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### Course: Get ready for Precalculus > Unit 4

Lesson 5: Graphing sinusoidal functions- Graph of y=sin(x)
- Intersection points of y=sin(x) and y=cos(x)
- Example: Graphing y=3⋅sin(½⋅x)-2
- Example: Graphing y=-cos(π⋅x)+1.5
- Graph sinusoidal functions
- Sinusoidal function from graph
- Construct sinusoidal functions
- Graph sinusoidal functions: phase shift

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# Sinusoidal function from graph

Sal finds the equation of a sinusoidal function from its graph where the minimum point (-2,-5) and the maximum point (2,1) are highlighted. Created by Sal Khan.

## Want to join the conversation?

- Can't this graph also be represented by a cosine function with a phase shift?(118 votes)
- Yes, any sine function can be represented by a cosine function with a shift of pi/2. And any cos function can be represented by a sin function with a shift of pi/2.(122 votes)

- Hi!

I don't know if it is my stupidity or not, but to me it seems like the "Graph sinusoidal functions: phase shift" exercise block in "Graphing sinusoidal functions" has exercises that are not covered by that point, only in this video, which is next in line. Can anyone else confirm or dismiss this?(111 votes)- I agree, fairly often I find that they threw an exercise in covering a topic we haven't learned yet, but is related. In other words skip it and the video, may or may not come later. Hope that's what you were looking for.(42 votes)

- at1:33, Sal immediately writes the amplitude as 3. How does he know that it's positive three that not negative three?(22 votes)
- Amplitudes are described by their absolute value. An amplitude of -3 or an amplitude of +3 will only flip the sinusoidal function over the x-axis. The actual height of the graph is unchanged.(28 votes)

- There is no video for constructing a sinusoidal function and the hints in the practice are not very helpful. I might be missing something. Exactly how can the midline be inferred from two points? What is an extremum?(16 votes)
- extremum are max and min values. If you are given the max and min then all you have do is average those values and you will get the midline. Since the curve oscillates from these points you will get the midline as a result. There is some stuff on graphing in algebra 2 that might also help.(9 votes)

- how do u calculate the phase shift using angles in sine and cosine(7 votes)
- If you are given the equation y = A sin(kx + c) + d, then the phase shift equals -c/k. The calculation is identical if you replace the sin with cos.(18 votes)

- I am trying to do the phase shift assignment for my class but i'm really not understanding how to find the period. I can get the mid-line and amp easily enough but i get confused on the sin or cosine part and the period(9 votes)
- In this answer, l am assuming the function involves a sine or cosine function.

If you are given the graph, then the period is the horizontal distance from one minimum to the next minimum (or from one maximum to the next maximum).

If you are instead given the equation, then the period is 2pi divided by the absolute value of the coefficient of x inside the sine or cosine function. This follows from the concept that increasing any angle by 2pi radians (a full revolution) has no effect on the angle’s sine and cosine.(10 votes)

- Is the period and the frequency the same thing? If not, what is the difference between them?(7 votes)
- Period and frequency are reciprocals of each other in Physics, i.e. P = 1/f and f = 1/P. When discussing the graphs of trig functions, the Period is the length of a cycle. The term "frequency" is not formally defined. For example, sin(x) has a period of 2pi, since sin(x) = sin(x + 2pi) and it is the smallest angle for which that is true. (Adding 2pi to an angle is equivalent to one full revolution around a circle.)

To find the period of sin(bx), calculate P = 2*pi/b. For example, sin(3x) has a period of 2pi/3. Note the inverse relationship between P and "b", just as between P and "f". You could THINK of "b" as being the frequency, but it isn't formally defined.(13 votes)

- Can someone please explain y=Asin(Bx+C)+D form to me? the problem set for this gives us 2 points (may even give the midline/max/min points) and tells us to give the equation, this sample problem from the video gives that graph which makes it so much easier. Is there an easy way to do these type of problems (find the equation given 2 points)? Thanks in advance(4 votes)
- f(x) = Asin(bx+c)+D — breaking it down:

A = Amplitude

B (assuming it's 1) = Period

C = Horizontal Shift

D = Vertical Shift/**midline**

Typically in "find the equation given 2 points," your first goal is finding the period. An useful graph @ the hints in the practice problems tremendously aided me, so I recommend using that. For simple:

If given mid --> minimum/maximum, take the x distance and multiply by**four**. Then take 2pi/(that #) to find B

If given min --> minimum, take the x distance and multiply by**two**. Then take 2pi/(tht #) to find B.

Finding A, if given mid --> mid/max, take the y distance

if given min/max --> max/min, take the y distance and**divide by 2**.

Finding D, find the average of the y-values of the coordinates given as max/min. Midline have might been already given.

I also recommend looking in the comments for each video, as many questions such as this has already been answered and could be more in depth than mine. Quite honestly this is objectively one of the more harder subjects in alg ii.(15 votes)

- Have been struggling to get the horizontal phase shift because still in April 2023 there have been no lessons/instructions on how to handle (bx + c) instead of just (kx). It is only found in the explanations to the quizzes.(9 votes)
- When you have sin(bx+c), you're doing two things:

1. You're magnifying the argument by a factor of b and hence, you're shrinking the "width" of the function (making it more congested)

2. You're shifting the argument by c units to the left (assuming c > 0). As to why the shift is to the left, read on:

Let's take a simpler function f(x) = 2x + 2. Now, this crosses the x axis at x = -1 (or, x = -1 is the value of x which makes the function 0). Now, if I make a new function as f(x+1), I get 2(x+1) + 2. Now, the value of x at which the function becomes 0 has changed (it was initially -1. Now it's -2). To compensate for this change, the function shifts by 1 unit to the left and now crosses the x axis at -2.(4 votes)

- I am very grateful for Sal Khan and all the people who work hard on Khan Academy; however, I have run into a problem. I have tried working on the exercises in this Lesson for two weeks to no success and only partial understanding of these topics. This video (Sinusoidal function from graph) does not seem to be completely consistent with helping to understand and solve the three exercises: graph sinusoidal functions, construct sinusoidal functions, and graph sinusoidal functions: phase shift. I have been using Khan Academy from Algebra 1 up until this point in Algebra 2 and I hope to continue, pursuing Calculus while completely understanding and appreciating the beauty of these Mathematical Topics. I wish to be able to completely understand these Trigonometric Topics and be able to move on further in my learning of Mathematics. I would greatly appreciate it if I get help on what to currently do in order to understand these topics and continue learning and using Khan Academy. Thank You.(8 votes)
- Thank you for your help. I was mainly struggling with Phase Shift; sorry if I was unclear, I was stating my problems with the exercises in general. Your example helped me understand the topic better, I will continue to try to work on these practice problems.(4 votes)

## Video transcript

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
period of sine of kx? Well, now, your x, your input
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.