005 Sample Final A, Question 16 - Revision history

MathAdmin at 18:38, 2 June 2015

2015-06-02T18:38:18Z

MathAdmin at 18:37, 2 June 2015

2015-06-02T18:37:45Z

MathAdmin at 18:27, 2 June 2015

2015-06-02T18:27:07Z

MathAdmin at 17:59, 2 June 2015

2015-06-02T17:59:45Z

MathAdmin: Created page with "'''Question ''' Graph the following, $-x^2+4y^2-2x-16y+11=0$ {| class="mw-collapsible mw-collapsed" style = "text-align:left;" ! Final Answers |-..."

2015-06-01T01:31:16Z

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 |-..."

New page

'''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
|-
|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>
|-
|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.
|-
|c) False. <math>y = x^2</math> does not have an inverse.
|-
|d) True. <math>cos^2(x) - cos(x) = 0</math> has multiple solutions.
|-
|e) True.
|-
|f) False.
|}

[[005 Sample Final A|'''<u>Return to Sample Exam</u>''']]

@@ Line 40: / Line 40: @@
 |'''Note:''' The asymptotes and box are not drawn in this graph. Most instructors require them to be drawn.
 |-
-|[[File:8A_Sample_Final,_Q_6.png]]
+|[[File:5_Sample_Final_16.png]]
 |}
 [[005 Sample Final A|'''<u>Return to Sample Exam</u>''']]

@@ Line 23: / Line 23: @@
 |-
 |'''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.
 |}
@@ Line 28: / Line 30: @@
 ! Step 2: &nbsp;
 |-
-|Now that we have the equation that looks like an ellipse, we can read off the center of the ellipse, (0, -1).
+|Now that we have the equation that looks like a hyperbola, we can read off the center, (-1, 2). Now we will draw the box by moving up and down from the center by 1, and draw the horizontal lines y = 3 and y = 1. Do the same for the vertical lines by moving left and right from the center by 2, and draw the vertical lines x = -3 and x = 1. Now we have a box. Draw the two dashed lines that form diagonals of the box. These are the asymptotes. Since the hyperbola opens up and down we draw the corresponding parabolas, making sure to not cross the asymptotes.
 |}
@@ Line 38: / Line 36: @@
 ! Final Answer: &nbsp;
 |-
-|The four vertices are: <math>(-3, -1), (3, -1), (0, 1) \text{ and } (0, -3)</math>
+|The four vertices are: <math>(-3, -1), (3, -1), (0, 1) \text{ and } (0, -3)</math>.
 |-
 |[[File:8A_Sample_Final,_Q_6.png]]

@@ Line 10: / Line 10: @@
 |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 <math>x^2</math> &nbsp; and &nbsp; <math>y^2</math> have the same sign, positive.
+|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.
 |-
-|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 <math>y^2</math> is positive, the hyperbola opens up and down.
 |}
@@ Line 20: / Line 20: @@
 ! Step 1: &nbsp;
 |-
-|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>.
+|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>
 |}