Difference between revisions of "004 Sample Final A, Problem 9"
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! Foundations | ! Foundations | ||
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|Answer: | |Answer: | ||
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! Step 1: | ! Step 1: | ||
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− | | | + | |Since <math>y=\frac{6}{(x-2)(x+1)}</math>, the vertical asymptotes are <math>x=2</math> and <math>x=-1</math>. |
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! Step 2: | ! Step 2: | ||
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− | | | + | |The horizontal asymptote is <math>y=0</math>. |
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! Step 3: | ! Step 3: | ||
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− | | | + | |There is no <math>x</math> intercept since <math>y\neq 0 </math> for any <math>x</math>. |
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− | | | + | |Plugging in <math>x=0</math>, we get <math>y=-3</math>. So, the <math>y</math> intercept is (0,3) |
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! Final Answer: | ! Final Answer: | ||
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− | | | + | | The vertical asymptotes are <math>x=2</math> and <math>x=-1</math>. The horizontal asymptote is <math>y=0</math>. There is no <math>x</math> intercept and the <math>y</math> intercept is (0,3). |
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[[004 Sample Final A|<u>'''Return to Sample Exam</u>''']] | [[004 Sample Final A|<u>'''Return to Sample Exam</u>''']] |
Revision as of 12:12, 6 May 2015
Graph the function. Give equations of any asymptotes, and list any intercepts.
Foundations |
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Answer: |
Solution:
Step 1: |
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Since , the vertical asymptotes are and . |
Step 2: |
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The horizontal asymptote is . |
Step 3: |
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There is no intercept since for any . |
Plugging in , we get . So, the intercept is (0,3) |
Step 4: |
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Final Answer: |
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The vertical asymptotes are and . The horizontal asymptote is . There is no intercept and the intercept is (0,3). |