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# Plotting maxima, minima and midline intersections of trig function

## Video transcript

Plot points corresponding
to the maxima, minima, and intersections with the
midline in the graph of y is equal to 2 sine
of 1/4 x minus 1. And they give us this
neat tool here where we can sit here and pick as
many points just by clicking. We can move the points
around, drag them around. If we want to get
rid of a point, we just have to drag
them to the trash. Now to work through
the math, I've copied and pasted this problem
right here on my scratch pad, and I rewrote the function. And I want to think
about, when does this hit its maximum
points, its minimum points, and its intersections
with the midline? Well this is going to hit
a maximum point whenever sine of 1/4 x is maximized. And when is sine
of 1/4 x maximized? Well the maximum value that the
sine function can take on is 1. So we're dealing with maximum
points whenever sine of 1/4 x is equal to 1. Now what about minimum points? Well we're dealing with minimum
points whenever this thing-- the minimum value that the
sine function can take on is negative 1. So we're hitting minimum
points whenever-- let me write it here-- so
minimum points whenever sine 1/4 x hits the minimum value
that the sine function can take on, hits the point negative 1. And of course, when
it hits this point, we're then going to
multiply it times this 2, so the maximum value of
this part of expression is going to be 2 times 1. The maximum value of
the whole function is going to be 2
times 1 minus 1. The minimum value is going to
be 2 times negative 1 minus 1. But we'll get to that. Now what about when we
intersect the midline? We intersect the
midline whenever this part of the
function is equal to 0. And we're essentially
just left with that part right over there. So let me write this-- midline. And this is going to be 0 when
sine of 1/4 x is equal to 0. So when do these things happen? When does the sine
function hit 1? When does the sine
function hit 0? When does the sine
function hit negative 1? And to think about that, I will
draw ourselves a little unit circle to help us
visualize things. So a little unit circle, I
don't want to make it too little so that we can-- so
a little unit circle right over there,
that's our y-axis. That is our x-axis. And it's a unit circle. It's centered at the origin, has
a radius 1-- so just like this. And this, of course, is
the point negative 1. Sorry, this is the
point 0, negative 1. This is the point 1 comma 0. This is the point 0 comma 1. This is the point
negative 1 comma 0. Now let's start with
an angle of 0 radians. So we're going to intersect
the unit circle right at that point, at
the point 1 comma 0. And remember, the
sine evaluated there by the unit circle
definition of trig functions is going to be the y value. So we see sine of 0 radians
is equal to 0-- and not just 0 radians, any integer multiple
of 2 pi plus 0 radians. So we could add 2 pi. We would go around again. We would get to the same point. Add 2 pi again-- well
now we've added 4 pi. We go again. We could subtract 2 pi,
and we're back there again. So this is going to
happen whenever-- whatever we input
into the sine function is equal to 0 radians plus
some multiple of 2 pi radians. So when is this going
to be equal to 0? Well, when 1/4 x
meets that constraint. So that's going
to happen whenever 1/4 x is equal to 0
plus or minus-- I guess I could say-- some
multiple of 2 pi. Note and remember,
all I'm saying here is, if you input anything like
this into the sine function, you're going to be at this
point on the unit circle. And when you evaluate the
sine of the function at any of those points, you
are going to get 0. Why did I write k here? Well this is just
for some integer k. So this is just saying 0 plus
2 pi, 0 plus 4 pi, 0 plus 6 pi, 0 minus 2 pi. This encapsulates all of that. So let's solve for x. Well to solve for x, we just
multiply both sides by 4. We get x is equal
to 4 times 0 is 0, plus or minus 4 times some
multiple of 2 pi would get us-- or 4 times 2 pi k is
going to give us 8 pi k. So when x is 0 plus or minus
any integer multiple of 8 pi, this whole function is going
to intersect the midline. Why? Because when x is any of these
things, when you input any of these x values there,
you multiply it by 1/4, you evaluate sine there,
then this whole expression is going to be 0. We're going to hit the
midline, y equals negative 1. So let's just go with
the simplest one. Let's just go with x equals 0. When x equals 0, all
of the stuff is 0. y is equal to negative 1. So we have the midline
right over here. x equals 0. y is equal to negative 1. And we're also hitting the
midline right over there. Now let's think about
a maximum value. So if we increase
our angle right over here all the way
to pi/2 2 radians, we get to this point
right over here. Sine hits its maximum
value whenever we're taking the sine of pi/2
plus or minus any integer multiple of 2 pi. So this is going to
hit a maximum whenever what we input into
the function is equal to pi/2 plus or minus
some integer multiple of 2 pi. Well once again,
multiply both sides by 4. You get x is equal to pi/2 times
4 is 2 pi plus or minus 8 pi k. Now which x value
would we want to do? Well and actually,
even in the last one, I picked x equals
0 for simplicity. But if I did x
equals 8 pi or 16 pi, those aren't even on this graph. Or if I want to do x is
equal to negative 8 pi, those aren't even on the graph. So this is actually the only
one that sits on this graph. Likewise over
here, x equals 2 pi is the only one that's
actually on this graph. If I added 8 pi, I
would go off the graph. If I did 2 pi minus 8
pi at negative 6 pi, that's off the graph. So I'm going to
do x equals 2 pi. When x equals 2 pi,
1/4 of 2 pi is pi/2. Sine of pi.2 we've
already saw is 1. 2 times 1 is 2, minus 1 is 1. So when x is 2 pi, we're at
the point y is equal to 1. That is a maximum point. And then finally-- same
logic for the minimum point. We need to see when
this 1/4 x hits not pi/2, but now negative pi/2. We could have also picked
3 pi over 2 if we wanted. So negative pi/2 is equal to
negative pi/2 plus or minus any integer multiple of 2 pi. So once again, multiply
both sides times 4. x is equal to negative
2 pi plus or minus 8 pi-- some integer
multiple of 8 pi. So once again, the only of these
that's actually on our graph is x is equal to negative 2 pi. When x is equal
to negative 2 pi, what does our function equal? Well, when x is equal
to negative 2 pi, 1/4 x is going to be equal
to negative pi/2. Sine of negative
pi/2 is negative 1. 2 times negative 1 is
negative 2, plus negative 1 gets us to negative 3. So it's that point
right over there. So those are our points. And if we actually
wanted to graph-- they just ask us to
graph these points-- but if we wanted to give a
little bit more texture here, we could draw the
midline, right over here, at y is equal to negative 1. And if we wanted to
actually try to plot this function, or at least
the part that we can graph, it would look
something like this. But actually, let's
just get the tool out, and it would keep going if
we had more real estate. We're not even plotting
a complete period here, but let's go back to our
tool and plot these points and make sure that
we got it right. So we want a
point-- that's where we intersect at the midline. This is our minimum point. That's our maximum point
based on what I just did. Check the answer. We got it right.