## unicwaan one year ago I need to find the domain and range, x and y intercepts, horizontal and vertical asymptotes. Can someone help me through it please?

1. anonymous

2. unicwaan

There should be a picture attached but it's: X^2 + x - 2 divided by x^2 - 3x - 4

3. anonymous

first thing, you should factor the num and denom of your function$f(x)=\frac{x^2+x-2}{x^2-3x-4}$have you tried to do so?

4. unicwaan

I did, the numerator factors into (x-1)(x+2) and the deniminator factors into (x-4)(x+1)

5. anonymous

very good, so what will be the domain considering the denom?

6. unicwaan

The domain would be all reals except x=4 and x=-1?

7. anonymous

quite right

8. anonymous

let's work on range of the function

9. unicwaan

alright

10. unicwaan

Is the range all reals?

11. anonymous

thats right, do you have a reasoning for that?

12. unicwaan

Well by the graph it extends infinitely up and down

13. unicwaan

and because quadratic functions' range will always be all reals

14. anonymous

that's right, but what if we don't know about the graph

15. anonymous

note that quadratics don't cover all reals|dw:1434182772266:dw|

16. unicwaan

Oh okay, then I do not a reasoning

17. unicwaan

do not have a reasoning*

18. anonymous

function is continuous on the interval (-1,4) and$\lim_{x \to -1^{+}} f(x)=\infty$$\lim_{x \to 4^{-}} f(x)=-\infty$therefore range is all of real numbers

19. anonymous

does that make sense?

20. unicwaan

So when you plugged in the ordered pair, the output was infinite basically?

21. anonymous

thats right

22. unicwaan

okay that makes sense

23. anonymous

for example$\lim_{x \to -1^{+}} f(x)=\frac{1 \times -2}{-5 \times 0^{+}}=\infty$

24. anonymous

ok what are x and y intercepts?

25. unicwaan

Alright so the y intercept = 1/2 because you plug in 0 for x to find y. The x intercept = -2,1 because they are the zeros of the numerator

26. anonymous

very right, thanks

27. anonymous

28. unicwaan

The horizontal asymptotes is 1 because the degree of each function are the same and therefore, must divide the leading coeeficients. The vertical asymptotes are y= -1,4 because they are the zeros of the denominator.

29. anonymous

right, you can put your words in mathematical phrases

30. anonymous

now you can graph your function

31. unicwaan

Okay so for the asympotes I put dotted lines to represent them, and then I plugged in points according to get this graph|dw:1434183896630:dw| Poor drawing but the basic jist of it :/

32. anonymous

we're done

33. unicwaan

Great! Thank you so much this really helped! :D

34. anonymous

no problem