Difference between revisions of "004 Sample Final A, Problem 2"

Revision as of 12:56, 6 May 2015

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-intercept.

Solution:

Step 1:
First, we put the equation into standard graphing form. Multiplying the equation $y={\frac {1}{3}}x^{2}+2x-3$ by 3, we get
$3y=x^{2}+6x-9$ .

Step 2:
Completing the square, we get $3y=(x+3)^{2}-18$ . Dividing by 3 and subtracting 6 on both sides, we have
$y+6={\frac {1}{3}}(x+3)^{2}$ .

Step 3:
From standard graphing form, we see that the vertex is (-3,-6). Also, to find the $x$ intercept, we let $y=0$ . So,
$18=(x+3)^{2}$ . Solving, we get $x=-3\pm 3{\sqrt {2}}$ .
Thus, the two $x$ intercepts occur at $(-3+3{\sqrt {2}},0)$ and $(-3-3{\sqrt {2}},0)$ .

Step 4:
To find the $y$ intercept, we let $x=0$ . So, we get $y=-3$ .
Thus, the $y$ intercept is (0,-3).

Final Answer:
The vertex is (-3,-6). The equation in standard graphing form is $y+6={\frac {1}{3}}(x+3)^{2}$ .
The two $x$ intercepts are $(-3+3{\sqrt {2}},0)$ and $(-3-3{\sqrt {2}},0)$ .
The $y$ intercept is (0,-3)

@@ 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
+|-
+|<math>y+6=\frac{1}{3}(x+3)^2</math>.
 |}
@@ 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>.
+|-
+|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)
 |}
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