## anonymous one year ago How do you identify the oblique asymptote for a function y=x^2-4/x+1

1. Nnesha

not all functions have oblique asymptote if the degree of the numerator one larger than the denominaotr degree then you can find slant(oblique)asy.

2. Nnesha

$\huge\rm y=\frac{ x^\color{ReD}{2}-4 }{ x^{\color{blue}{1}}+1 }$ highest degree of the numerator is 2 and the denominator is one so we can find oblique asy divide x^2-4 by x+1 using long division or synthetic division btw for this question you can factor x^2 -4

3. anonymous

so the degree of the numerator is bigger than that which is the denominator so this one will be a slant how do you find a slant

4. anonymous

so factoring x^2 - 4 would be x- 2?

5. Nnesha

no that wouldn't work sorry we should divide

6. anonymous

oh okay thats fine

7. Nnesha

synthetic division or long which one is easy for u?

8. anonymous

synthetic division does the one become -1 or stay the same

9. Nnesha

yes right we should solve for x x+1 = 0 x=-1 |dw:1444312769154:dw| x^2 -4 is same as x^2 +0x - 4 we should write highest degree to lowest so that's why i wrote 0 there

10. Nnesha

carry down the leading coefficient and then multiply bye -1 |dw:1444312872330:dw| combine them 0 -1 = -1 and then repeat steps

11. anonymous

ah I see one more thing real quick a negative * a negative is a positive right?

12. Nnesha

yes negative times negative = positive :=)

13. anonymous

so the final answer should be 1x^2-x-3?

14. Nnesha

hmm when we divide by synthetic division the degree of the answer should be one less than the original equation

15. Nnesha

|dw:1444313248979:dw|

16. Nnesha

does it make sense ? hmm

17. anonymous

okay so its x-3 yeah!

18. Nnesha

|dw:1444313357036:dw|

19. anonymous

ohh Okay so its x-1 with a remainder of -3

20. Nnesha

|dw:1444313566542:dw|

21. Nnesha

yes right it's y =x-1 that's our oblique asy.

22. anonymous

AHH Mrs Nnesha you are awesome and if I had a medal to give you I would Thank You so much

23. Nnesha

np :=) hope that helpd