anonymous
  • anonymous
Using complete sentences, explain how to find the y-intercept for each function and determine which function has the largest y-intercept. f(x) = 2 tan(3x + π) g(x) = 3 sin(4x - π) - 2 I have absolutely no idea how to solve for this without just graphing the equations, and I am pretty sure that I have to explain how to find the y-intercepts without graphing them. I have tried solving for "x = 0", but I didn't get the correct values for the y-intercepts (I graphed these equations to check).
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
you plug in 0 for x to get the y-intercepts. f(x) = 2 tan (3x + π) f(0) = 2 tan (π) f(0) = 0 y-intercept (0, 0)
anonymous
  • anonymous
Wouldn't it be f(0) = 2 tan( (3*0) + π ) f(0) = 2 tan(0 + π) f(0) = 2 tan(π)
anonymous
  • anonymous
yes I just skipped those steps

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anonymous
  • anonymous
Oh, I see. So then: f(0) = 2 tan(π) f(0) = 2 (0.0548861508) f(0) = 0.1097723016 This isn't exactly 0 :( This is where I keep struggling :/
anonymous
  • anonymous
And on the graph, it doesn't display as "0.10977..". It says (0, 0). :((
anonymous
  • anonymous
your calculator is in degree mode. change it to radian :)
anonymous
  • anonymous
OHHHHHHHHHHHHHHHHHHHHHHHHH
anonymous
  • anonymous
Thank you so much, omg. I would have never thought to do that. I don't even know how to, since I was using my phone. I'll use my computer calculator. So tan(π) = 0 radians?
anonymous
  • anonymous
tan (π radians) = 0 radians is just another way to measure the angle. (π radians = 180°)
anonymous
  • anonymous
but yeah, just put tan π in the calc
anonymous
  • anonymous
It still gives me "0.0548861508" D:
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=tan+0
anonymous
  • anonymous
OOHHHHH http://www.wolframalpha.com/input/?i=tan+%CF%80
anonymous
  • anonymous
So my calculators were reading it as "3.141592..." and not as "pi radians", then. This makes things so much clearer now. I really appreciate all of your time and help.
anonymous
  • anonymous
no problem

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