Difference between revisions of "008A Sample Final A, Question 6"

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|1) Since both x and y are squared it must be a hyperbola or an ellipse. Since the coefficients of the <math>x^2</math> and <math>y^2</math> terms are both positive the graph must be an ellipse.
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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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|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.
 
|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.

Revision as of 11:21, 23 May 2015

Question: Sketch . Give coordinates of each of the 4 vertices of the graph.

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 an ellipse since both   and   have the same sign, positive.
2) Since the coefficient of the term is smaller, when we divide both sides by 36 the X-axis will be the major axis.

Solution:

Step 1:
We start by dividing both sides by 36. This yields .
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:
8A Sample Final, Q 6.png

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