008A Sample Final A, Question 6 - Revision history

MathAdmin at 06:52, 26 May 2015

2015-05-26T06:52:33Z

MathAdmin at 19:25, 23 May 2015

2015-05-23T19:25:14Z

MathAdmin at 19:21, 23 May 2015

2015-05-23T19:21:25Z

MathAdmin at 19:19, 23 May 2015

2015-05-23T19:19:09Z

MathAdmin at 21:45, 28 April 2015

2015-04-28T21:45:48Z

MathAdmin: Created page with "'''Question: ''' Sketch $4x^2 + 9(y + 1)^2 = 36$ . Give coordinates of each of the 4 vertices of the graph. {| class="mw-collapsible mw-collapsed" style = "text-ali..."

2015-04-28T21:08:37Z

Created page with "'''Question: ''' Sketch <math>4x^2 + 9(y + 1)^2 = 36</math>. Give coordinates of each of the 4 vertices of the graph. {| class="mw-collapsible mw-collapsed" style = "text-ali..."

New page

'''Question: ''' Sketch <math>4x^2 + 9(y + 1)^2 = 36</math>. Give coordinates of each of the 4 vertices of the graph.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations
|-
|1) What type of function is this? What type of graph 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. Since the coefficients of the <math>x^2</math> and <math>y^2</math> terms are both positive the graph must be an ellipse.
|-
|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.
|}

Solution:

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Step 1:
|-
|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>.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! 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.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Step 3:
|-
|Now draw an oval through the four points you just drew.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Final Answer:
|-
|Now draw an oval through the four points you just drew.
|}

[[008A Sample Final A|<u>'''Return to Sample Exam</u>''']]

@@ Line 2: / Line 2: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Foundations
+!Foundations: &nbsp;
 |-
 |1) What type of function is this?
@@ Line 18: / Line 18: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-! Step 1:
+! 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>.
@@ Line 24: / Line 24: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-! Step 2:
+! 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).
@@ Line 34: / Line 34: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-! Final Answer:
+! Final Answer: &nbsp;
 |-
 |The four vertices are: <math>(-3, -1), (3, -1), (0, 1) \text{ and } (0, -3)</math>

@@ Line 10: / Line 10: @@
 |Answer:
 |-
-|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.
+|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.
 |-
 |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.

@@ Line 4: / Line 4: @@
 !Foundations
 |-
-|1) What type of function is this? What type of graph is this?
+|1) What type of function is this?
 |-
 |2) What can you say about the orientation of the graph?

@@ Line 31: / Line 31: @@
 |-
 |Then mark the two points that are 2 units up, and two units down from the center.
 |}
@@ Line 42: / Line 36: @@
 ! Final Answer:
 |-
-|Now draw an oval through the four points you just drew.
+|[[File:8A_Sample_Final,_Q_6.png]]
 |}
 [[008A Sample Final A|<u>'''Return to Sample Exam</u>''']]