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Tangent Properties Symmetry, domain, asymptotes, zeroes

  • Thread starter aisha
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[tex] y=-2\tan (3x+\pi) +3 [/tex]

state the following

period = [tex] \frac {\pi} {3} [/tex]

y-intercept=3

Range = {y: yER}

Domain= {x: ?????????}

Symmetry???

Vertical Asymptotes?

Zeros???

Can someone please tell me how to figure out what is left there are so many zeroes and i cant read them off of my graphing calculator and same with vertical asymptotes aren't zeroes and this the same thing? I'm a little confused can someone please help me out? Im not sure how to find symmetry either .
 
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Answers and Replies

G01
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Peope here won't do your homework for you. If you can prove to someone you gave the problem decent effort before coming here for help then maybe someone will POINT YOU IN THE RIGHT DIRECTION but not solve it for you.
 
HallsofIvy
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G01 said:
Peope here won't do your homework for you. If you can prove to someone you gave the problem decent effort before coming here for help then maybe someone will POINT YOU IN THE RIGHT DIRECTION but not solve it for you.
That's just a little bit harsh- The original post DID include answers to the the first part of the problem- period, y-intercept, range.

Aisha- the "domain" is the set of all x values for which the function CAN be calculated. The basic function here is tan(x). For what values of x is it NOT possible for find tan(x)? Looking at a graph of y= tan(x) in your textbook might help you see that.


Symmetry should be obvious from a graph- or from the fact that tan(-x)= ?


Vertical asymptotes- closely associated with the problem of finding the domain! Look at graph- perhaps using a large "window".

zeros- where is [tex] y=-2\tan (3x+\pi) +3= 0 [/tex]? Solve the equation!
 
584
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ok i still dont think I get symmetry but for the range ive got
{x: x cannot = -90,90,270,...,XER}
this is from tan(theta) not from the equation I wrote in the first post.

If i look at that equation on my graphing calculator i get
{x: x cannot= 0.47, 1.53, -3.76, -4.80.. XER}

the vertical asymptotes will be the same numbers I assume

Please help me out a little more thanks :yuck:
 
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im not sure how to solve the equation equal to zero
 
HallsofIvy
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Aisha- did you notice that [itex]\pi[/tex] in your function? When trig functions are used as functions, rather than to solve problems about right triangles, the argument is always in radians, not degrees!

tan(x) is not defined (and so the domain does not include) [itex]\frac{\pi}{2}[/tex], [itex]-\frac{\pi}{2}[/itex], or generally any odd multiple of [itex]\frac{\pi}{2}[/tex]- that is, any number that can be written [itex](2n+1)\frac{\pi}{2}[/itex] where n represents any integer.
However, your function involves [itex]tan(3x+\pi)[/tex]. For what values of x is [itex]3x+\pi= (2n+1)\frac{\pi}{2}[/itex]?

Of course, you can solve [itex]-2\tan (3x+\pi) +3 = 0[/itex]!
That's the same as saying [itex]tan(3x+\pi)= \frac{3}{2}[/itex].
Can you find [itex]\theta[/itex] so that [itex]tan(\theta)= \frac{3}{2}[/itex]?
You might need to use a calculator for that. Be sure to put it in radian mode!

Once you found that [itex]\theta[/itex], finish by finding x so that [tex]3x+\pi= \theta[/itex].
 
G01
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sry when I posted that I was having a bad day... sry to take it out on all of you guys
 

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