Writing Quadratic Equations from Tables Without 0 for X

[@repatasgallahar3138]

👏 Amazing! Helped me to learn to help my daughter. We both watched. I guess you can teach an old dog a new trick.

[@thebielecmethod4884]

Haha thank you so much for this comment. This is exactly why I make the videos I do. Love hearing you and your daughter are learning together!!! Tell her good luck in class and continue to be the great dad you are!

[@evapaetsch3912]

this helped me so much! i have a geometry test next period and now i’m confident about it!!

[@thebielecmethod4884]

Love to hear it! Good luck

[@mickeryx3594]

Have a test tomorrow, thanks for this advice.

[@thebielecmethod4884]

Good luck!

[@dagecko365]

Thank you so much. Can u make a video making a vertex equation from a table of values?

[@thebielecmethod4884]

Sure can. When do you need it by? I have a few other videos I’m making this next week and then I can add that one to my list.

[@highgglsgf]

hello, I've been struggling with this one x | y 0.01 0.06 > 0.18 0.02 0.24 > 0.12 > 0.3 0.03 0.54 > 0.12 > 0.43 0.04 0.96 also when x = 0, y = 0 using your reverse method. 2nd difference of this table is a = 0.12 ÷ 2 = 0.06 y = ax² + bx + c and using algebra to solve for b gives me different b values. b1 = 5.99 (0.01, 0.06) b2 = 11.9988 (0.02, 0.24) b3 = 17.9982 (0.03, 0.54) b4 = 23.99 (0.04, 0.96) thus, i cant pinpoint the precise quadratic formula on hand so i used an online quadratic regression calculator and got this formula y = 600(x)² + 1.0516e^-14(x) + -9.7145e^-17 Another way is to solve for a in y = ax2, which gives the expression y = 600(x)² - the simplest formula yet Does this imply that this procedure cannot be applied to some other table of values? would you please explain this one?

[@thebielecmethod4884]

Yes precisely. This method becomes more difficult when dealing with more complex numbers as your dealing with here. I can make another video showing you a different route for this particular problem when I get back from spring break.

[@sara-fb9oi]

Quick question, lets say the first points on the table were (5,36), then how would you find c because it would be difficult to keep going back to 5,4,3,2,1,&0th term? Is there a formula or any way to find it out?

[@thebielecmethod4884]

Yea, you can write 3 equations using y=ax^2+bx+c and replacing the x and y. Your example of (5,36) would be 36=a(5)^2+b(5)+c 36=25a+5b+c Use your other 2 points to make 2 more equations. Then solve the system of equations for a, b, c. This is another way that can be done fairly quickly.

[@gogzog]

what about if the numbers are skipping around??

[@thebielecmethod4884]

Great question! There are other methods I can make a video on if you have the question your trying to answer.

[@gogzog]

@@thebielecmethod4884 you've helped me more than any math teacher

[@8bitquinn647]

What happens if my y-intercept numbers are irregular? For example it goes down -3, and then -3 again, and then -9, and then -15?

[@thebielecmethod4884]

Could you send me a list of x and y values to look at as an example? Having -3 twice shouldn’t happen and I’d have to see the x values.

[@mohammadmahrous]

what id zero term si not at the same distance?

[@thebielecmethod4884]

Do you mind sending me the table you have an I can give you some other routes you can take

[@stickimation7846]

Are the relation is the same with equation?

[@thebielecmethod4884]

Yes if the numbers were provided to you in a relation instead of a table the result would be the same equation.

[@superdude9006]

Thanks a lot!

[@thebielecmethod4884]

Anytime!

[@neel3135]

thank you

[@thebielecmethod4884]

Anytime!

[@alanakrantz4607]

what if it doesn’t have a 2nd difference?

[@thebielecmethod4884]

What are your x and y values? Do you mean the 2nd difference isn’t constant?

[@giljhonherbias1336]

@@thebielecmethod4884 yeah what if the second difference is not constant?

[@thebielecmethod4884]

Then you aren’t dealing with a quadratic.

[@Satgamer80d]

quadratics always have a constant second difference, the same way lines always have a constant first difference.

[@sara-fb9oi]

Then its either linear or exponential not quadratic

[@FelicityGonzalez-e2d]

Sooooo helpful thank you so much!

[@thebielecmethod4884]

Anytime!

[@alymiuu]

what if my x and y are both 0???

[@thebielecmethod4884]

Then your c term would be 0. This equation would be written out as y=ax^2+bx, with no value written down for c.

[@alymiuu]

@@thebielecmethod4884thank u!!