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The explanation is written with the assumption that you meant
Use the distributive property :
Substitute x for
Square both sides:
Add
Remove a factor of the 3 from the first 3 terms:
Use the middle in the right side of the pattern
Substitute the left side of the pattern into the equation:
Substitute
Divide both sides by
Write the denominators as squares:
The is standard Cartesian form of the equation of an ellipse with a center at
Here is the graph of the function (it's not drawn to scale):
I will explain how I graph it below:
Your given equation is
I would write the given equation in the standard form according to
You get:
This way, it is easier to define your amplitude, phase shift, vertical shift, and period multiplier.
In order to graph, we need to know the following:
Let's get all our required values from our equation:
It is important to note that B is not apparent in this equation because it is equal to 1. The actual equation should really be
Our period multiplier is important for determining our period, and it also affects our phase shift. However, we do not have to worry about it for this equation since it's just equal to 1.
Since we have both an amplitude and a vertical shift, we get new minimum and maximum y values for the graph:
Maximum Y Value = Vertical Shift + Amplitude =
Minimum Y Value = Vertical Shift  Amplitude =
Most importantly, this graph is a positive cosine graph because our equation is 5cos.. and not 5cos.. . A positive cosine graph always starts at the maximum value.
Also, our vertical shift is +1, so our new xaxis start has shifted up by 1. We are starting at y = 1 and not y = 0.
As with our phase shift, it is
So.. with all this information, we can now actually graph the equation of
Here are some visual cues.
My art is pretty bad, sographing the graph on a website should be much more different.
I hope this helps.
Here is the graph of
Step 1: Sketch the graph of
Step 2: Sketch the line
Step 3: Reflect the graph of
Q1
Let
Again
So
Otherwise without using exact values
Hence
Now
Q2
Now
So
Note that,
We see that,
Due to the periodicity of
Define
Then we have
Look at the function
graph{arcsin(x) [10, 10, 5, 5]}
it takes in an argument between 0 and 1 and outputs the principle value of
So, we have that
I.e.,
And,
As
Notice that for
We can use
We need to solve the same inequalities as before,
Then we see that for
Then,
In order to answer this I have assumed a vertical shift of
The standard cos function
If we want a period of
That is
To get an amplitude of
There is to be no horizontal shift, so the argument for
In order to achieve the vertical shift (that I assumed would be
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