Ask your own question, for FREE!
Mathematics 14 Online
OpenStudy (anonymous):

Choose the graph of y = tan x/2.

OpenStudy (anonymous):

OpenStudy (anonymous):

OpenStudy (anonymous):

OpenStudy (anonymous):

OpenStudy (anonymous):

@satellite73

OpenStudy (anonymous):

@terenzreignz

OpenStudy (anonymous):

its the third one right?

OpenStudy (anonymous):

the third on is wrong for sure

OpenStudy (anonymous):

Consider how at "pi", you will have a vertical asymptote because the tan x/2 function has cos x/2 in the denominator and that (cos x/2) will be "0" when "x" = pi (because cos pi/2 = 0), making tan x/2 "infinite" at pi and multiples of pi, hence, the vertical asymptote. So, no, #3 is not the right choice, but with the above hint, you should be able to get it. Also, you'll need to know that the slope of the derivative of the tan function (sec^2) is positive. If you are not in calculus yet, then take this to mean that the graph will look similar in shape to the y=x^3 function. It can't be "c" because that graph shows a function equaling "0" at x=pi where tan x/2 is infinite at x/2. It can't be "d" because that graph is infinite at x/2. That would hold for the function tan x, but not for the function tan x/2. It can't be "b" because right off we see that the shape of the graph is wrong because the slope of the derivative is negative. That leaves . . .

OpenStudy (anonymous):

A!

OpenStudy (anonymous):

You got it! But do go over the whole explanation because there is a lot of good info there.

OpenStudy (anonymous):

I have one more for you since your the only one who understands this

OpenStudy (anonymous):

Good luck in all of your studies and thx for the recognition! @SBrashears1

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!