Determining the equation of a trig function | Graphs of trig functions | Trigonometry | Khan Academy

Write the equation of the
function f of x graphed below. And so we have this
clearly periodic function. So immediately you
might say, well, this is either going to be a sine
function or a cosine function. But its midline
and its amplitude are not just the plain vanilla
sine or cosine function. And we can see that
right over here. The midline is halfway
between the maximum point and the minimum point. The maximum point right over
here, it hits a value of y equals 1. At the minimum points,
it’s a value of y is equal to negative 5. So halfway between
those, the average of 1 and negative 5, 1 plus
negative 5 is negative 4. Divided by 2 is negative 2. So this right over
here is the midline. So this is y is
equal to negative 2. So it’s clearly shifted down. Actually, I’ll talk
in a second about what type of an expression
it might be. But now, also, let’s
think about its amplitude. Its amplitude–
that’s how far it might get away from the
midline– we see here. It went 3 above the midline. Going from negative 2 to 1,
it went 3 above the midline at the maximum point. And it can also go 3 below the
midline at the minimum point. So this thing clearly
has an amplitude of 3. So immediately, we
can say, well, look. This is going to have
a form something like f of x is equal to
the amplitude 3. We haven’t figured
out yet whether this is going to be a cosine
function or a sine function. So I’ll write “cosine” first. Cosine maybe some coefficient
times x plus the midline. The midline– we
already figured out– was minus 2 or negative 2. So it could take
that form or it could take f of x is equal to 3
times– it could be sine of x or sine of some
coefficient times x. Sine of kx minus 2 plus
the midline– so minus 2. So how do we figure
out which of these are? Well, let’s just think about
the behavior of this function when x is equal to 0. When x is equal to
0, if this is kx, then the input into the
cosine is going to be 0. Cosine of 0 is 1. Whether you’re talking about
degrees or radians, cosine of 0 is 1. While sine of 0– so
if x is 0, k times 0 is going to be
0– sine of 0 is 0. So what’s this thing doing
when x is equal to 0? Well, when x is equal to
0, we are at the midline. If we’re at the midline, that
means that all of this stuff right over here evaluated to 0. So since, when x equals 0, all
of this stuff evaluated to 0, we can rule out the
cosine function. When x equals 0 here, this
stuff doesn’t evaluate to 0. So we can rule out this
one right over there. And so we are left with this. And we just really
need to figure out– what could this
constant actually be? And to think about
that, let’s look at the period of this function. Let’s see. If we went from
this point– where we intersect the midline–
and we have a positive slope, the next point that we do
that is right over here. So our period is 8. So what coefficient
could we have here to make the period of
this thing be equal to 8? Well, let us just
remind ourselves what the period of sine of x is. So the period of
sine of x– so I’ll write “period” right
over here– is 2pi. You increase your angle by
2 pi radians or decrease it. you’re back at the same
point on the unit circle. So what would be the
is increasing k times faster. So you’re going to get to the
same point k times faster. So your period is going
to be 1/k’th as long. So now your period is
going to be 2 pi over k. Notice, as x increases, your
argument into the sine function is increasing k times as fast. You’re multiplying it by k. So your period is
going to be short. It’s going to take you less
distance for the whole argument to get to the same point
on the unit circle. So let’s think
about it this way– so if we wanted to say 2
pi over k is equal to 8, well, what is our k? Well, we could take the
reciprocal of both sides. We get k over 2 pi
is equal to 1/8. Multiply both sides by 2pi. And we get k is
equal to– let’s see. This is 1. This is 4. k is equal to pi/4. And we are done. And you can verify that by
trying out some of these points right over here. This function is equal to 3
sine of pi over 4x minus 2.