8A F11 Q3 - Revision history

Matt Lee at 22:22, 6 April 2015

2015-04-06T22:22:58Z

Matt Lee at 22:19, 6 April 2015

2015-04-06T22:19:22Z

Matt Lee at 08:17, 26 March 2015

2015-03-26T08:17:05Z

Grad at 06:04, 21 March 2015

2015-03-21T06:04:54Z

Grad at 05:45, 21 March 2015

2015-03-21T05:45:46Z

Grad at 05:35, 21 March 2015

2015-03-21T05:35:11Z

Grad at 05:23, 21 March 2015

2015-03-21T05:23:26Z

Grad at 04:50, 21 March 2015

2015-03-21T04:50:41Z

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

2015-03-21T04:31:54Z

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

New page

'''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.

{| 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:
|-
|Replace any occurrence of x by 5, so <math>f(5) = \log_3(5 + 3) - 1 = \log_3(8) - 1</math>
|}

← Older revision		Revision as of 22:22, 6 April 2015
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	\|}		\|}

−	[[8AF11Final\|<u>'''Return to Exam</u>''']]	+	[[8AF11Final\|<u>'''Return to Sample Exam</u>''']]

← Older revision		Revision as of 22:19, 6 April 2015
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	\|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>		\|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>
	\|}		\|}
		+
		+	[[8AF11Final\|<u>'''Return to Exam</u>''']]

@@ Line 62: / Line 62: @@
 |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 if (2, 0).
+|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).
 |}
@@ Line 79: / Line 79: @@
 |-
 |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>
 |}

@@ Line 4: / Line 4: @@
 b) Sketch the graph. Provide the focus and directrix.
-Note: In this problem, what is referred to as standard graphing form is vertex form, in case you search on the internet.
+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;"
@@ Line 49: / Line 49: @@
 |-
 |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>
 |}

← Older revision		Revision as of 05:35, 21 March 2015
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	b) Sketch the graph. Provide the focus and directrix.		b) Sketch the graph. Provide the focus and directrix.

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

	{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"		{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 55: / Line 55: @@
 |-
 |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>
 |}

@@ Line 47: / Line 47: @@
 |Complete the square to get <math>\frac{-1}{3}(x - 2) + 1 = (y^2 + 2y + 1) = (y + 1)^2</math>
 |-
-|Subtract one from both sides and multiply both sides by -3 so <math>(x - 2) = -3(y + 1)^2 + 3</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>
 |}

@@ Line 37: / Line 37: @@
 ! Step 1:
 |-
-|Replace any occurrence of x by 5, so <math>f(5) = \log_3(5 + 3) - 1 = \log_3(8) - 1</math>
+|There are two ways to obtain the standard graphing form.
 |}