Ok, we have y equals the sine of 2x minus pi. Now this is how I am going to have my
kids do this so hopefully you appreciate this and it works for you. We are going to list
off all the information that we know about this graph then make a t table and sketch
it. Ok, the amplitude is set by "a" in the beginning of your equation and mine is equal
to one so the amplitude is one. The amplitude it the absolute value of a. The period of
sine, cosine, secant, and cosecant is 2pi over divided by "b" which is in front of the
x. It is 2pi divided by 2 so the period is pi. The phase shift, which I am going to just
write ps, the phase shift is your left or right horizontal movement, is c divided by
b so it is going to be pi over two. Because we have a minus sign in our parenthesis which
is part of our standard form, a*trigfunction(bx-c) + d. That minus sign is there in the original
formula so the phase shift is the movement to the right. The graph is going to be moved
pi over two units to the right. Our vertical shift, which I am going to write as VS, our
vertical shift is from a plus or minus of a constant at the end of our equation and
there is none. Our vertical shift is zero. Now if we are doing this without a calculator,
we need to know what we are counting by, that is going to be one quarter of the period.
So we are going to count by one quarter of the period of pi, so we get pi over four.
I want to minimize the amount of mental math I have to do, I am going to write my phase
shift, my count value, and my c value all with matching denominators... common denominators
to alleviate the mental math. We have pi over four, pi over two, and pi. The common denominator
between all of these values is four. I am going to write this phase shift as 2pi over
four and when I write this equation in the t table I am going to write c as 4p over four.
Again count by, phase shift, and c should all be written with a common denominator to
make the mental math a little bit easier. Here is my t table. X... and Y equals sine
of 2x minus 4pi over 4. As long as you know how to deal with fractions and you know just
the quadrantal angles of the unit circle, the thinking parts of this problem are done.
We are going to start at the phase shift which is 2p over four. Don't bother reducing this,
that is why/how I want to do all this mental math easily. Now I want to count by one quarter
of the period so two over four plus one over four is three over four, so 3pi over 4, 4pi
over 4, 5pi over 4, and I need five points so 6pi over 4. Again we are counting by one
quarter of the period. Now I am going to show all the steps in one of these lines but after
that I am going to do mental math so I can get through two examples in fifteen minutes
or at least attempt to. The sine of two times 2pi over four minus 4pi over four. Well, two
times two is four so that becomes the sine of 4pi over four minus 4pi over four. That
comes out to be zero and as you are doing this work, if you start at the phase shift
and count by one quarter of the period, if you do your fraction work correctly, you should
you be getting your quadrantal angles zero, pi over two, pi, three pi over two and back
to two pi. If you do not get a trig function giving you an answer of one, zero, or negative
one you have done something wrong. So check your work. That is zero and the sine of zero
is zero. Now I am just going to verbally say this. The sine of 2 times 3p/4 minus 4pi/4.
Two times three is six over four. So six over four minus four over four, six minus four
is two and two over four is one half. So this is the sine of pi/2 and the sine of pi over
two is one. Plug in 4p/4. The sine of 2 times 4pi/4 minus 4p/4. That is eight and eight
minus four is four over our common denominator of four. Four pi divided by four is pi and
the sine of pi is zero, so we are back to 0. The sine of 2 times 5pi/4 minus 4pi/4.
Two times five is ten and ten minus four is six over the common denominator of four. Six
over four reduces down to three over two, so this is the sine of 3pi/2 which is negative
one. When you plug in the last one you are going to get a value of zero. Here we have
our t table set up and we are going to graph two periods of this graph. Lay down the x
axis which again is representing theta. We have the values of 2pi/4, 3pi/4, 4pi/4, 5pi/4,
and 6pi/4. Now those are the values that came off my t table. They are all positive and
I don't know where zero is so I am going to count backwards until I get to zero or I skip
over it to a negative number. So 3, 2, 1, ok 0, negative pi/4, negative 2p/4. Now that
I know where zero is I am going to place my y axis down here and mark off the values of
one and negative one. Now I am going to start graphing my sine function. Pi over two zero.
