Question

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:

Answer 1

In general, the sin function is an odd function. cos and sec are even and the other 4 trig functions are odd. The y-intercept is just found by plugging in x = 0.
3sin(2*0) = 3sin(0) = 0. So you get a y-intercept of (0,0)
The x-intercepts 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?

Answer 2

I thought to find the y intercept you had to replace the x with y

Answer 3

I'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 y-intercept is a spot where you cross the y-axis and will always be when x = 0. Any y-intercept will look like (0,y). So to find a y-intercept in any equation, its always best to just replace x with 0 and solve.

Answer 4

Yeah I see what you mean now, my bad

Answer 5

No worries. But yeah, when it comes to solving for x-intercepts, I would start off with solving sinx = 0. Do you know the solutions of that?

Answer 6

Pi and 2pi?

Answer 7

Are we considering only the interval [0,2pi), or all values?

Answer 8

umm I have no idea

Answer 9

I think it was 0,2pi

Answer 10

Ah, 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?

Answer 11

Yes I think so

Answer 12

Yeah. 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?

Answer 13

Yes

Answer 14

Yep. So everything kind of make sense then?

Answer 15

One more, I forgot how to figure out what it's symmetry is

Answer 16

Well, 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.

Answer 17

but 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)

Answer 18

Freckles 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 :)

Answer 19

you are doing awesome :)

Answer 20

Im stil glad you caught the mistake, lol. On a random note, any good with geometry, freckles?