For the given sine function
we're asked to find the vertical shift,
midline, amplitude, period, and phase shift.
In order to do this we want the given function
to be in this form here so that we can easily
identify the values of A, B, D, and C.
Notice right now it's not quite in this form.
We need to factor 11 pi over six
from this binomial here,
and let's show some work for this.
Let's write the binomial as 11 pi
over six X plus, because we need to factor
11 pi over six from 55 pi over six,
let's write this fraction as five times 11 pi
over six, and now we can see the factor
of 11 pi over six in both terms.
If we factor out 11 pi over six,
we're left with X plus five.
So let's write the given function as Y equals.
Notice here we have the constant on the left.
In form it's on the right.
So let's write the function as negative two sine
of the quantity 11 pi over six.
Because we want this to be in the form of X minus D,
let's write X plus five as X minus negative five,
which is equivalent,
and then we have plus eight on the right.
In this form notice how A is equal to negative two,
B is equal to 11 pi over six,
D is equal to negative five,
again, because it has to be in the form
of X minus D, and C is equal to positive eight.
Now let's find the vertical shift
and the midline.
The vertical shift is indicated
by the value of C.
Because C is positive eight,
the vertical shift is up C units,
or up eight units.
For the basic sine function,
the midline is Y equal zero or the X axis.
Because the vertical shift is up eight units,
the midline is now going to be Y equals eight.
The amplitude is indicated by the absolute value of A.
Because A is equal to negative two
and the absolute value of negative two
is positive two, the amplitude is positive two,
but because A is negative,
the graph will be reflected across the midline.
The period is equal to two pi divided by B
where B is 11 pi over six.
So the period is two pi divided by 11 pi over six,
which is equal to two pi over one
times the reciprocal of 11 pi over six,
which is six over 11 pi.
Simplifying before multiplying,
there's one pi and pi here and here.
Multiplying and we have 12/11.
So the period is 12/11,
and finally we're asked to find the phase shift
or horizontal shift, which is indicated
by the value of D.
Notice in this form D is equal to negative five.
So because D is equal to negative five
or because D is less than zero,
the function is translated left
the absolute value of D units,
or in this case left five units.
And now let's verify these results
by looking at the graph.
Because the graph was shifted up eight units,
we know the midline is Y equals eight,
which is this vertical line here.
Normally when graphing the basic sine function,
we start at zero and graph to two pi radians
because the period is two pi radians,
but because have a phase shift
of five units to the left,
we will graph one period of the given function
starting at X equals negative five,
which is this vertical line here.
Because the period is 12/11,
if we add 12/11 to negative five,
we get X equals negative three and 10/11,
which is indicated by this red vertical line.
So over this interval we have one complete graph
of the given sine function.
Because the amplitude is two
and the midline is positive eight,
eight plus two is equal to 10.
So the maximum function value is 10,
and eight minus two is equal to positive six.
The minimum function value is positive six.
To graph the basic sine function over one period,
we divide the period into four equal subintervals,
and the pattern to graph the basic sine function
is midline, maximum, midline, minimum, and midline.
In our case though, because A is negative,
we also have a reflection across the midline.
So if we were to partition this period
into four equal parts or pieces,
because we have a reflection across the midline,
the pattern is midline, minimum, midline, maximum, midline.
Once we have this piece of the graph,
we can copy and paste the graph over a larger interval.
This graph does verify our results are correct.
I hope you found this helpful.