Three pi over four and one. Pi, because four divided by four is one, so pi zero. Five pi
over four we are at negative one and at six pi over four we are back to zero. Now here
is what I have so far. Now do you remember that the regular sine graph I did in the last
video? The regular y equals the sine of x, the parent function. It's oscillation at zero
zero. Well why is this one starting at 2pi over four. Well it is because there is a phase
shift of pi over two and 2pi/4 reduces down to pi over two. That has pushed the graph
to the right a little bit. What is the pattern now? What if I wanted to graph two periods
and not just one which is what I have got here. Zero, negative one, zero, one, zero,
and I have got these values on the x axis to match my t table so the pattern is very
easy now. 0,-1,0,1,0, ok so this point is at pi over 4 negative one, back to zero, back
to one, back to zero and then you see the exact same pattern re-stamped basically from
what I had over here. There is two periods of y equals sine of two x minus pi. WHOOO!!!
Let's do another one. I think I have time. I will just go until my fifteen minutes are
up or do whatever I can manage to get done. One more graphing problem. We have got y equals
negative three cosine of one half x plus four. No phase shift but we have a fractional b
value which will increase the period. Amplitude is the absolute value of a so it is going
to be three. The period for sine, cosine, secant, and cosecant is 2pi divided by b and
when you pigtail that up, when you do 2 pi divided one half that is the same as 2pi times
two which is equal to 4pi. So since my b value is less than one, it is actually going to
stretch out the period and instead of oscillating between zero and 2pi, it is going to take
4pi steps on the x axis to complete that period. What else do we have. The phase shift, there
is no plus or minus in my parenthesis so this is zero. My vertical shift is my constant
at the end, not inside my the math function so it is four. We are going to count by a
quarter of the period. What is the period? The period is 4pi so we are going to count
by one quarter of 4pi which is pi. Again, if we are counting by a quarter of the period
we are going to get those quadrantal angles around the unit circle and get the easy values
of zero, one, and negative one. T table. We have x and y equals negative three cosine
of one half x plus four. Books generally don't use parenthesis around this but I like to.
This negative in front of the three is going to reflect the graph over the cosine function
over x axis. So instead of the function starting at one it is going to start at negative one...
maybe. Let's see what the three does. We are going to start our t table at the phase shift
and count by a quarter of the period. And we do not need to worry about common denominators
because our denominator is one. We are starting at the phase shift, which you are going to
hear me say in all these videos now, and count by a quarter of the period. So, pi, 2pi, 3pi,
and we need five points so 4pi. Let's plug each one of these in. How much time do I have
by the way? I better start moving! One half of zero is zero and the cosine of zero, that
is the x coordinate so that is one. So one times negative three is negative three plus
four. Negative three plus four is one. Plug in pi. One half of pi is pi over two and the
cosine of pi over two is zero, times negative three is zero plus four which is four. Now
are paying attention to what I am doing here? I am plugging these in, I am working out,
I am working my way back through the equation until I get to leading coefficient and then
I worry about the plus or minus at the end. So 2pi times one half is 1pi. The cosine of
ONE pi is negative one. This is negative one here at 2pi and that times negative three
is positive three plus four is seven. Let's plug in 3pi. The cosine of 3pi over 2 is,
one, two, three pi over two, the cosine of 3p/2 is zero times negative three is zero
plus four is back to four. Don't.... Get the five points before you start guessing what
you think the pattern is by the way. And 4pi times one half is 2pi. So the cosine of 2pi
is again one. So the cosine of 2pi is one, one times negative three is negative three
and then plus four is back to one. So I have all my points now and thankfully I have not
run out of time. I have an x value of zero so I don't have to worry about where the y
axis crosses. So we have zero, pi, 2pi, 3pi, 4pi, and this will give me one period. If
I want I to go backwards or forwards I could go 5pi, 6pi, 7pi, 8pi and so on and just start
plotting my points. I don't have to think about the amplitude, the flip, the period
change, or the vertical shift. All that is just going to happen. We have zero one...we
need a y value of 7 so 1,2,3,4,5,6,7. So zero one, pi four, 2pi seven, 3p at four, then
we are back down to one and that is one period of cosine. And now I could do another pattern,
another period and make if I like a second period. You will probably be asked to graph
two periods of these trig functions. Also, in the minute that I have left here now that
we have two periods of cosine. Instead of starting at one, we sort of started at the
bottom of the pattern, and in this case it is not negative one, but it is at the bottom
of the cycle. The amplitude. The overall height is six from the top to bottom and if you take
half of that you get the actual amplitude of three. And by the way, as well, if you
average these numbers of seven and one, if you find the average of those numbers and
get four. That happens to match the vertical shift. So, if are ever given a graph and asked
to find the vertical shift you average the maximum and minimum values to find that vertical
shift. ALL RIGHT. I am Mr. Tarrou. Thank you for watching. GO DO YOUR HOMEWORK!!!