Difference between revisions of "005 Sample Final A, Question 16"

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|1) Since both x and y are squared it must be a hyperbola or an ellipse. We can conclude that the graph is an ellipse since both <math>x^2</math> &nbsp; and &nbsp; <math>y^2</math> have the same sign, positive.
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|1) Since both x and y are squared it must be a hyperbola or an ellipse. We can conclude that the graph is a hyperbola since <math>x^2</math> &nbsp; and &nbsp; <math>y^2</math> have the different signs, one negative and one positive.
 
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|2) Since the coefficient of the <math>x^2</math> term is smaller, when we divide both sides by 36 the X-axis will be the major axis.
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|2) Since the <math>y^2</math> is positive, the hyperbola opens up and down.
 
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|We start by dividing both sides by 36. This yields <math>\frac{4x^2}{36} + \frac{9(y + 1)^2}{36} = \frac{x^2}{9} + \frac{(y+1)^2}{4} = 1</math>.
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|We start by completing the square twice, once for x and once for y. After completing the squares we end up with <math>-(x + 1)^2 +4(y - 2)^2 = 4</math>
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|'''Common Mistake:''' When completing the square we will end up adding numbers inside of parenthesis. So make sure you add the correct value to this other side. In this case we add -1, and 16 for completing the square with respect to x and y, respectively.
 
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Revision as of 11:27, 2 June 2015

Question Graph the following,

Foundations:  
1) What type of function is this?
2) What can you say about the orientation of the graph?
Answer:
1) Since both x and y are squared it must be a hyperbola or an ellipse. We can conclude that the graph is a hyperbola since   and   have the different signs, one negative and one positive.
2) Since the is positive, the hyperbola opens up and down.

Solution:

Step 1:  
We start by completing the square twice, once for x and once for y. After completing the squares we end up with
Common Mistake: When completing the square we will end up adding numbers inside of parenthesis. So make sure you add the correct value to this other side. In this case we add -1, and 16 for completing the square with respect to x and y, respectively.
Step 2:  
Now that we have the equation that looks like an ellipse, we can read off the center of the ellipse, (0, -1).
From the center mark the two points that are 3 units left, and three units right of the center.
Then mark the two points that are 2 units up, and two units down from the center.
Final Answer:  
The four vertices are:
8A Sample Final, Q 6.png

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