008A Sample Final A, Question 3 - Revision history

MathAdmin at 06:51, 26 May 2015

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MathAdmin at 06:09, 23 May 2015

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MathAdmin at 06:08, 23 May 2015

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MathAdmin at 21:44, 28 April 2015

2015-04-28T21:44:13Z

MathAdmin: Created page with "'''Question:''' a) Find the vertex, standard graphing form, and X-intercept for $x = -3y^2-6y+2$ ..."

2015-04-28T21:06:01Z

Created page with "'''Question:''' a) Find the vertex, standard graphing form, and X-intercept for <math>x = -3y^2-6y+2</math> ..."

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'''Question:''' a) Find the vertex, standard graphing form, and X-intercept for <math>x = -3y^2-6y+2</math>

                 
b) Sketch the graph. Provide the focus and directrix.

Note: In this problem, what is referred to as standard graphing form is the vertex form, in case you search on the internet.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Foundations
|-
|1) What type of function are we asking you to graph (line, parabola, circle, etc.)?
|-
|2) What is the process for transforming the function into the standard graphing form?
|-
|3) After we have the standard graphing form how do you find the X-intercept, and vertex?
|-
|4) Moving on to part b) How do we find a <math>3^{rd}</math> point on the graph?
|-
|5) From the standard graphing form how do we obtain relevant information about the focus and directrix?
|-
|Answers:
|-
|1) The function is a parabola. Some of the hints: We are asked to find the vertex, and directrix. Also only one variable, of x and y, is squared.
|-
|2) First we complete the square. Then we divide by the coefficient of x.
|-
|3) To find the X-intercept, replace y with 0 and solve for x. Since the parabola is in standard graphing form, the vertex of <math>x - h = a(y - k)^2</math> is (h, k).
|-
|4) To find a <math>3^{rd}</math> point, we can either use the symmetry of a parabola or plug in another value for x.
|-
|5) From the equation <math>x - h = a(y - k)^2</math>, we use the equation <math>a = \frac{1}{4p}</math> to find p. P is both the distance from the vertex to the focus and the distance from the vertex to the directrix.
|}

Solution:

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
! Step 1:
|-
|There are two ways to obtain the standard graphing form.
|-
|Regardless of the method the first step is the same: subtract 2 from both sides to yield <math>x - 2 = -3y^2 - 6y</math>
|-
|Method 1:
|-
|Divide both sides by -3 to make the coefficient of <math>y^2</math>, 1. This means <math>\frac{-1}{3}(x - 2) = y^2 + 2y</math>
|-
|Complete the square to get <math>\frac{-1}{3}(x - 2) + 1 = (y^2 + 2y + 1) = (y + 1)^2</math>
|-
|Multiply both sides by -3 so <math>(x - 2) - 3 = -3(y + 1)^2</math>, and simplify the left side to yield <math>x - 5 = -3(y + 1)^2</math>
|-
|Method 2:
|-
|Instead of dividing by -3 we factor it out of the right hand side to get <math>x - 2= -3(y^2 + 2y)</math>.
|-
|Now we complete the square inside the parenthesis and add -3 to the left hand side resulting in <math>x - 5 = -3(y^2 + 2y +1) = -3(y + 1)^2</math>
|}

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
! Step 2:
|-
|Since the parabola is in standard graphing form we can read off the vertex, which is (2, -1).
|-
|We get the X-intercept by replacing y with 0 and solving for x. So <math>x - 5 = -3(1)^2</math>, and the X-intercept is (2, 0).
|}

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
! Step 3:
|-
|Now we need the value of p using the relation <math>a = \frac{1}{4p}</math>, where a = -3.
|-
|So <math>-3p = \frac{1}{4}</math>, and <math>p = \frac{1}{-12}</math>.
|}

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
! Step 4:
|-
|Since <math>a < 0</math>, the parabola opens left. Since the focus is inside the parabola, and p tells us the focus is from the vertex, the focus is at <math>(2 - \frac{1}{12}, -1) = (\frac{23}{12}, -1)</math>.
|-
|We also know that the directrix is a vertical line on the outside of the parabola with the distance from the directrix to the vertex being p. Thus the directrix is <math>x = 2 + \frac{1}{12} = \frac{25}{2}</math>
|}

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
! Final Answer:
|-
|Vertex: (2, -1), standard graphing form: <math>x - 5 = -3(y + 1)^2</math>, X-intercept: (2, 0), focus: <math>(\frac{23}{12}, -1)</math>, directrix: <math>x = \frac{25}{2}</math>
|}

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

@@ Line 80: / Line 80: @@
 |Since <math>a < 0</math>, the parabola opens left. Since the focus is inside the parabola, and p tells us the focus is from the vertex, the focus is at <math>(5 - \frac{1}{12}, -1) = (\frac{59}{12}, -1)</math>.
 |-
-|We also know that the directrix is a vertical line on the outside of the parabola with the distance from the directrix to the vertex being p. Thus the directrix is <math>x = 5 + \frac{1}{12} = \frac{61}{2}</math>
+|We also know that the directrix is a vertical line on the outside of the parabola with the distance from the directrix to the vertex being p. Thus the directrix is <math>x = 5 + \frac{1}{12} = \frac{61}{12}</math>
 |}
@@ Line 86: / Line 86: @@
 ! Final Answer:
 |-
-|Vertex: (5, -1), standard graphing form: <math>x - 5 = -3(y + 1)^2</math>, X-intercept: (2, 0), focus: <math>(\frac{59}{12}, -1)</math>, directrix: <math>x = \frac{61}{2}</math>
+|Vertex: (5, -1), standard graphing form: <math>x - 5 = -3(y + 1)^2</math>, X-intercept: (2, 0), focus: <math>(\frac{59}{12}, -1)</math>, directrix: <math>x = \frac{61}{12}</math>
 [[File:8A_Sample_Final_A,_Q_3.png]]
 |}
 [[008A Sample Final A|<u>'''Return to Sample Exam</u>''']]

@@ Line 7: / Line 7: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-! Foundations
+! Foundations: &nbsp;
 |-
 |1) What type of graph is this? (line, parabola, circle, etc.)
@@ Line 36: / Line 36: @@
 {| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
-! Step 1:
+! Step 1: &nbsp;
 |-
 |There are two ways to determine the standard graphing form.
@@ Line 60: / Line 60: @@
 {| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
-! Step 2:
+! Step 2: &nbsp;
 |-
 |Since the parabola is in standard graphing form we can read off the vertex, which is (5, -1).
@@ Line 68: / Line 68: @@
 {| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
-! Step 3:
+! Step 3: &nbsp;
 |-
 |Now we need the value of p using the relation <math>a = \frac{1}{4p}</math>, where a = -3.
@@ Line 76: / Line 76: @@
 {| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
-! Step 4:
+! Step 4: &nbsp;
 |-
 |Since <math>a < 0</math>, the parabola opens left. Since the focus is inside the parabola, and p tells us the focus is from the vertex, the focus is at <math>(5 - \frac{1}{12}, -1) = (\frac{59}{12}, -1)</math>.
@@ Line 84: / Line 84: @@
 {| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
-! Final Answer:
+! Final Answer: &nbsp;
 |-
 |Vertex: (5, -1), standard graphing form: <math>x - 5 = -3(y + 1)^2</math>, X-intercept: (2, 0), focus: <math>(\frac{59}{12}, -1)</math>, directrix: <math>x = \frac{61}{12}</math>