In this video, we will graph a transformation
of the sine function when the equation is
in the form y equals a times sine
of the quantity bx plus c plus d.
The absolute value of a is equal to the amplitude.
And if a is less than zero or negative,
the graph is reflected across the midline.
Two pi divided by b is equal to the period.
And when graphing by hand,
we always divide the period into four equal subintervals
to make it easier to make an accurate graph.
Negative c divided by b is equal
to the phase shift or horizontal shift,
where if this value is positive, the shift is right.
If this value is negative, the shift is left.
And then finally, d is equal to the vertical shift.
If d is positive, the shift is up.
If d is negative, the shift is down.
Let's look at an example.
Let's graph the function x equals two times sine
of the quantity negative pi x plus two pi plus 1/2.
For this function, the first thing we have to recognize is
that b is negative.
Here, b is negative pi.
And two pi divided by b is equal to the period.
We know the period must be positive,
and therefore we are going to have
to change the form of this function here.
Let's first factor our a negative
from the quantity negative pi x plus two pi.
This would give us y equals two times sine.
And now we'll factor out a negative.
So we have negative,
or the opposite of the quantity pi x minus two pi.
And then we still have plus 1/2.
From here, we are going to use the negative angle identity.
Sine negative x equals negative sine x.
So notice here the angle is negative
or the opposite of pi x minus two pi,
and therefore this function is equal
to y equals negative two sine
of the quantity pi x
minus two pi plus 1/2.
So now that we've transformed the equation into a form
where b is positive,
let's now find the values of a, b, c, and d
and describe the transformation.
Notice a is equal to negative two.
B is equal to pi.
C is equal to negative pi over two.
And d is equal to positive 1/2.
Because a is equal to negative two, the amplitude is equal
to the absolute value of negative two,
which is positive two.
So we know that the amplitude is equal to positive two.
Because a is negative, though,
the graph is reflected across the midline.
So let's list that as well.
We have a reflection across the midline.
Next, b is equal to pi.
And two pi divided by b is equal to the period.
So two pi divided by b is equal
to two pi divided by pi, which simplifies to two.
So the period is equal to two units.
Negative c divided by b is equal to the phase shift.
And notice that c is negative pi over two.
So negative c divided by b is equal to negative
or the opposite of negative two pi divided by b.
And again, b is equal to pi.
And therefore, this simplifies to two.
Because this value is positive, we have a horizontal shift
or phrase shift of right two units.
And then finally, d is equal to positive 1/2.
So because d is equal to positive 1/2,
we have a vertical shift of up 1/2 of a unit.
And now let's set up our graph to sketch the function.
Because we have a vertical shift of up 1/2 of a unit,
the midline is not going
to be the x-axis or y equals zero.
It will be y equals 1/2,
which is this horizontal line here.
So this is the midline.
Next, because the phase shift is right two units,
we will begin our graph here at x equals two.
And from here, because the period is also two units,
we will go right two more units.
Two plus two is equal to four.
We will have one complete graph of our sine function
over this period, which is again two units.
And when graphing by hand,
we always want to divide this period
into four equal subintervals.
So we'll divide this in half and then in half again.
So over this interval here,
we'll graph one complete graph of our sine function.
Because the amplitude is positive two
and this is the midline,
the maximum function value is going
to be 0.5 plus two or 2.5, which is here on the y-axis.
And then 1/2 minus two is negative 1.5, which is here.
This will be the minimum function value.
And now we'll use our knowledge
of the basic sine function shown here
to graph the given sine function,
keeping in mind we do have a reflection across the midline.
Looking at the basic sine function shown here,
we're one period from zero to two pi radians.
If we divide this into four subintervals,
we have five key points.
We start at the midline here,
and then we go up to a maximum, back to the midline,
down to a minimum, and back up to the midline.
We will follow this same pattern for our graph,
except we do have a reflection across the midline,
and therefore our pattern is going to be
midline, minimum, midline, maximum, midline.
Starting at x equals two, we're at the midline.
And then because we have a reflection across the midline,
we now go down to a minimum function value here
of negative 1.5, back to the midline,
back up to the maximum function value of 2.5,
and then back down to the midline.
And now I can graph one period of the given sine function,
which looks like this.
And once we have one period,
we can simply copy and paste this graph
to the left and right
or follow the pattern from left to right.
Continuing to graph on the right,
we would be down at a minimum function value here
when x is 4.5.
The graph continues to the right, looking like this.
And now graphing to the left of x equals two,
let's go ahead and divide the interval for zero to two
into four equal subintervals
to make it easier to graph an accurate graph.
So again, following a pattern,
starting at two and moving to the left,
at x equals 1.5, we'd be at a maximum function value.
Then back down to the midline,
back down to a minimum, and then back to the midline.
We can now make a nice graph of the given function,
all the way from x equals zero to x equals 4.5.
The final graph would look like this.
I hope you found this helpful