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anonymous
 one year ago
F(x)= 3sin(2x) Can someone help me find these and explain how they did it. Will FAN and MEDAL!
even/odd:
symmetric about:
x intercepts:
y intercepts:
anonymous
 one year ago
F(x)= 3sin(2x) Can someone help me find these and explain how they did it. Will FAN and MEDAL! even/odd: symmetric about: x intercepts: y intercepts:

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0In general, the sin function is an odd function. cos and sec are even and the other 4 trig functions are odd. The yintercept is just found by plugging in x = 0. 3sin(2*0) = 3sin(0) = 0. So you get a yintercept of (0,0) The xintercepts are a bit trickier. We want to solve the equation 3sin(2x) = 0. To make this easier, we can divide by 3 and just solve sin(2x) = 0. To solve this, lets first consider the equation sinx = 0. Can you tell me the x values that make sinx = 0?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I thought to find the y intercept you had to replace the x with y

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm not actually sure what doing that would give you. Replacing x with y sounds like what you would do for finding an inverse. But think of it this way: A yintercept is a spot where you cross the yaxis and will always be when x = 0. Any yintercept will look like (0,y). So to find a yintercept in any equation, its always best to just replace x with 0 and solve.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah I see what you mean now, my bad

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No worries. But yeah, when it comes to solving for xintercepts, I would start off with solving sinx = 0. Do you know the solutions of that?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Are we considering only the interval [0,2pi), or all values?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think it was 0,2pi

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ah, okay. Well, if its not all values, then the usual interval is [0,2pi) with 0 included, 2pi not included. In this case, the solutions would be 0 and pi. So the solutions are x = 0 and x = pi. But our problem was sin(2x) = 0. Thus we actually have 2x = 0 and 2x = pi. This implies x = 0 is a solution and x = pi/2 is a solution. Can you see thaT?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah. Thats usualy my strategy to solving such problems. Like, if I have sin(3x) = 1 for example, I first consider sinx = 1, which is true for \(x = \pi/2 + 2k\pi\). Then I replace x with the actual angle, in this case 3x, giving me \(3x = \pi/2 + 2k\pi\) \(x = \pi/6 + 2k\pi/3\) Kinda see the pattern and what I would be doing each time?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yep. So everything kind of make sense then?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0One more, I forgot how to figure out what it's symmetry is

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, you kind of have to know that cosx and secx are both even functions and the other 4 are odd functions. The normal process to show a function is odd is to show that f(x) = f(x) and the normal process to show a function is even is to show f(x) = f(x). Doing that method is a bit trickier with trig functions, so its simply best to just know that cosx, secx = even, the other 4 = odd.

freckles
 one year ago
Best ResponseYou've already chosen the best response.0but if x=2pi is a solution of sin(x)=0 then 2x=2pi is a solution of sin(2x)=0 and pi is also in the interval [0,2pi) So I think if you are solving sin(2x)=0 on [0,2pi) and you decide to solve sin(u)=0 well notice we replaced 2x with u so 2x=u and if 2x=u then dividing both sides by 2 gives us x=u/2 so instead of having x is between 0 and 2pi you have u/2 is between 0 and 2pi and so u is between 0 and 4pi so you actually need to solve sin(u)=0 on [0,4pi) to find all the solutions of sin(2x)=0 on [0,2pi)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Freckles is right, just overlooked that. Because sin(2x) has a period of pi, need to consider solutions in an interval twice as long, so [0,4pi) my bad, thanks, freckles :)

freckles
 one year ago
Best ResponseYou've already chosen the best response.0you are doing awesome :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Im stil glad you caught the mistake, lol. On a random note, any good with geometry, freckles?
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