anonymous

So we can get rid of the denominator of 12

23. Michele_Laino

yes!

24. anonymous

:) YAY! Thanks!

25. Michele_Laino

:)

26. anonymous

Last one?

27. anonymous

What is the oblique asymptote of the function f(x) = the quantity x squared minus 5x plus 6 over the quantity x minus 4?

28. Michele_Laino

ok! :)

29. anonymous

I got x-1 as the oblique asymptote

30. Michele_Laino

your function is like this: $\frac{{\sqrt {{x^2} - 5x + 6} }}{{x - 4}}$

31. anonymous

No. It is without the radical sign

32. Michele_Laino

no, sorry, is like this? $\frac{{{x^2} - 5x + 6}}{{x - 4}}$

33. anonymous

Yup:)

34. anonymous

I got x-1.

35. Michele_Laino

If I write the asymptote like this: y=mx+q then we have: $\Large \begin{gathered} m = \mathop {\lim }\limits_{x \to \infty } \frac{{f\left( x \right)}}{x} = \mathop {\lim }\limits_{x \to \infty } \frac{{{x^2} - 5x + 6}}{{x\left( {x - 4} \right)}} = \hfill \\ \hfill \\ = \mathop {\lim }\limits_{x \to \infty } \frac{{{x^2}\left( {1 - \frac{5}{x} + \frac{6}{{{x^2}}}} \right)}}{{{x^2}\left( {1 - \frac{4}{{{x^2}}}} \right)}} = \hfill \\ \end{gathered}$

36. anonymous

Ohh okay:) Kinda confused but I'm getting it

37. anonymous

can't you just use synthetic divison?

38. Michele_Laino

that limit is 1, so your slope is right!

39. anonymous

Thanks:)

40. anonymous

I have to go but I need more help. When will you be back?

41. Michele_Laino

I stay here for at least 2 hours

42. anonymous

Ok:) Thank you. U gtg now!

43. Michele_Laino

wait I have to compute the quantity q

44. Michele_Laino

45. Michele_Laino

we have this formula: $\large \begin{gathered} q = \mathop {\lim }\limits_{x \to \infty } \left\{ {f\left( x \right) - mx} \right\} = \mathop {\lim }\limits_{x \to \infty } \left\{ {f\left( x \right) - x} \right\} = \hfill \\ \hfill \\ = \mathop {\lim }\limits_{x \to \infty } \left\{ {\frac{{{x^2} - 5x + 6}}{{x - 4}} - x} \right\} = \mathop {\lim }\limits_{x \to \infty } \left\{ {\frac{{{x^2} - 5x + 6 - {x^2} + 4x}}{{x - 4}}} \right\} = \hfill \\ \hfill \\ = \mathop {\lim }\limits_{x \to \infty } \left\{ {\frac{{6 - x}}{{x - 4}}} \right\} = \mathop {\lim }\limits_{x \to \infty } \left\{ { - 1 + \frac{2}{{x - 4}}} \right\} = ...? \hfill \\ \end{gathered}$

46. Michele_Laino

that limit is -1, so I confirm your answer, namely the equation of your asymptote is: y=x-1