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This is a quadratics question and can someone help me graph this

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If there's \(x^2\) it's a parabola.
I know but How Do I graph It

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Other answers:

How do I make a proper table of values?
Do I just guess or is there a way of finding good numbers for x
You could make a table and pick some points. Say, -2,-1,0,1,2 ?
Ill try that again but last time i did it i was not getting a curve
The values I did were -20 -10 0 10 20
For example For -10 i am getting -9 for y and for 10 i am getting 7.6 for y. This means the graph is not symmetrical
that just means its not symmetrical about the y axis choose x=0,20,40,60,80,100,120 you'll find it symmetrical about x=60
you are looking at the graph
I have another question. SO this was not symmetrical along the y axis and when I was making my table of values, I was not able to get a cororect parabola. How can I fnid out wether I should make the x values all positive or negative so I do get the symmetrical point in my table of values?
Oh, We jsut learned quadratic relations and we have not come accross this formula i guess we will learn this soon
Thanks for he help
it says "use graphing technology" so you are allowed to cheat^2%2F144%2B5%2F6x++domain+0..120
you can find the x coordinate of the vertex of the parabola by x=-b/2a
i meant multiply by 144
\[-\frac{1}{144}x^2+\frac{5}{6}x=0\] multiply by -144 to get \[x^2-120x=0\] factor as \[x(x-120)=0\]
gives the two zero highest point is smack dab in the middle

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