MathAdmin at 16:58, 2 June 2015

2015-06-02T16:58:43Z

MathAdmin: 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''..."

2015-06-01T05:46:40Z

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

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a) Find the vertex, standard graphing form, and ''x''-intercepts for <math>f(x) = \frac{1}{3}x^2 + 2x - 3</math> 
b) Sketch the graph. Provide the ''y''-intercept.
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Foundations
|-
|1) What is the standard graphing form of a parabola?
|-
|2) What is the vertex of a parabola?
|-
|3) What is the <math>y</math>-intercept?
|-
|Answer:
|-
|1) Standard graphing form is <math>y-h=a(x-k)^2</math>.
|-
|2) Using the standard graphing form, the vertex is <math>(h,k)</math>.
|-
|3) The <math>y</math>-intercept is the point <math>(0,y)</math> where <math>f(0)=y</math>.
|}

Solution:

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

{|class = "mw-collapsible mw-collapsed" style = "text-align:left;"
! 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
|-
|<math>y+6=\frac{1}{3}(x+3)^2</math>.
|}

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

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

{|class = "mw-collapsible mw-collapsed" style = "text-align:left;"
! Final Answer:
|-
|The vertex is (-3,-6). The equation in standard graphing form is <math>y+6=\frac{1}{3}(x+3)^2</math>.
|-
|The two <math>x</math> intercepts are <math>(-3+3\sqrt{2},0)</math> and <math>(-3-3\sqrt{2},0)</math>.
|-
|The <math>y</math> intercept is (0,-3)
|}

[[004 Sample Final A|'''Return to Sample Exam''']]

@@ Line 68: / Line 68: @@
 |-
 |The <math>y</math> intercept is (0,-3)
 |}
 [[004 Sample Final A|<u>'''Return to Sample Exam</u>''']]

004 Sample Final A, Problem 2 - Revision history

MathAdmin at 16:58, 2 June 2015

MathAdmin: 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''..."

MathAdmin: 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''..."