So we have got y equals to four tangent pi x. So our fake amplitude is four. If you answers
in your t table are not separated by four units you have made a mistake. The period
is pi divided by b. hmm, that is kind of weird. Don't worry about it. You have done it right.
It is pi over b. If the pi's cancel out and you just get 1, then the period is 1. The
phase shift. There is no plus or minus in the function so there is no phase shift. There
is no plus or minus of a constant at the end of the equation, or outside of the function,
so there is no vertical shift either. And we always count by a quarter of the period
and the period of this graph is one. So we are counting by a quarter. Alright. Lets set
up our t table. Shall we:) We set our t table by once again starting at the phase shift
and counting by a quarter of the period. Ok, no pi's in there. But the pi is right there
in the equation. We are going to take that x and multiply by pi first, then take the
tangent of that value, and finally multiply by four. So if we take zero and plug it in,
zero times pi of course is zero. The tangent of zero... remember that the tangent of theta
is y over x... and at zero that is y is zero and x is one. So our answer is zero. One fourth
pi. One fourth for x times pi. So pi over four. The tangent of pi over four is one,
and one times four is four. Two over four or one half. We don't have any other fractions
in this problem to worry about so this is one half. So this is one half time pi. One
half times pi it pi over two. Tangent of pi over two... y over x... you are dividing by
zero so this is undefined. Plug in 3 pi over four. Three pi over four, you plugging in
three fourths, Three pi over four. The tangent of 3 pi over four is negative one, so we have
negative four. And then we are going to have zero again when we plug in negative one. So
our graph, our x axis. Zero, one fourth, one half, three fourths, and four fourths... or
simply one. And we can go negative one fourth as well. Now draw your vertical axis, your
y axis through where zero is. We have y values of 1, 2, 3, 4, 5 and negative 1,2,3,4,5. At
zero we have zero. Pi over four we are at four. At one fourth we have four, excuse me.
At two over four we have an undefined value. We have a vertical asymptote here. At three
fourths we have the value of negative four. And at one or four over four we have zero
again. Now we have this sort of broken period. If you want a more complete smooth graph,
not something with a break in the middle, you know the pattern now. One tick the left
of zero we have negative four. One more tick to the left we have a vertical asymptote.
So we have now graphed one and a half periods of tangent. I am out of time so I am Mr. Tarrou,
Go Do Your Homework. Except I am not really out of time:) Alright. Lets do our second
example. Y equals the cotangent of x plus pi over four minus one. For this cotangent
graph our fake amplitude is two. Our period again tangent and cotangent have a period
of pi not 2 pi. So the period of this graph is pi divided by b, the value in front of
the x. This is going to simply be pi. Phase shift is c over b. Don't forget the standard
form of a trig function like this is "a" times trig function parenthesis bx minus c closed
parenthesis plus d. The standard form already has a negative sign there not a plus sign.
So that plus sign means a sign change must have happened because c was negative. Our
phase shift is going to be negative pi over four. The vertical shift is your plus or minus
a constant at the end of your equation outside of the actual math function, so that is going
to be minus one. Our vertical shift is down one. When we set up our t table counting by
one quarter of the period. One quarter of pi is...well.. pi over four. This is going
to be a little bit close, let me move it over. Now don't forget your phase shift... You always
start with your phase shift and then count by a quarter of the period. Our phase shift
is negative pi over four and we are going to start with that. Now start counting by
one quarter of the period. When you make these sketches, pardon me. When you make these sketches
make sure your county-by-value, your phase shift, and your value of c all have a common
denominator. That is going to make the work inside the t table a little bit easier to
do mentally since we are not using calculators. Negative pi over four plus pi over four is
zero. Then pi over four, two pi over four, and three pi over four. Remember we only need
five points to make the first period of our trig function. Off our t table. Again we are
trying to graph these as easy as possible. So we are looking for the zeros, the ones,
and the negative ones. You plug in negative pi over four, plus pi over four is zero. The
cotangent of zero radians is... remember cotangent is x over y... so that is going to be one
over zero. So the first value is going to be undefined. And then at zero. When you plug
into zero into this equation we are going to first add pi over four. The cotangent of
pi over four is going to be one, times two is two, and then minus one. Two minus one
is one. Do that vertical shift last. Now pi over four plus pi over four is two pi over
four. The cotangent of pi over two, x over y, is zero over one which is zero. Times two
is zero, and then minus one. Then at two pi over four plus one pi over four is three pi
over four. The cotangent of three pi over four, one, two, three pi over four is negative
one. Now times two is negative two and minus one is negative three. Now three pi over four
plus one pi over four is four pi over four... or one. The cotangent of one pi with the y
value being zero again is going to be undefined. Ok. Now that we have our t table set up, lets
go ahead and set up the x axis and find where zero is so we know where to place the y axis.
So we have negative pi over four, zero, pi over four, two pi over four, and three pi
over four. Find out where that zero is and put the y axis there. If you happen to skip
over zero, just place the y axis between the negative and positive x values. Now lets see
what we've got. We have that vertical asymptote at negative pi over four. At zero we have
a y value of one. At pi over four we have negative one. At two pi over four we are at
negative three. And at three pi over four we have a vertical asymptote again. So here
we have for y equals two times cotangent of x plus pi over four minus one we have our
first period. And then if your teachers are requiring you like am for most of these questions
to graph more than one period... usually two periods... just keep this pattern going along.
You have got negative one fourth, zero, one fourth, two fourths, three fourths, four pi
over fourths. The point there, would be after the vertical asymptote we have a y value of
one, and next we have negative one. I am running out of space. There is a period and a half.
The range of this graph is going to be all real numbers. The domain is... Lets just pick
an asymptote. The domain is going to be all real numbers where x is not equal to three
pi over four plus or minus pi times n where n is an integer. plus or minus n time the
period. Every period you are going to have this vertical asymptote show up again. So
the domain is all real numbers where x is not equal to three pi over four plus or minus
n pi. I am Mr. Tarrou. That is your last example. BAM!!! Go do your homework:D