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

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(Created page with "'''Question ''' Graph the following, <center><math> -x^2+4y^2-2x-16y+11=0</math></center> {| class="mw-collapsible mw-collapsed" style = "text-align:left;" ! Final Answers |-...")
 
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! Final Answers
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!Foundations: &nbsp;
 
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|a) False. Nothing in the definition of a geometric sequence requires the common ratio to be always positive. For example, <math>a_n = (-a)^n</math>
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|1) What type of function is this?
 
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|b) False. Linear systems only have a solution if the lines intersect. So y = x and y = x + 1 will never intersect because they are parallel.
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|2) What can you say about the orientation of the graph?
 
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|c) False. <math>y = x^2</math> does not have an inverse.
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|Answer:
 
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|d) True. <math>cos^2(x) - cos(x) = 0</math> has multiple solutions.
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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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|e) True.
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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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Solution:
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! Step 1: &nbsp;
 
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|f) False.  
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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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! Step 2: &nbsp;
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|Now that we have the equation that looks like an ellipse, we can read off the center of the ellipse, (0, -1).
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|From the center mark the two points that are 3 units left, and three units right of the center.
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|Then mark the two points that are 2 units up, and two units down from the center.
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! Final Answer: &nbsp;
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|The four vertices are: <math>(-3, -1), (3, -1), (0, 1) \text{ and } (0, -3)</math>
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|[[File:8A_Sample_Final,_Q_6.png]]
 
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[[005 Sample Final A|'''<u>Return to Sample Exam</u>''']]
 
[[005 Sample Final A|'''<u>Return to Sample Exam</u>''']]

Revision as of 10:59, 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 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:  
The four vertices are:
8A Sample Final, Q 6.png

Return to Sample Exam