004 Sample Final A, Problem 2 - Revision history

Kayla Murray at 21:45, 11 May 2015

2015-05-11T21:45:52Z

Kayla Murray at 19:56, 6 May 2015

2015-05-06T19:56:01Z

Kayla Murray at 04:17, 1 May 2015

2015-05-01T04:17:35Z

Kayla Murray: Created page with " a) Find the vertex, standard graphing form, and ''x''-intercepts for $f(x) = \frac{1}{3}x^2 + 2x - 3$
b) Sketch the graph. Provide the ''y''..."

2015-04-28T22:07:24Z

Created page with "<span class="exam"> a) Find the vertex, standard graphing form, and ''x''-intercepts for <math>f(x) = \frac{1}{3}x^2 + 2x - 3</math><br> b) Sketch the graph. Provide the ''y''..."

New page

<span class="exam"> a) Find the vertex, standard graphing form, and ''x''-intercepts for <math>f(x) = \frac{1}{3}x^2 + 2x - 3</math><br>
b) Sketch the graph. Provide the ''y''-intercept.

@@ Line 4: / Line 4: @@
 ! Foundations
 |-
+|1) What is the standard graphing form of a parabola?
 |-
 |Answer:
 |-
+|1) Standard graphing form is <math>y-h=a(x-k)^2</math>.
 |}

@@ Line 17: / Line 17: @@
 ! Step 1:
 |-
+|First, we put the equation into standard graphing form. Multiplying the equation <math>y=\frac{1}{3}x^2 + 2x - 3</math> by 3, we get
 |-
+|<math>3y=x^2+6x-9</math>.
 |}
@@ Line 25: / Line 25: @@
 ! Step 2:
 |-
+|Completing the square, we get <math> 3y=(x+3)^2-18</math>. Dividing by 3 and subtracting 6 on both sides, we have
 |}
@@ Line 31: / Line 33: @@
 ! Step 3:
 |-
+|From standard graphing form, we see that the vertex is (-3,-6). Also, to find the <math>x</math> intercept, we let <math>y=0</math>. So,
 |-
+|<math>18=(x+3)^2</math>. Solving, we get <math>x=-3\pm 3\sqrt{2}</math>.
 |-
+|Thus, the two <math>x</math> intercepts occur at <math>(-3+3\sqrt{2},0)</math> and <math>(-3-3\sqrt{2},0)</math>.
 |-
+|
@@ Line 43: / Line 45: @@
 ! Step 4:
 |-
+|To find the <math>y</math> intercept, we let <math>x=0</math>. So, we get <math>y=-3</math>.
 |-
+|Thus, the <math>y</math> intercept is (0,-3).
 |-
+|
@@ Line 53: / Line 55: @@
 ! Final Answer:
 |-
+|The vertex is (-3,-6). The equation in standard graphing form is <math>y+6=\frac{1}{3}(x+3)^2</math>.
 |}
 [[004 Sample Final A|<u>'''Return to Sample Exam</u>''']]

← Older revision		Revision as of 04:17, 1 May 2015
Line 1:		Line 1:
	<span class="exam"> a) Find the vertex, standard graphing form, and ''x''-intercepts for <math>f(x) = \frac{1}{3}x^2 + 2x - 3</math><br>		<span class="exam"> a) Find the vertex, standard graphing form, and ''x''-intercepts for <math>f(x) = \frac{1}{3}x^2 + 2x - 3</math><br>
	b) Sketch the graph. Provide the ''y''-intercept.		b) Sketch the graph. Provide the ''y''-intercept.
		+	{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"
		+	! Foundations
		+	\|-
		+	\|
		+	\|-
		+	\|Answer:
		+	\|-
		+	\|
		+	\|}
		+
		+
		+	Solution:
		+
		+	{\| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
		+	! Step 1:
		+	\|-
		+	\|
		+	\|-
		+	\|
		+	\|}
		+
		+	{\|class = "mw-collapsible mw-collapsed" style = "text-align:left;"
		+	! Step 2:
		+	\|-
		+	\|
		+	\|}
		+
		+	{\|class = "mw-collapsible mw-collapsed" style = "text-align:left;"
		+	! Step 3:
		+	\|-
		+	\|
		+	\|-
		+	\|
		+	\|-
		+	\|
		+	\|-
		+	\|
		+	\|}
		+
		+	{\|class = "mw-collapsible mw-collapsed" style = "text-align:left;"
		+	! Step 4:
		+	\|-
		+	\|
		+	\|-
		+	\|
		+	\|-
		+	\|
		+	\|}
		+
		+	{\|class = "mw-collapsible mw-collapsed" style = "text-align:left;"
		+	! Final Answer:
		+	\|-
		+	\|
		+	\|}
		+
		+	[[004 Sample Final A\|<u>'''Return to Sample Exam</u>''']]

004 Sample Final A, Problem 2 - Revision history

Kayla Murray at 21:45, 11 May 2015

Kayla Murray at 19:56, 6 May 2015

Kayla Murray at 04:17, 1 May 2015

Kayla Murray: Created page with " a) Find the vertex, standard graphing form, and ''x''-intercepts for f(x) = \frac{1}{3}x^2 + 2x - 3 b) Sketch the graph. Provide the ''y''..."

Kayla Murray: Created page with " a) Find the vertex, standard graphing form, and ''x''-intercepts for $f(x) = \frac{1}{3}x^2 + 2x - 3$
b) Sketch the graph. Provide the ''y''..."