Difference between revisions of "004 Sample Final A, Problem 4"

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! Step 1:  
 
! Step 1:  
 
|-
 
|-
|First we replace the inequalities with equality. So <math>y = \vert x\vert + 1</math>, and <math>x^2 + y^2 = 9</math>.
+
|First we replace the inequalities with equality. So <math>y = 2x - 3</math>, and <math>y = 4 - x^2</math>.
 
|-
 
|-
 
|Now we graph both functions.
 
|Now we graph both functions.
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|point satisfies the inequality or not. For both equations we will pick the origin.
 
|point satisfies the inequality or not. For both equations we will pick the origin.
 
|-
 
|-
|<math>y < \vert x\vert + 1:</math> Plugging in the origin we get, <math>0 < \vert 0\vert + 1 = 1</math>. Since the inequality is satisfied shade the side of  
+
|<math>y > 2x - 3:</math> Plugging in the origin we get, <math> 0 > 2(0) - 3 = -3 </math>. Since the inequality is false, we shade the side of  
 
|-
 
|-
|<math>y < \vert x\vert + 1</math> that includes the origin. We make the graph of <math>y < \vert x\vert + 1</math>, since the inequality is strict.
+
|<math>y = 2x - 3</math> that does not include the origin. We make the graph of <math>y < \vert x\vert + 1</math> dashed, since the inequality is strict.
 
|-
 
|-
|<math>x^2 + y^2 \le 9:</math> <math>(0)^2 +(0)^2 = 0 \le 9</math>. Once again the inequality is satisfied. So we shade the inside of the circle.
+
|<math>y \le 4 - x^2:</math> Plugging in the origin we get <math>0 \le 4 - (0)^2 = 4</math>. Since this inequality is true, we shade the side of <math>y = 4 - x^2</math> that includes the origin. Here we make the graph of <math> y = 4 - x^2 </math> solid since the inequality sign is <math>\le</math>
|-
 
|We also shade the boundary of the circle since the inequality is <math>\le</math>
 
 
|}
 
|}
  
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! Final Answer: &nbsp;
 
! Final Answer: &nbsp;
 
|-
 
|-
|The final solution is the portion of the graph that below <math>y = \vert x\vert + 1</math> and inside <math> x^2 + y^2 = 9</math>
+
|The final solution is the portion of the graph that below <math>y = 4 - x^2</math> and above <math> y = 2x - 3</math>
 
|-
 
|-
 
|The region we are referring to is shaded both blue and red.
 
|The region we are referring to is shaded both blue and red.
 
|-
 
|-
|[[File:8A_Final_5.png]]
+
|[[File:4_Sample_Final_4.png]]
 
|}
 
|}
  
  
 
[[004 Sample Final A|<u>'''Return to Sample Exam</u>''']]
 
[[004 Sample Final A|<u>'''Return to Sample Exam</u>''']]

Latest revision as of 10:08, 2 June 2015

Graph the system of inequalities. Solution:

Step 1:  
First we replace the inequalities with equality. So , and .
Now we graph both functions.
Step 2:  
Now that we have graphed both functions we need to know which region to shade with respect to each graph.
To do this we pick a point an equation and a point not on the graph of that equation. We then check if the
point satisfies the inequality or not. For both equations we will pick the origin.
Plugging in the origin we get, . Since the inequality is false, we shade the side of
that does not include the origin. We make the graph of dashed, since the inequality is strict.
Plugging in the origin we get . Since this inequality is true, we shade the side of that includes the origin. Here we make the graph of solid since the inequality sign is
Final Answer:  
The final solution is the portion of the graph that below and above
The region we are referring to is shaded both blue and red.
4 Sample Final 4.png


Return to Sample Exam