## anonymous 5 years ago Sketch the graph of f and use it to find the domain, range, and to determine whether the function is one-to-one. f(x)=root16-x^2

1. amistre64

$\sqrt{16-x^2}$ is this the equation?

2. anonymous

Yes.

3. amistre64

the graph looks like angel wings.... going off in both directions

4. amistre64

its not a one to one function

5. amistre64

http://www.wolframalpha.com/input/?i=sqrt%2816-x^2%29 lol... ok, not angel wings, but still not one to one

6. anonymous

it is the upper half of a circle with radius 4

7. anonymous

the equation for a circle centered at the origin with radius 4 is $x^2+y^2=4^2$

8. anonymous

Yeah, that's what I have on my calculator. I don't remember how to do the rest though....

9. anonymous

if you solve this equation for y you get $y=\pm\sqrt{1-x^2}$

10. anonymous

excuse me i mean $y=\pm\sqrt{16-x^2}$

11. anonymous

if you only take the positive part you are getting just the stuff above the x axis.

12. anonymous

it is in fact a one to one funciton

13. anonymous

Okay, that makes sense why it only shows up half on my graph.

14. anonymous

oops no it isn't sorry

15. anonymous

doesn't pass horizontal line test.

16. amistre64

when x = -4 OR +4 we get the same y

17. anonymous

domain is [-4,4] range is [0,4]

18. anonymous

of course i was being stupid

19. myininaya

satellite!

20. anonymous

55 to go i am in awe. but might have to wait until tomorrow yes?

21. anonymous

hello myininaya how goes it?

22. amistre64

maybe, with the holidays its hard to get good clickers lol

23. anonymous

This is true, lol. Thanks guys<333

24. myininaya

i need help this guy is totally confusing me about stuff i dont what the frick hes talking about

25. anonymous

yeah really. who does math on memorial day?

26. anonymous

what question?

27. anonymous

I do, apparently :P

28. myininaya