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

Question

amistre64 · Answer

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

anonymous · Answer

Yes.

amistre64 · Answer

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

amistre64 · Answer

its not a one to one function

amistre64 · Answer

anonymous · Answer

it is the upper half of a circle with radius 4

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

it is in fact a one to one funciton

anonymous · Answer

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

anonymous · Answer

oops no it isn't sorry

anonymous · Answer

doesn't pass horizontal line test.

amistre64 · Answer

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

anonymous · Answer

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

anonymous · Answer

of course i was being stupid

myininaya · Answer

satellite!

anonymous · Answer

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

anonymous · Answer

hello myininaya how goes it?

amistre64 · Answer

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

anonymous · Answer

This is true, lol. Thanks guys<333

myininaya · Answer

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

anonymous · Answer

yeah really.  who does math on memorial day?

anonymous · Answer

what question?

anonymous · Answer

I do, apparently :P

myininaya · Answer

follow me