Good day /sci/!
I need your help. I was really sick for the past 2 weeks and had to miss most of my classes. My math professor (precalc) did a quiz and let me to it at home since I couldn't attend. I wasn't in class while they covered this material, and forgot it since a while back in highschool. Can anyone please solve this one and if possible explain briefly how they did, so I can use this as an example? Thanks anons
Underage & b8. Gtfo homework faggot
>>7705695
I'm 21, and I covered this material when I was 17
>>7705703
You could always try googling how to do it? This is easy stuff
http://www.bing.com/search?q=phase+shifts+of+trigonometric+functions&FORM=AWRE&PC=CY06
Honestly why even post on 4chan? Would it literally not be easier and quicker to just Google it?
>>7705710
Thant is true, And I'm doing it, even while I posted here, but I figured maybe someone has the few minutes to explain, maybe will be more clear than websites I found on google.
And if it was just me jerking off while trying to get others do my homework, I'd post more than one exercise, I posted this one to use the solution as a reference
Ok faggot I feel nice.
The number in front of the trig function is the amplitude. If it's negative, you start on the opposite side of the x axis.
The period is how long it takes for the function to go through an entire oscillation. For sin and cos, it's 2pi. However this can be modified. If you had sin(x/4), you'd multiply 2pi by 4.
The phase shift is the part a number you add or subtract to the period. It just changes your initial position.
For example, 4cos(3x+1/9pi) has an amplitude of 4, period of 2pi/3, and a shift of 1/9 pi.
That should help you enough
I learned this in Physics, not Calc.
But, listen to this guy >>7705734 he's right
>>7705716
To graph the function -5cos(4x), consider this:
1. Start with cos(x)
2. The negative sign makes the cosine flip.
3. The 5 makes the peak and trough 5 times larger (similar to something like 5x^2).
4. The 4 in the "4x" term makes the period 4 times shorter. This is the frequency part of the equation, the actual period is calculated by doing "2pi/b," where "b" is whatever is multiplied by the x term.
5. The first question has no phase shift, so I'll look at the second question. Since the phase shift for the second one is -pi/6, that means that instead of starting the normal sine function at 0 (as you would normally do), you have to start it at pi/6, and it will end at 4pi+pi/6.
>>7705734
I also should have specified that a positive number in the shift part actually moves the graph back.
So having cos(x+pi/4) moves the graph pi/4 to the left, yielding a negative shift.
>>7705734
Thanks anon!
>>7705681
sin(x) and cos(x) range between -1 and +1, so when you have something like y = k * sin(x) or y = k * cos(x), the y ranges between -k and +k. If you have something like y = k * sin(x) + h, y ranges between -k+h and +k+h.
The period of sin(x) and cos(x) is 2 pi. When you have a function in the form sin(k * x) (k now is in the argument of the function) the period becomes (2 pi)/k. If you ADD something (like y = cos(x+h)) the period does not change, but the whole function is translated by a vector (-h; 0).
