SOLUTION: {{{ (2/3)x^2-3x+6=y }}} I need to find the vertex and axis of symmetry to graph the quadratic form, I understand how to graph quadratic forms but the fraction in this specific pro

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Question 341749: +%282%2F3%29x%5E2-3x%2B6=y+ I need to find the vertex and axis of symmetry to graph the quadratic form, I understand how to graph quadratic forms but the fraction in this specific problem has me all jumbled up. I tried solving it and got x= 9/4 for the axis of symmetry and (9/4,87/16)for the vertex but im almost 100% certain that is wrong. I would really appreciate some help (PLEASE HELP ME). Thanks in advance.
-Kay
Found 2 solutions by jim_thompson5910, Alan3354:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
To find the vertex and axis of symmetry, we need to complete the square for +%282%2F3%29x%5E2-3x%2B6


%282%2F3%29x%5E2-3x%2B6 Start with the given expression.


%282%2F3%29%28x%5E2-%289%2F2%29x%2B9%29 Factor out the x%5E2 coefficient 2%2F3. This step is very important: the x%5E2 coefficient must be equal to 1.


Take half of the x coefficient -9%2F2 to get -9%2F4. In other words, %281%2F2%29%28-9%2F2%29=-9%2F4.


Now square -9%2F4 to get 81%2F16. In other words, %28-9%2F4%29%5E2=%28-9%2F4%29%28-9%2F4%29=81%2F16


%282%2F3%29%28x%5E2-%289%2F2%29x%2Bhighlight%2881%2F16-81%2F16%29%2B9%29 Now add and subtract 81%2F16 inside the parenthesis. Make sure to place this after the "x" term. Notice how 81%2F16-81%2F16=0. So the expression is not changed.


%282%2F3%29%28%28x%5E2-%289%2F2%29x%2B81%2F16%29-81%2F16%2B9%29 Group the first three terms.


%282%2F3%29%28%28x-9%2F4%29%5E2-81%2F16%2B9%29 Factor x%5E2-%289%2F2%29x%2B81%2F16 to get %28x-9%2F4%29%5E2.


%282%2F3%29%28%28x-9%2F4%29%5E2%2B63%2F16%29 Combine like terms.


%282%2F3%29%28x-9%2F4%29%5E2%2B%282%2F3%29%2863%2F16%29 Distribute.


%282%2F3%29%28x-9%2F4%29%5E2%2B21%2F8 Multiply.


So after completing the square, %282%2F3%29x%5E2-3x%2B6 transforms to %282%2F3%29%28x-9%2F4%29%5E2%2B21%2F8. So %282%2F3%29x%5E2-3x%2B6=%282%2F3%29%28x-9%2F4%29%5E2%2B21%2F8.


So y=%282%2F3%29x%5E2-3x%2B6 is equivalent to y=%282%2F3%29%28x-9%2F4%29%5E2%2B21%2F8.


So the equation y=%282%2F3%29%28x-9%2F4%29%5E2%2B21%2F8 is now in vertex form y=a%28x-h%29%5E2%2Bk where a=2%2F3, h=9%2F4, and k=21%2F8


Remember, the vertex of y=a%28x-h%29%5E2%2Bk is (h,k).


So the vertex of y=%282%2F3%29%28x-9%2F4%29%5E2%2B21%2F8 is since h=9%2F4 and k=21%2F8


Also, remember that the axis of symmetry is simply x=h. So the axis of symmetry is x=9%2F4


If you need more help, email me at jim_thompson5910@hotmail.com

Also, feel free to check out my tutoring website

Jim

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
+%282%2F3%29x%5E2-3x%2B6=y+ I need to find the vertex and axis of symmetry to graph the quadratic form, I understand how to graph quadratic forms but the fraction in this specific problem has me all jumbled up. I tried solving it and got x= 9/4 for the axis of symmetry and (9/4,87/16)for the vertex but im almost 100% certain that is wrong. I would really appreciate some help (PLEASE HELP ME). Thanks in advance.
------------------------
axis of symmetry is x = -b/2a
x = 3/(4/3) = 9/4
-------------------
The vertex is (9/4,f(9/4))
f(9/4) = (2/3)*(81/16) - 27/4 + 6 = 27/8 - 27/4 + 6 = -27/8 + 6 = 21/8
--> (9/4,21/8)